Basically, I want to retrieve a certain product after clicking on an element. I'm using AJAX to pass the variable, and PHP to display the SQL query. I will use the product id on a WHERE statement, to retrieve and display the correct product.
Here is part of my code so far:
<html>
<head>
<title>Left Frame</title>
<link href='http://fonts.googleapis.com/css?family=Indie+Flower' rel='stylesheet' type='text/css'>
<link href="stylesheets/main.css" rel="stylesheet" type="text/css">
<script src="javascripts/jquery-1.11.2.js">
</script>
</head>
<body>
<div class="container">
<div id="bottomHalf">
<img id="blank" src="assets/bottom_half.png" style= "z-index: 5" >
<img src="assets/frosen_food.png"
usemap="#Map2"
border="0"
id="frozen"
style="z-index: 0;
visibility:hidden;" >
<map name="Map2">
<area shape="rect" coords="7,95,126,146" alt="Hamburger Patties" href="#" id="hamburgerPatties">
</div>
</div>
<script language="javascript">
$("#hamburgerPatties").click(function(){
var hamburgerPatties = "1002";
$.ajax({
type:"GET",
url: "topRightFrame.php",
data: "variable1=" + encodeURIComponent(hamburgerPatties),
success: function(){
//display something if it succeeds
alert( hamburgerPatties );
}
});
});
</script>
</body>
</html>
Part of my PHP code:
<?php
$product_id =(int)$_GET['variable1'];
$servername = "************";
$username = "************";
$password = "*************";
$dbname = "poti";
$tableName = "products";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM $tableName ";
$result = $conn->query($sql);
// Display all products on the database
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "Product ID: " . $row["product_id"]. " - Product Name: " . $row["product_name"]. " Unit Price:" . $row["unit_price"]. "<br>";
}
} else {
echo "0 results";
}
// Display product I clicked on the other frame
if (isset($GET['variable1'])) {
$sql = "SELECT * FROM $tableName WHERE product_id = " . $_GET['variable1'];
$result = $conn->query($sql);
if ($result) {
echo '<table>';
while ($row = $result->fetch_assoc())
{
echo '<tr>';
echo '<td>', "Product ID: " . $row["product_id"]. " - Product Name: " . $row["product_name"]. " Unit Price:" . $row["unit_price"]. "<br>";
echo '</tr>';
}
echo '</table>';
}
}
$conn->close();
?>
I'm able to display all the products. But starting from the ifsset statement, the code no long works. I get no error message or anything. How can I solve this? I'm pretty new to PHP.
EDIT: Ok, I managed to get the product I want when I hard code the product id. Now I need to get this variable using javascript.
You have syntax errors in your if statements.
When you're setting if, you should enclose the statement in braces:
instead of if something : do if(something): and then end it with endif; (when using colons :)
// Display product I clicked on the other frame
if (isset($GET['variable1'])):
$query = "SELECT * FROM $tableName WHERE product_id = " . $_GET['variable1'];
$result = mysqli_query($conn, $query);
if ($result):
echo '<table>';
while ($row = $result->fetch_assoc()) {
echo '<tr>';
echo '<td>', $row['product_id'], '</td>'; // and others
echo '</tr>';
}
echo '</table>';
endif;
endif;
Or, use {} braces instead of colons :
// Display product I clicked on the other frame
if (isset($GET['variable1'])){
$query = "SELECT * FROM $tableName WHERE product_id = " . $_GET['variable1'];
$result = mysqli_query($conn, $query);
if ($result){
echo '<table>';
while ($row = $result->fetch_assoc()) {
echo '<tr>';
echo '<td>', $row['product_id'], '</td>'; // and others
echo '</tr>';
}
echo '</table>';
}
}
You have syntax error in if block. use {} for if block also use echo instead of pure html tag(table). This error prevents successful state of ajax request. also don't forget to close table.
if ($result)
{
echo '<table>';
while ($row = $result->fetch_assoc())
{
echo '<tr>';
echo '<td>', $row['product_id'], '</td><td>'; // and others
echo '</tr>';
}
echo '</table>';
}
and the ajax code to display php result instead of test alert would be this:
<script language="javascript">
$("#hamburgerPatties").click(function(){
var hamburgerPatties = "1002";
$.ajax({
type:"GET",
url: "topRightFrame.php",
data: "variable1=" + encodeURIComponent(hamburgerPatties),
success: function(data){
alert(data);
}
});
});
</script>
Related
I've been hanging on this problem for 3 days and just can't solve it. It seems relatively simple, but unfortunately, I can't do it, because I'm a noob.
