I'm newbie in php and JavaScript, and I want to print all rows in my table
I use this code, it works but no output have come from it, and no errors occurs
function show(){//function get data and send it to Ajax, and I can make 1 function get 3 parameters
// var formData = new FormData();
// formData.append('result', $(("#a").val()));//php elem , html element
// formData.append('contactUsEmail', $("#contactUsEmail").val());
// formData.append('contactUsMessage', $("#contactUsMessage").val());
if(false){
swal("Attention !", "Please fill in all fields", "error");
}else {
//Ajax
if (window.XMLHttpRequest) {
ajax = new XMLHttpRequest();
} else {
ajax = new ActiveXObject("Microsoft.XMLHTTP");
}
ajax.onreadystatechange = function () {
if (this.readyState == 4 && this.status == 200) {
var text =this.responseText; // 0 unvalid , 1 true
// document.getElementById("a").innerHTML = this.responseText;
}
}
ajax.open("GET", "one.php", true);
ajax.send(formData);
}
}
php code here :
<?php
function OpenCon()
{
$dbhost = "localhost";
$dbuser = "root";
$dbpass = "";
$db = "chat";
$conn = new mysqli($dbhost, $dbuser, $dbpass,$db) or die("Connect failed: %s\n". $conn -> error);
return $conn;
}
function show(){
$sql = "SELECT id FROM masseges";
$result = OpenCon()->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "id: ". $row["id"]."bb";
}
} else {
echo "0 results";
}
OpenCon()->close();
}
No errors in page, but a blank page. How can I fix it ?
You are not actually returning any html tags back to your website.
You should be able to see the result from the php echo in the network tab of your development console.
If you want to fill a table, I suggest you create the table structure on your website and add the data from php as a new table row.
HTML:
<table>
<tr>
<th>id</th>
</tr>
</table>
PHP:
echo "<tr><td>id: ". $row['id']."bb</td></tr>";
Another word of advide:
Have a look at jquery! Gonna save you a a lot of time!
I have been where you are half a year ago :-)
Related
I am having problems creating a PHP session following a successful AJAX call. Here is the AJAX code:
function onSignIn(googleUser) {
var profile = googleUser.getBasicProfile();
var id = profile.getId();
var em = profile.getEmail();
var name = profile.getName();
var pic = profile.getImageUrl();
var xhttp = new XMLHttpRequest();
xhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
document.getElementById('confirm-login').style.display = 'block';
}
};
xhttp.open("GET", "./assets/inc/profile.php?id="+id+"&e="+em+"&n="+name+"&p="+pic, true);
xhttp.send();
}
This part works perfectly. I only include it for completeness sake.
Here's the contents of profile.php
<?php
$id = $_GET["id"];
$email = $_GET["e"];
$name = $_GET["n"];
$pic = $_GET["p"];
require_once("db.php");
$result = $mysqli->query("SELECT googleid FROM user_tbl WHERE googleid = '$id' LIMIT 1");
if($result->num_rows == 0) {
$sql = "INSERT INTO user_tbl (googleid, email, fullname, pic, loc) VALUES ('$id', '$email', '$name', '$pic', '')";
if (mysqli_query($mysqli, $sql)) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "" . mysqli_error($mysqli);
}
} else {
echo "already exists";
}
$mysqli->close();
session_start();
$_SESSION['gid'] = $id;
?>
All of this code works except for session_start(); and $_SESSION['gid'] = $id; when I return to another PHP page (which correctly has session_start(); at the very top of the page) the variable has not been created in profile.php
Any help as to what I'm doing wrong would be much appreicated.
You can't start a session after the script has sent output. (There should have been output to that effect; if there wasn't, try changing PHP's warnings.) The session_start() call must come before any echo call that is actually executed.
On an unrelated topic, you will want to learn how to escape your database parameters.
I have a js function that calls in an xml request to fetch data from a separate php file. I can get a returned data through echoing it from the separate php file.
