Send variable to PHP file - javascript

In a WordPress post I'm trying to send data to a PHP file stored in the root folder of my website with this code:
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js" type="text/javascript">
console.log('hi');
var cen = document.getElementById("centro").value;
$.ajax({
url: 'centroUser.php',
type: "POST",
data: { 'cen': cen },
success: function(data){
console.log(data);
}
});
</script>
centroUser.php:
<?php
$uid = $_POST['cen'];
echo($uid);
?>
The problem is that I can't get it to work, the variable $uid doesn't get echoed and even the console.log('hi') doesn't get called. I'm new to AJAX so I don't really know what I'm doing wrong, I have tried looking for other answers but I couldn't find something that worked.

Your <script> tag has a src and a body.
Try:
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js" type="text/javascript"></script>
<script type="text/javascript">
console.log('hi');
var cen = document.getElementById("centro").value;
$.ajax({
url: 'centroUser.php',
type: "POST",
data: { 'cen': cen },
success: function(data){
console.log(data);
}
});
</script>
If the src has a URI value, user agents must ignore the element's contents and retrieve the script via the URI - see here.

Related

send variable with ajax when load page

try setting my code for send variable from view to controller,but my code is not run and show error
Uncaught ReferenceError: $ is not defined
view
<script type="text/javascript">
$(document).ready(function() {
var st = $('#st').val();
var postdata = { st: st };
$.ajax({
type: 'POST',
url: '<?= base_url();?>task_tickets/tes',
data: postdata,
success: function(response) {
console.log(response);
}
});
});
</script>
controller
function cek(){
echo $st;
}
Looks like either you have not included jQuery plugin or added after your JavaScript code. $ is alias to jQuery.
In your view before script tag, include this
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.4.0/jquery.js"></script>

how to update a value in database using jquery when a link is clicked in php

I want to increment count field in database when a link is clicked in php file.
So, I've added the following jquery part to my .php file. But still, it doesn't work. Please help!
<body>
click here
<script>
$(function ()
{
$('#click').click(function()
{
var request = $.ajax(
{
type: "POST",
url: "code.php"
});
});
}
</script>
</body>
code.php:
<?php
$conn=mysqli_connect("localhost","root","","sample");
mysqli_query($conn,"UPDATE e set count=(count+1) WHERE sid='1'");
mysqli_close($connection);
?>
You made a several mistakes in your code.
Update
You can send your SID input type text from ajax with data and you can get the value in your php file with the $sid = $_POST['sid'].
<body>
click here
<input type="text" value="" name="sid" id="sid">
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>
<script>
$(document).ready(function(e){
$('#click').click(function(event)
{
var sidvalue = $("#sid").val(); /*from here you get value of your sid input box*/
event.preventDefault();
var request = $.ajax(
{
type: "POST",
url: "code.php",
data: 'sid='+sidvalue ,
success: function() {
window.location.href = 'https://www.google.com/';
}
});
});
});
</script>
After the ajax success response you can make redirect to your desire location.
Code.php
<?php
$conn=mysqli_connect("localhost","root","","sample");
$sid = $_POST['sid']; // use this variable at anywhere you want.
mysqli_query($conn,"UPDATE e set count=(count+1) WHERE sid='1'");
mysqli_close($conn);
?>
in code.php in mysqli_close you should use $conn not $connection.
Go with this code. It might help you. I have just tested all the things in localhost. This is working perfect.
use preventDefault() when click event is called.check jquery :
<body>
click here
<script>
$(function ()
{
$('#click').click(function(e)
{
e.preventDefault();
var request = $.ajax(
{
type: "POST",
url: "code.php"
});
});
}
</script>
</body>
Redirect your link on ajax success. Like this -
<body>
click here
<script>
$(function ()
{
$('#click').click(function()
{
var request = $.ajax(
{
type: "POST",
url: "code.php",
success: function(response){
window.location="http://www.google.com";
}
});
});
}
</script>

