Just learning ajax today and am stock somewhere. Trying to link my php links to work with ajax so the data with that link Id would be displayed without refreshing the page. Below are the codes of what I have tried.
index.php
<div class="page">
<ul>
<?php
$sql = <<<EOF
SELECT * FROM addcategory ;
EOF;
$ret = $db->query($sql);
While ($row = $ret->fetchArray(SQLITE3_ASSOC)) {
$catname = $row['catname'];
$catid = $row['catid'];
echo "<li><a href='index.php?category_id=$catid'>$catname</a></li>";
}
?>
</ul>
</div>
<div class="cat_question">response displays here</div>
php script
require_once ("db.php");
$db = new MyDb();
$catid = (int)$_GET['category_id'];
$csql = <<<EOF
SELECT * FROM questions WHERE category_id = {$catid};
$cret = $db->query($csql);
While ($crow = $cret->fetchArray(SQLITE3_ASSOC)) {
$catquestion = $crow['question'];
$catans = $crow['answer'];
echo "<div>$catquestion</div><p>$catans</p>";
}
JavaScript
$('.page a).click(function(e) {
e.preventDefault();
If (window.XMLHttpRequest) {
var xhttp = new XMLHTtpRequest();
} else {
xhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xhttp.onreadystatechange = function() {
If (xhttp.readyState == 4 && xhttp.status == 200) {
var response = xhttp.responseText;
$('.cat_question').html(response);
}
};
xhttp.open("GET", "category.php?category_id=$catid", true);
xhttp.send();
});
I know if i change the xhttp URL to category.php?category_id=1 or any id it works but that would mean in would have to write ajax request for each a tags. Please what is the way to solve this block. Thanks in advance.
Related
since I wrote this code and it turned out that it is not entering the if statement ,My target was to bring the array from php(db) and put it in array in js,what to do?Here is the code ,is the problem that it is located in an extension js file ?Moreover php file getusercatpref.php is returning ["violen","ios","programming","nodjs"].
var arr = new Array();
var xmlhttp = new XMLHttpRequest();
xmlhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
var myArr = JSON.parse(this.responseText);
arr = myArr;
}
};
xmlhttp.open("GET", "getusercatpref.php", true);
xmlhttp.send();
PHP Code
<?php
$name="root";
$pas="";
$dbname="jobpursuit";
$con=mysqli_connect("localhost", $name, $pas,$dbname);
$arrayCategories = array();
$sql="Select * FROM preferences where username='12' ";
$result=mysqli_query($con,$sql);
$index=0;
while($row = mysqli_fetch_assoc($result)){
$arrayCategories[$index] = $row['categoryname'];
$index=$index+1;
}
echo json_encode($arrayCategories);
?>
How to get different variable value in different ids from ajax response.
In my programming get all responses from test.php file but i want to differently for both variable.
Index.php File:-
<html>
<span> Picture : </span>
<!-- In picture hint i want to print $varbilefirst value from test.php file -->
<span id="picturehintget"></span>
<input type="button" id="<?php echo "8"; ?>" class="submitnone" name="ansclick" onclick="QuestionId(this.id)" value="Submit">
<div> Demo : <p id="demoquiz"></p> </div>
<!--In Demo i want to print $varbilesec value from test.php file -->
<script type="text/javascript">
function QuestionId(obj)
{
var id = obj;
var xhttp = new XMLHttpRequest();
xhttp.onreadystatechange = function ()
{
if (this.readyState == 4 && this.status == 200)
{
document.getElementById("demoquiz").innerHTML = this.responseText;
document.getElementById("picturehintget").innerHTML = this.responseText;
}
};
xhttp.open("GET", "test.php?id=" + id, true);
xhttp.send();
}
</script>
</html>
test.php file
<?php
$id = $_GET['id'];
$varbilefirst = "For Picture Hint";
echo $varbilefirst;
$varbilesec = "For Demo Hint";
echo $varbilesec;
?>
I want exactly different varible value for different ids.
<span id="picturehintget">For Picture Hint</span>
<p id="demoquiz">For Demo Hint</p>
You can try with in php file using array
<?php
$id = $_GET['id'];
$varbilefirst = "For Picture Hint";
$varbilesec = "For Demo Hint";
echo json_encode(["varbilefirst"=>$varbilefirst, "varbilesec"=>$varbilesec]);
?>
In Javascript
<script type="text/javascript">
function QuestionId(obj)
{
var id = obj;
var xhttp = new XMLHttpRequest();
xhttp.onreadystatechange = function ()
{
if (this.readyState == 4 && this.status == 200)
{
document.getElementById("demoquiz").innerHTML = this.responseText.varbilesec;
document.getElementById("picturehintget").innerHTML = this.responseText.varbilefirst;
}
};
xhttp.open("GET", "test.php?id=" + id, true);
xhttp.send();
}
</script>
If I understood well your request you can return JSON from test.php like this:
<?php
$id = $_GET['id'];
$myObj->demo= $varbilefirst;
$myObj->picture = $varbilesec;
$myJSON = json_encode($myObj);
echo $myJSON;
?>
Inside if of Javascript:
if (this.readyState == 4 && this.status == 200)
{
var myObj = JSON.parse(this.responseText);
document.getElementById("demoquiz").innerHTML = myObj.demo;
document.getElementById("picturehintget").innerHTML = myObj.picture;
}
So I have a code, where a number is retrieved from a text file and is displayed through HTML.