I would like to use the button's onclick function to get its 2 pieces of information
```html
<input id="'.$row['id'].'" type="button" value="Pause" name="'.$row['name'].'" onclick="pushDataToDB()">```
name = "'. $ row [' name '].'" [the name reference from a database
table]
and value="Pause"
this two informations i would like to store in some other database table.
I have tried to solve this problem in different ways.
I have tried to store in different ways the event.target.name & event.target.event javascript output in PHP Variables and then use it in the mysql insert line but i failed.
I have tried to post the value to the ajax.php then store the value in a PHP variable and use this as Value to push it to the database, but this also doesn't work
index.php
<div class="button2">
<form action="ajax.php" method="POST">
<input id="'.$row['id'].'" type="button" value="Pause" name="'.$row['name'].'" onclick="pushDataToDB()">
</Form>
</div>'
script.js:
function pushDataToDB() {
$.get("ajax.php");
return false;
}
ajax.php
<?php
include_once ('dbh.php');
if(isset($_POST['name'])){
$sqlAufPause = "INSERT INTO aktuellaufpause (name, pausenart) VALUES ('$name', 'BP')";
}
if ($conn->query($sqlAufPause) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
?>
I would say this is the reason, why the name Value is empty,
but i don't know how to fix ist..
emptyNameValue
You can use JQuery + AJAX to perform asynchronous POST
<?php
$con = mysqli_connect("localhost", "root", "", "testdb");
$query = mysqli_query($con, "SELECT * FROM mytable WHERE id = 1");
$row = mysqli_fetch_assoc($query);
print_r($row);
?>
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<title>Document</title>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.5.1/jquery.min.js"></script>
</head>
<body>
<div class="button2">
<form id="frm">
<input id="<?php echo $row['id'] ?>" type="button" value="Pause" name="<?php echo $row['name'] ?>" onclick="pushDataToDB()">
</Form>
</div>
</body>
<script type="text/javascript">
function pushDataToDB() {
var name = "<?php echo $row['name'] ?>";
var value = "Pause";
$.ajax({
url: "ajax.php",
type: "POST",
data: {
"name": name,
"value": value
},
success: function(e) {
if (e == "1") {
alert("Success!");
} else {
alert("error!");
}
}
});
}
</script>
</html>
ajax.php
<?php
$con = mysqli_connect("localhost", "root", "", "testdb");
$name = $_POST['name'];
$value = $_POST['value'];
$query = "INSERT INTO mytable(name, value) values('$name','$value')";
$result = mysqli_query($con, $query);
if ($result) {
echo "1";
} else {
echo "Error!";
}
EDITTTT****
I changed
var name = "<?php echo $row['id'] ?>";
to
var name = "<?php echo $row['name'] ?>";
since it is the "name" you want stored in the database
I solved the problem this way.
Thank you so much!!!
index.php
<?php
$sql = "SELECT * FROM `DB-TABLENAME` ORDER BY `VALUE1`, `VALUE2`";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
while ($row = mysqli_fetch_assoc($result)){
echo "<tr>";
echo "<td>";
echo '<input style="margin: 0 auto 6px 17px;" type="checkbox" id="scales" name="scales">';
echo "</td>";
echo "<td>";
echo $row['VALUE1'];
echo "</td>";
echo "<td>";
echo $row['VALUE2'];
echo "</td>";
echo "<td>";
echo '<div class="button1">
<input id="'.$row['id'].'" type="button" value="BP" name="'.$row['VALUE2'].'" onclick="pushDataToDB()">
</div>';
echo "</td>";
echo "<td>";
echo '<div class="button2">
<form id="frm">
<input id="'.$row['id'].'" type="button" value="Mittagspause" name="'.$row['name'].'" onclick="pushDataToDB()">
</form>
</div>';
echo "</td>";
echo "</tr>";
}
} else {
echo "there are no comments";
}?>
<script type="text/javascript">
function pushDataToDB() {
var name = event.target.name;
var value = event.target.value;
$.ajax({
url: "ajax.php",
type: "POST",
data: {
"name": name,
"value": value
},
success: function(e) {
if (e == "1") {
alert("Success!");
} else {
alert("error!");
}
}
});
}
</script>
ajax.php
<?php
include_once('dbconnectioncredentials.php');
$con = mysqli_connect($servername, $username, $password, $dbname);
$name = $_POST['VALUE1'];
$value = $_POST['VALUE2'];
$query = "INSERT INTO aktuellaufpause (VALUE1, VALUE2) VALUES ('$name','$value')";
$result = mysqli_query($con, $query);
if ($result) {
echo "1";
} else {
echo "Error!";
}
?>