Here's my current code:
var xhttp = new XMLHttpRequest();
xhttp.onreadystatechange = function()
{
if(this.readyState == 4 && this.status == 200)
{
//On Data Receive
countryHeader.innerHTML = this.responseText;
}
};
xhttp.open("GET", "country.php?c=" + countryName, true);
xhttp.send();
And on my php:
include("conn.php");
$c = htmlentities($_GET["c"]);
$sec_country = mysqli_real_escape_string($con, $c);
//Searches the db
$sql = "SELECT * FROM countries WHERE country_code = '$sec_country' LIMIT 1";
$result = mysqli_query($con, $sql);
$count = mysqli_num_rows($result);
if($count == 1)
{
//Get Data
$row = mysqli_fetch_assoc($result);
$countryName = $row['country_name'];
$countryPrice = $row['country_price'];
echo $countryName." is worth $".$countryPrice;
}
else
{
//Invalid Code/No Data
echo "No Country Found";
}
If I send in a country code for example like rus, it would return Russia is worth $1B mainly from the echo $countryName." is worth $".$countryPrice;
But what if I want to separately send $countryName and $countryPrice?
For example responseText.a and responseText.b
You can send JSON response from PHP. Here is a reference -> https://www.w3schools.com/js/js_json_php.asp
I am trying to pass data back to the server and then use the reply to update the browser page.
My code for a SELECT input is as follows;
<select id ="MatchCaptain" name="MatchCaptain" onchange="findTeleNo(this.value)"
<?php
$MC = $_SESSION["MatchCapt"];
player_load($MC);
?>
>
</select>
The script code is as follows;
<script>
function findTeleNo(that){
alert("I am an alert box!" + that);
var xhttp;
if (window.XMLHttpRequest) {
// code for modern browsers
xhttp = new XMLHttpRequest();
} else {
// code for old IE browsers
xhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
document.getElementById("TeleNo").value = this.responseText;
}
}
};
xhttp.open("GET", "findTeleNo.php?q=" + that, true);
xhttp.send();
</script>
The purpose of the script is to take the value selected in the dropdown (variable "that") and submit it to the php file as variable q.
And the PHP file is as follows;
<?php
$MatchCaptain = $_REQUEST["q"];
$teleNo = "";
$db_handle = mysqli_connect(DB_SERVER, DB_USER, DB_PASS );
$database = "matchmanagementDB";
$db_found = mysqli_select_db($db_handle, $database);
if ($db_found) {
$SQL = "SELECT * FROM `playerstb` ORDER BY `Surname` ASC, `FirstName` ASC";
$result = mysqli_query($db_handle, $SQL);
$ufullName = split_name($MatchCaptain);
while ( $db_field = mysqli_fetch_assoc($result) ) {
$uName = $db_field['FirstName'];
$uName = trim($uName);
$Surname = $db_field['Surname'];
$Surname = trim($Surname);
$fullName = $uName." ".$Surname;
if ($fullName == $ufullName )
{
$teleNo = $db_field['TeleNo'];
break;
}
}
}
echo $teleNo;
function split_name($name) {
$name = trim($name);
$last_name = (strpos($name, ' ') === false) ? '' : preg_replace('#.*\s([\w-]*)$#', '$1', $name);
$first_name = trim( preg_replace('#'.$last_name.'#', '', $name ) );
$ufullName = $first_name." ".$last_name;
return $ufullName;
}
?>
The php file requests the q variable from the url and makes it $MatchCaptain.
This will be a name like Joe Bloggs. The next piece of code connects to a MySQL table to extract players first names surnames and telephone numbers. The first names and surnames are concatenated to form the fullname which is compared with the $MatchCaptainWhen a match is made the variable $teleNo is set to the Telephone Number of that player. The echo statement rerurns the value to the script.
The field id I am trying to update is;
<p><b>Telephone Number: </b> <span id="TeleNo"> <?php echo $_SESSION["TeleNo"]; ?></span></p>
The alert in the script function findTeleNo shows me that I have entered the function but nothing happens after that.
Any help as to how I get this working would be grateful.
I have changed my script to
<script>
function findTeleNo(that){
var xhttp;
if (window.XMLHttpRequest) {
// code for modern browsers
xhttp = new XMLHttpRequest();
} else {
// code for old IE browsers
xhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xhttp.open("GET", "findTeleNo.php?q=" + encodeURIComponent(that), true);
xhttp.send();
xhttp.onreadystatechange = function() {
if (xhttp.readyState === 4) {
if (xhttp.status === 200) {
// OK
alert('response:'+xhttp.responseText);
document.getElementById("TeleNo").innerHTML = this.responseText;
// here you can use the result (cli.responseText)
} else {
// not OK
alert('failure!');
}
}
};
};
</script>
The response shown by alert('response:'+xhttp.responseText); is correct and the line of code
document.getElementById("TeleNo").innerHTML = this.responseText;
does print the response to the web page.