The thing is that it only retrieves the number when I refresh the page. If I change the number on other page, it doesn't change on the other page.
I've seen similar questions before, most of them said to use AJAX, but I don't understand much about AJAX and I only need it for this bit of code, it's not like I'm going to constantly use it. I would like to know if there is any other besides AJAX and if there is only AJAX, please present examples of code.
PHP code:
<?php
$file = "num.txt"; //Path to your *.txt file
$contents = file($file);
$num = implode($contents);
?>
JavaScript code. Basically it gets the PHP value and outputs to the html.
document.getElementById("num").innerHTML = '<?php echo $num; ?>';
Keep in mind, I don't want to refresh the page. Just the variable.
EDIT: PHP code - Getting an error - Notice: Undefined index: keynum in C:\xampp\htdocs\num.php on line 3
Here is the code of the PHP
<?php
$keynum = $_POST['keynum'];
$post = "INSERT INTO post (keynum) VALUES ($keynum)";
$fileLocation = getenv("DOCUMENT_ROOT") . "/num.txt";
$file = fopen($fileLocation,"w");
$content = $keynum;
fwrite($file,$content);
fclose($file);
echo 'Response';
die();
?>
You can achieve what you want using XMLHttpRequest, like this:
function updateData() {
var xhttp = new XMLHttpRequest();
xhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
document.getElementById('num').innerHTML = this.responseText;
}
};
xhttp.open('GET', '/num.php', true);
xhttp.send();
}
setInterval(updateData, 2000);
If you want to refresh the variable without refreshing the page, then you need to use an asynchronous request. It doesn't need to be AJAX, you can use the native XMLHttpRequest() function.
If you want more info: https://developer.mozilla.org/en-US/docs/Web/API/XMLHttpRequest/Synchronous_and_Asynchronous_Requests
or with jquery:
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<script>
setInterval(function(){
$.ajax({url:"num.php", success:function(result){
$("#num").html(result);
}});
}, 3000);
</script>
<textarea id="num"></textarea>
get_nbr.php
<?php
$file = "num.txt"; //Path to your *.txt file
$contents = file($file);
$num = implode($contents);
echo $num;
?>
index.php
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<!---trigger get_nbr() on page load on an infinite loop every two seconds -->
<body onload="setInterval(get_nbr, 2000)">
<script>
function get_nbr() {
var xhttp = new XMLHttpRequest();
xhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
document.getElementById("num").innerHTML = this.responseText;
}
};
xhttp.open("GET", "get_nbr.php", true);
xhttp.send();
}
</script>
<p id="num"></p>
like this.
let url = ""; //whatever url your requesting.
setInterval(function(){
var x = new XMLHttpRequest();
x.open("GET" , url , true);
x.onreadystatechange = function(){
if(x.readyState == 4 && x.status == 200)
{
document.getElementById("num").innerHTML = x.responseText;
}
}
x.send();
} , 2000);
-- UPDATE , added post --
setInterval(function(){
var x = new XMLHttpRequest();
x.open("POST" , url , true);
x.setRequestHeader("Content-Type" , "application/x-www-form-urlencoded");
x.onreadystatechange = function(){
if(x.readyState == 4 && x.status == 200)
{
document.getElementById("num").innerHTML = x.responseText;
}
}
x.send("postVar1=123");
} , 2000);
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 created a Live Search using AJAX,PHP and mysql.here when I click on search result ,redirecting me to a particular page it works perfectly. Now I need a small change.
All I need is:
When I click on the search result, that particular result should be display in the input field.
Here is my AJAX code:
<script type="text/javascript">
END OF AJAX CODE
PHP CODE
<?php
ob_start();
session_start();
include("Base.php");
$dbase=new Base();
#$userID=$_SESSION['userID'];
$createdDate=date("Y-m-d");
$createdTime=date("h:i:s A");
$partialStates=mysql_escape_string($_REQUEST['q']);
$qryy="SELECT * from `gon_pro` WHERE `pro_name` LIKE
'%$partialStates%' ";
$pser=$dbase->execute($qryy);
$ser_nums=mysqli_num_rows($pser);
while($co[]=mysqli_fetch_array($pser)){
}
?>
<?php
foreach ($co as $key => $namo) {
$cv=$namo['pro_name'];
$cv_id=$namo['id'];
$cv_p=$namo['price'];
?>
<a href="pro_det.php?prolod=<?php echo $cv_id; ?>"><p class="res
col-md-6"><?php echo $cv; ?></p></a>
<?php
}
?>
END OF PHP CODE
function getStates(str) {
if (str.length == 0) {
document.getElementById("row").innerHTML = "";
document.getElementById("results").innerHTML =""
} else {
var xmlhttp = new XMLHttpRequest();
xmlhttp.onreadystatechange = function() {
if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
document.getElementById("results").innerHTML =
xmlhttp.responseText;
}
};
xmlhttp.open("GET", "support/getStates.php?q=" + str, true);
xmlhttp.send();
}
}
</script>