So I have a user page in which I include several tables, one of which uses a pagination script that is handled by javascript (so that I dont have to refresh the page when looking through the pages) now I'm trying to include this on my user.php page but it doesnt seem to work, any help appreciated
User.php page
<?php require 'session.php';
require 'header.php';
require 'includes/database.php';
$userid = $_GET['id'];
$con = mysql_connect($servername, $username, $password);
if(!$con) {
die('Could not connect: ' . mysql_error());
}
mysql_select_db($dbname);
//CLIENT INFO TABLE
include 'user/clientinfo.php';
echo "<br>";
// XLRSTATS TABLE
include 'user/xlrstatsinfo.php';
echo "<br>";
//ALIAS TABLE
include 'user/aliases.php';
//IP TABLE
include 'user/ipaliases.php';
//PENALTY TABLE
include 'user/penalties.php';
//PENALTY TABLE
include 'user/userchatlog.php';
?>
userchatlog.php page
<?php
require '../session.php';
?>
<div id='retrieved-data'>
<!--
this is where data will be shown
-->
<img src="../images/ajax-loader.gif" />
</div>
<link rel="stylesheet" type="text/css" href="http://144.76.158.173/ech/includes/css/main.css" />
<script type = "text/javascript" src = "../includes/js/jquery-1.7.1.min.js"></script>
<script type = "text/javascript">
$(function() {
//call the function onload
getdata( 1 );
});
function getdata( pageno ){
var targetURL = '../includes/pagination/userchatlog/search_results.php?page=' + pageno;
$('#retrieved-data').html('<p><img src="../images/ajax-loader.gif" /></p>');
$('#retrieved-data').load( targetURL ).hide().fadeIn('slow');
}
</script>
and my search_results.php
<!-- include style -->
<link rel="stylesheet" type="text/css" href="http://144.76.158.173/ech/includes/css/main.css" />
<?php
//open database
include '../../database.php';
//include our awesome pagination
//class (library)
include 'ps_pagination.php';
//connect to mysql server
$mysqli = new mysqli($servername, $username, $password, $dbname);
//check if any connection error was encountered
if(mysqli_connect_errno()) {
echo "Error: Could not connect to database.";
exit;
}
function _ago($tm,$rcs = 0) {
$cur_tm = time(); $dif = $cur_tm-$tm;
$pds = array('second','minute','hour','day','week','month','year','decade');
$lngh = array(1,60,3600,86400,604800,2630880,31570560,315705600);
for($v = sizeof($lngh)-1; ($v >= 0)&&(($no = $dif/$lngh[$v])<=1); $v--); if($v < 0) $v = 0; $_tm = $cur_tm-($dif%$lngh[$v]);
$no = floor($no); if($no <> 1) $pds[$v] .='s ago'; $x=sprintf("%d %s ",$no,$pds[$v]);
if(($rcs == 1)&&($v >= 1)&&(($cur_tm-$_tm) > 0)) $x .= time_ago($_tm);
return $x;
}
//query all data anyway you want
$sql = "SELECT * FROM chatlog WHERE msg NOT LIKE '%RADIO%' ORDER BY id DESC";
//now, where gonna use our pagination class
//this is a significant part of our pagination
//i will explain the PS_Pagination parameters
//$conn is a variable from our config_open_db.php
//$sql is our sql statement above
//3 is the number of records retrieved per page
//4 is the number of page numbers rendered below
//null - i used null since in dont have any other
//parameters to pass (i.e. param1=valu1¶m2=value2)
//you can use this if you're gonna use this class for search
//results since you will have to pass search keywords
$pager = new PS_Pagination( $mysqli, $sql, 45, 4, null );
//our pagination class will render new
//recordset (search results now are limited
//for pagination)
$rs = $pager->paginate();
//get retrieved rows to check if
//there are retrieved data
$num = $rs->num_rows;
if($num >= 1 ){
//creating our table header
echo "<table id='my-tbl'>";
echo "<tr>";
echo "<th>Time</th>";
echo "<th>Name</th>";
echo "<th>Message</th>";
echo "</tr>";
//looping through the records retrieved
while( $row = $rs->fetch_assoc() ){
$msg=$row['msg'];
$team=$row['client_team'];
$team = str_replace("42","<img src=images/green.png>",$team);
$team = str_replace("3","<img src=images/red.png>",$team);
$team = str_replace("2","<img src=images/blue.png>",$team);
$team = str_replace("1","<img src=images/grey.png>",$team);
$msg = str_replace("^0","</font><font color=black>",$msg);
$msg = str_replace("^1","</font><font color=red>",$msg);
$msg = str_replace("^2","</font><font color=lime>",$msg);
$msg = str_replace("^3","</font><font color=yellow>",$msg);