I have a web page with a large number of select boxes plus a few other types of inputs. I want to populate these select boxes and input boxes with values held in a MySQL database based on the values selected in two of the select boxes.
I have tried to use AJAX to request the data but cannot see how to return the data held in a database to a specific select box from the server php script.
jscript used to request the data is as follows
<script>
function venueChange(str) {
if (str == "") {
document.getElementById("Venue").innerHTML = "";
return;
} else {
if (window.XMLHttpRequest) {
// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp = new XMLHttpRequest();
} else {
// code for IE6, IE5
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
document.getElementById("Player1").innerHTML = this.responseText;
}
};
var opponent_name = document.getElementById("Opponents").value
var variables = "venue_name=str&opponent_name=Hello World";
xmlhttp.open("GET","getTeamPHP?" + variables,true);
xmlhttp.send();
}
}
</script>
The getTeamPHP files is as follows;
<?php
require '../../configure.php';
$q = intval($_GET['venue_name']);
$v = $_GET['opponent_name'];
$database = "matchmanagementdb";
$con = mysqli_connect(DB_SERVER, DB_USER, DB_PASS,$database);
if (!$con) {
die('Could not connect: ' . mysqli_error($con));
}
//mysqli_select_db($con,$database);
$sql="SELECT * FROM user WHERE Opponents = '".$q."' AND Venue = '".$v."'";
$result = mysqli_query($con,$sql);
while($row = mysqli_fetch_array($result)) {
echo document.getElementById("Player1").innerHTML = $row['FirstName'] ;
}
mysqli_close($con);
?>
I have decided that the solution lies in creating three scripts to;
1) read the search details
2) search the database and populate the form
3) allow updates.
As per http://www.dynamicdrive.com/forums/showthread.php?45895-How-to-populate-php-html-form-with-MySQL-data.
So closing this question.
I have a HTML page JavaScript which send a GET request data to PHP file to return all datas saved in the database . PHP replies with a HTML-table - that works fine!
But: When if i click a button (which calls the same JavaScript function) to update my table in order to display the new data, i get the same result (and i have definitely new data on table).
If I call the PHP manually via the browser it'll show me the new results immediately and at this moment it is also working with JavaScript (but only once).
Here is a part of my code.
HTML/JS:
<button onclick="GetData()"></button>
<div id="test"></div>
<script>
function GetData(){
var xhttp = new XMLHttpRequest();
document.getElementById("test").innerHTML = "";
xhttp.onreadystatechange = function(){
if (xhttp.readyState == 4 && xhttp.status == 200){
document.getElementById("test").innerHTML = xhttp.responseText;
}
};
xhttp.open("GET", "../GetData.php", true);
xhttp.send();
}
</script>
PHP:
//DB details
$dbHost = 'localhost';
$dbUsername = 'lalalala';
$dbPassword = 'lalalalal';
$dbName = 'lalalala';
//Create connection and select DB
$db = new mysqli($dbHost, $dbUsername, $dbPassword, $dbName) or die ("UUUUPS");
$sql = "select name, beschreibung, image, video from data";
$result = $db->query($sql);
if ($result->num_rows > 0) {
$return = '<table class ="table table-hover"><thead><tr><th scope="col">Name</th><th scope="col">Beschreibung</th><th scope="col">Bilddatei</th><th scope="col">Video-Url</th></tr></thead><tbody>';
// output data of each row
while($row = $result->fetch_assoc()) {
$return .= "<tr><td>".$row["name"]."</td><td>".$row["beschreibung"]."</td><td><img src=data:image/png;base64,".base64_encode($row["image"])."/></td><td>".$row["video"]."</tr>";
}
$return .= "</tbody></table>";
$db->close();
echo $return;
} else {
echo "0 results";
}
Thank you for your help!
It seems your browser is caching your result, that's why you see data.
You can test it like this:
var random = Math.floor(Math.random() * 100);
xhttp.open("GET", "../GetData.php?"+random, true);
If this helps, look into expire headers in your PHP script. Also, the way you're doing queries in quite outdated. It's a very PHP4 way. Have a look here: http://php.net/manual/en/book.mysqli.php
I guess you probably know this, but just in case. Have you had a look in your browsers inspector, when testing you html page? especially the network tab within that inspector. There you can see the actual response from the server and you can see if it is served from cache or fetched (you can even disable cache there), maybe this helps.
Kind regard,
Mark