How to upload a HTML variable to a PHP file using AJAX

On my website I am trying to basically generate a random code (which I will set up later) and then pass that code into a PHP file to later retrieve it when the client needs it. But my code just isn't working.
Here is the code:
Javascript/HTML:
function init() {
var code = "12345";
$.ajax({
type: 'POST',
url: 'codes.php',
data: { code: code},
success: function(response) {
$('#result').html(response);
}
});
}
PHP:
<?php
$code = $_POST['code'];
echo $code
?>
So what I understand that is supposed to happen is that the code is uploaded or 'posted' to the php file and then the #result is the echo $code. None of that happens and I have no idea.
Your code working perfect with some basic changes.
You need a html element with id 'result'.
And then you need to call your init() as per requirement.
<div id="result"></div>
<script>
function init() {
var code = "12345";
$.ajax({
type: 'POST',
url: 'codes.php',
data: { code: code},
success: function(response) {
$('#result').html(response);
}
});
}
init();
</script>
I tried this on my server in the head of my document, and it worked :)
I used on complete instead of on success.
<script type="text/javascript" src="https://code.jquery.com/jquery.min.js"></script>
<script>
function init() {
$.ajax({
type: "POST",
url: "codes.php",
data: {
'code': '12345'
},
complete: function(data){
document.getElementById("result").innerHTML = data.responseText
},
});
}
init();
</script>
with codes.php the same as you have :)
just a few notes:
Make sure you point your url to the correct file. You can check it by using the console network. Or you can simply print anything out, not just the $_POST data. e.g:
echo 'Test info';
Open browser developer panel, to see if is there any client code issue. For example, document with id 'result' existed, or you have not included jquery in. The developer console will tell you everything on the client side. For Chrome, check it out here https://developer.chrome.com/devtools
Have you actually called init() ?

Javascript variable is not visible

When I access the call.php page:
The variable name JSON is not accessible to param.js as below:
This is the code:
init.js
$("document").ready(function(){
var data = {
"action": "init"
};
data = $(this).serialize() + "&" + $.param(data);
var json;
$.ajax({
type: "POST",
dataType: "json",
url: "response.php", //Relative or absolute path to response.php file
data: data,
success: function(data){
json = JSON.parse(data["json"]);
}
});
});
response.php
<?php
if (is_ajax()) {
if (isset($_POST["action"]) && !empty($_POST["action"])) { //Checks if action value exists
$action = $_POST["action"];
switch($action) { //Switch case for value of action
case "init": test_function(); break;
}
}
}
function is_ajax() {
return isset($_SERVER['HTTP_X_REQUESTED_WITH']) && strtolower($_SERVER['HTTP_X_REQUESTED_WITH']) == 'xmlhttprequest';
}
function test_function(){
$return = $_POST;
$return["favorite_beverage"] = "Coke";
$return["favorite_restaurant"] = "McDonald's";
$return["json"] = json_encode($return);
echo json_encode($return);
}
?>
param.js
alert(json.favorite_beverage);
call.php
<html>
<head>
<title></title>
</head>
<script src="http://code.jquery.com/jquery-1.11.3.min.js"></script>
<script src="http://code.jquery.com/jquery-migrate-1.2.1.min.js"></script>
<script type="text/javascript" src="init.js"></script>
<script type="text/javascript" src="param.js"></script>
<body>
</body>
</html>
Did I miss something? Your help is kindly appreciated.
You have several issues here.
The variable json is declared within a function (inside the document.ready() callback) so it is only available within that function. You cannot access it from outside that function. That is how variable scoping works in Javascript.
The ajax success handler is called sometime later (its asynchronous) so you cannot reliably set data from that and use it in other parts of your code. Any code that wants to use the result of an ajax call should be put into the success handler directly or in a function that you call from the success handler and then pass the data to that function.
Here's an example of the fix described in the second point:
$("document").ready(function(){
var data = {
"action": "init"
};
data = $(this).serialize() + "&" + $.param(data);
var json;
$.ajax({
type: "POST",
dataType: "json",
url: "response.php", //Relative or absolute path to response.php file
data: data,
success: function(data){
json = JSON.parse(data["json"]);
// call a function and pass it the ajax data
myFunction(json);
}
});
});
// function located elsewhere
function myFunction(data) {
// process the data here
}
FYI, you can learn a lot more about dealing with data returned from ajax calls here: How do I return the response from an asynchronous call?