$msg = str_replace("^4","</font><font color=blue>",$msg);
$msg = str_replace("^5","</font><font color=aqua>",$msg);
$msg = str_replace("^6","</font><font color=#FF00FF>",$msg);
$msg = str_replace("^7","</font><font color=white>",$msg);
$msg = str_replace("^8","</font><font color=white>",$msg);
$msg = str_replace("^9","</font><font color=gray>",$msg);
$name=$row['client_name'];
$name=htmlentities($name);
$time=$row['msg_time'];
echo "<tr class='data-tr' align='center'>";
echo "<td align=center>";
echo _ago($time);
echo "</td>";
echo "<td align=center>";
echo $team;
echo "<a href='http://144.76.158.173/ech/user.php?id=".$row["client_id"]."' > $name </a></td>";
echo "<td align=left> $msg </td>";
echo "</tr>";
}
echo "</table>";
}else{
//if no records found
echo "No records found!";
}
//page-nav class to control
//the appearance of our page
//number navigation
echo "<div class='page-nav' align='center'>";
//display our page number navigation
echo $pager->renderFullNav();
echo "</div>";
?>
Hi I have a search function in my index.php page. When a user types in a field and enters 'search' it goes to jobsearch.php and prints out the job summary from a mysqli database.
I've created links that go on the end of the results. So when the user sees the job summary the user clicks on the link (click here) then it directs the user to a pop up view which shows more information (job _description) about the job. Is there a way of doing this? I don't want to have to create a html page for every job as there could be 100s of jobs in the database.
I would like the click-here link to take the ID of the desired job and print out the job description on a pop up page.
It looks like this:
this is jobs.php which prints out a list of jobs..
<?php
$servername = "*****";
$username = "root";
$password = "*****";
$dbname = "jobslist";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT id, job_title, job_description, job_location, job_category FROM jobs_list";
$result = $conn->query($sql);
//display table
echo "<table border='1'>
<!--<tr>
<th>ID</th>
<th>Title</th>
<th>Description</th>
<th>Location</th>
<th>Category</th>
</tr>-->";
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "<tr>";
echo "<td min-height:'200' ><h2>". $row["id"] ."</h2></td>";
echo "<td>". $row["job_title"] . "</td>";
echo "<td min-width:'700' >". $row["job_description"] . "</td> " ;
echo "<td>". $row["job_location"] . "</td>";
echo "<td>". $row["job_category"] . "</td> " ;
//this prints out a clickable link for every job
echo "<td><a href='" . $row['id'] . "'>Click Here</a></td>";
echo "</tr>";
}
}
else {
echo "0 results";
}
echo "</table>";
$conn->close();
?>
Change
//this prints out a clickable link for every job
echo "<td><a href='" . $row['id'] . "'>Click Here</a></td>";
To
echo "<td><a target='jobdescriptionwindow' href='jobdescription.php?id=" . $row['id'] . "'>Click Here</a></td>";
This will open a new window (called 'jobdescriptionwindow') when you click on the link passing the ID of the job clicked through to jobdescription.php. You will also have to make sure that your browser isn't stopping this window from opening.
Create a new PHP document called jobdescription.php with these contents:
<?php
if (isset($_GET['id'])) { // Check ID is is present in parameters
$servername = "*****"; // <-- Enter your details
$username = "root";
$password = "*****"; // <-- Enter your details
$dbname = "jobslist";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
// I do: (int)$_GET['id'] to type cast the value of $_GET['id'] to protect against injection. http://php.net/manual/en/language.types.type-juggling.php
$sql = "SELECT id, job_title, job_description, job_location, job_category FROM jobs_list WHERE id = ".(int)$_GET['id'];
$result = $conn->query($sql); // <-- Added new line
$row = $result->fetch_assoc();
echo $row['job_title'] . '<br />';
echo $row['job_description']; //<-- Does now show your description?
var_dump($result); // Lay out this data as required
// Outputs object(mysqli_result)#2 (5) { ["current_field"]=> int(0) ["field_count"]=> int(5) ["lengths"]=> array(5) { [0]=> int(1) [1]=> int(23) [2]=> int(102) [3]=> int(4) [4]=> int(2) } ["num_rows"]=> int(1) ["type"]=> int(0) }
} else {
echo "No ID provided";
}
This will get the details for the job ID sent in the GET parameters.
You do not need to generate a page for every job. You could generate one php that gets a job id as a GET parameter (getJob.php?id=5), fetches the data from data base on that id and returns the data as HTML page.
In addition to beingalex's code, you could also use fancybox to load content in the popup via ajax instead of a new window.
See this page:
http://fancyapps.com/fancybox/
Take a look at examples specifically ajax.
Search for 'ajax' on this page.This is what you need exactly.
jobs.php href-->
echo "<td><a href='something.php?id=".$row['id']. "'>Click Here</a></td>";
Something.php:
if(isset($GET_['id'])){
$getid = $_GET['id'];
$sql = "SELECT job_title, job_description, job_location, job_category FROM jobs_list WHERE id = '".$getid."'";
$result = $conn->query(sql);
while($row = $result->fetch_assoc()){
echo $row['job_title'].'<br />';
echo $row['job_description'].'<br />;
echo $row['job_location'].'<br />;
echo $row['job_category'];
}
}
But this code is attackable with SQL Injection! You need to escape the $_GET param, or you need to learn / use PDO.
Include jQuery,Bootstrap and show information on modal popup. The modal windows opens on job title link click.
<?php
$servername = "*****";
$username = "root";
$password = "*****";
$dbname = "jobslist";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT id, job_title, job_description, job_location, job_category FROM jobs_list";
$result = $conn->query($sql);
$tbdata = "";
$modals = "";
if ($result->num_rows > 0) {
// display table if results exist
echo "<table class=\"table table-striped\"><tr><th>Job Title</th><th>Location</th><th>Category</th></tr>";
while($row = $result->fetch_assoc()) {
$tbdata.="<tr><td>". $row["job_title"] . "</td><td>". $row["job_location"] . "</td><td>". $row["job_category"] . "</td></tr>";
$modals.="<div class=\"modal fade\" id=\"#jbinfo_". $row["id"] . "\" tabindex=\"-1\" role=\"dialog\" aria-labelledby=\"jbLabel". $row["id"] . "\" aria-hidden=\"true\"><div class=\"modal-dialog\"><div class=\"modal-content\"><div class=\"modal-header\"><h4 class=\"modal-title\" id=\"jbLabel". $row["id"] . "\">". $row["job_title"] . "</h4></div><div class=\"modal-body\">". $row["job_description"] . "</div><div class=\"modal-footer\"><button type=\"button\" class=\"btn btn-default\" data-dismiss=\"modal\">Close</button></div></div></div></div>";
}
echo $tbdata;
echo "</table>";
echo $modals;
} else {
echo "No results";
}
$conn->close();
?>
Before your closing body tag place the following:
<!-- Latest compiled and minified CSS -->
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.4/css/bootstrap.min.css">
<!-- Optional theme -->
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.4/css/bootstrap-theme.min.css">
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.3/jquery.min.js"></script>
<!-- Latest compiled and minified JavaScript -->
<script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.4/js/bootstrap.min.js"></script>
Im looking to pass one variable from my PHP code to my Javascript code to validate a figure. I'm creating a shopping cart scenario whereby i have a variable 'in_stock' which specifies the amount of stock available for a product. If I enter an amount that is greater than the 'in_stock' available i need an error to be thrown. This is currently what I have:
Java Script
<script language="javascript">
function numCheck()
{
var enteredChar = document.getElementById('add_value').value;
var stockavailable ="<?php echo $stock?>";
//To check if a non-numerical value is entered
if (isNaN(enteredChar))
{
alert("Not a Number!");
return false;
}
//To check if the input is empty
if (enteredChar=="")
{
alert("Empty!");
return false;
}
//To check if the amount entered is not greater than the amount in stock
if (enteredChar > stockavailable)
{
alert("Not enough in stock");
return false;
}
}
</script>
This is all occurring whilst my PHP has called the database and returned the value on the screen. As follows:
PHP
$product_id =(int)$_GET['product_id'];
$username = "potiro";
$password = "pcXZb(kL";
$hostname = "rerun";
//connection to the database
$dbhandle = mysql_connect($hostname, $username, $password)
or die("Unable to connect to MySQL");
//select a database to work with
$selected = mysql_select_db("poti",$dbhandle)
or die("Could not select examples");
//execute the SQL query and return records
$result = mysql_query("SELECT * FROM products where product_id=$product_id");
echo '<form name="form1">';
echo '<table class="Grocery-table">';
while($row = mysql_fetch_array($result))
{
$stock =$row['in_stock'];
$product_id = $row['product_id'];
echo "<tr><td><b>Product ID</b></td>";
echo "<td>";
echo $product_id;
echo "</td></tr>";
echo "<tr><td><b>Product Name</b></td>";
echo "<td>";
echo $row['product_name'];
echo "</td></tr>";
echo "<tr><td><b>Unit Price</b></td>";
echo "<td>";
echo $stock;
echo "</td></tr>";
echo "<tr><td><b>Unit Quantity</b></td>";
echo "<td>";
echo $row['unit_quantity'];
echo "</td></tr>";
echo "<tr><td><b>In Stock</b></td>";
echo "<td>";
echo $row['in_stock'];
echo "</td></tr>";
echo '<tr><td><b>Add</b></td><td><Input type="text" id="add_value" name="cart"></input></td></tr>';
echo '<tr><td></td><td><input type="submit" value="Submit" onclick="return numCheck()"></td></tr>';
}
echo "</table>";
echo "</form>";
mysql_close($dbhandle);
?>
add your stock value to a hidden input value stock_value
echo '<tr><td><b>Add</b></td><td><Input type="text" id="add_value" name="cart" ></input><Input type="hidden" id="stock_value" name="stock_value" value="'.trim($stock).'"></input></td></tr>';
echo '<tr><td></td><td><input type="submit" value="Submit" onclick="return numCheck()"></td></tr>';
in javascript
<script language="javascript">
function numCheck()
{
var enteredChar = document.getElementById('add_value').value;
var stockavailable =document.getElementById('stock_value').value;
I dont know how to use ajax in my problem:
I have a function in php (assign) that update a temporary table in database, I want to when user user click on a button (feedback function that is defined in javascript) this function (assign) run, what should I do?
<script>
function feedback(){
var boxes = document.getElementsByClassName('box');
for(var j = 0; j < boxes.length; j++){
if(boxes[j].checked) {
assign(1);
}
else{
assign(0);
}
}
}
</script>
<?php
$con = mysql_connect("localhost", "root", "")
or die(mysql_error());
if (!$con) {
die('Could not connect to MySQL: ' . mysql_error());
}
mysql_select_db("project", $con)
or die(mysql_error());
$result = mysql_query("select * from words");
echo "<table border='1'>
<tr>
<th>word</th>
<th>meaning</th>
<th>checking</th>
</tr>";
while($row = mysql_fetch_array($result)) {
echo "<tr>";
echo "<td>" . $row['word'] . "</td>";
$idd= $row['id'] ;
echo "<td>". "<div class='hiding' style='display:none'>".$row['meaning']."</div>"."</td>";
echo "<td>";
echo "<input class=\"box\" name=\"$idd\" type=\"checkbox\" value=\"\"> ";
echo "</td>";
echo "</tr>";
}
echo "</table>";
function assign($checkparm){
//mysql_query("update words set checking=$checkparm ");
mysql_query("create TEMPORARY TABLE words1user1 as (SELECT * FROM words) ");
mysql_query("update words1user1 set checking=$checkparm ");
}
mysql_close($con);
?>
<button onclick="ShowMeanings()">ShowMeanings</button>
<button onclick="feedback()">sendfeedback</button>
There is only one way to call a php function after the page is loaded:
1.ajax:
function callPHP() {
$.ajax ({
url: "yourPageName.php",
data: { action : assign }, //optional
success: function( result ) {
//do something after you receive the result
}
}
in your PHP, write
if ($_POST["action"] == "assign")
{
assign(your parameters); //You need to put the parameters you want to pass in
//the data field of the ajax call, and use $_POST[]
//to get them
}
There are many great guides on the internet. I will however suggest you get too know JQuery. It will help you on your learning curve.
function ajaxCall(){
$.ajax({
type: "GET",
url: "scripts/on/serverside.php"
});
};