json import breaks down but from same server source

I am exploring the ajax synchronous json import into my javascript code.
The JSON source link I want to use is
http://www.nusantech.com/hendak/default.php?m=galaksi&galaksi=1&viewID=1&t=json
But to keep server loads down, a week ago or so I created a static page showing the same data at
http://www.nusantech.com/hendak/noobjson.php
My javascript import is as below:
<head>
<title>Nusantech</title>
<script src="\OpenLayers213\OpenLayers.js"></script>
<script type="text/javascript" src="http://code.jquery.com/jquery.min.js"></script>
<script type="text/javascript">
var jsonData = {};
$.ajax({
url: "http://hendak.seribudaya.com/noobjson.php",
async: false,
dataType: 'json',
success: function(data) {
jsonData = data;
}
});
alert("Galaksi value retrieved from JSON (expected: 1) : "+jsonData.galaksi);
</script>
<script type="text/javascript">
function kemasMaklumat(id,content) {
var container = document.getElementById(id);
container.innerHTML = content;
}
</script>
</head>
From there I retrieve the values I want on jsonData, eg, (x,y) coordinates as
(jsonData.planets[7].coordinates[0].x,jsonData.planets[7].coordinates[0].y)
It works fine with the noobjson.php link, but when I point it back to default.php, nothing appears. The page took a while to load which make it seem like its loading the json values, but the alert("Galaksi value retrieved") returns undefined.
I copy & pasted the output from the default.php page on a JSON verifier on the web and it showed OK. I don't know why the static link works but the $_GET based link doesn't.
Can someone suggest me what is happening?
EDIT
I have tried:
<script type="text/javascript">
var jsonData = {};
$.ajax({
// url: "http://hendak.seribudaya.com/noobjson.php",
url: "http://hendak.seribudaya.com/default.php?"+encodeURIComponent("galaksi=1&viewID=1&m=galaksi&t=json"),
// url: "http://hendak.seribudaya.com/default.php?galaksi=1&viewID=1&m=galaksi&t=json",
async: false,
dataType: 'json',
type: 'GET',
contentType: "application/json",
success: function(data) {
jsonData = JSON.parse(JSON.stringify(eval("("+data+")")));
alert("Success");
},
error: function(data) {
alert("Failed to download info." + data);
}
});
</SCRIPT>
enter code here
I always get the Failed to download info unless I use the noobjson URL.
It is as if that URL with the GET doesn't exist.
You have to encode the URL component before sending the request. Try:
$.ajax({
url: "http://www.nusantech.com/hendak/default.php?" + encodeURIComponent('m=galaksi&galaksi=1&viewID=1&t=json'),
async: false,
dataType: 'json',
success: function(data) {
jsonData = data;
}
});
Reference: encodeURIComponent()
I have solved it.
In the default.php, what I have done was:
if ($_GET["t"]=="json") {
$viewID=$_GET["viewID"];
$galaksi=$_GET["galaksi"];
$con=mysqli_connect($server, $user, $password, $database);
$sql="SELECT Hari FROM berita WHERE Galaksi=".$galaksi;
$hari=1;
$result = mysqli_query($con,$sql); while(($row = mysqli_fetch_array($result)) ){$hari=$row['Hari']; }
$lb="";
if ($_GET["t"]!="json") { echo "<PRE>\n"; $lb="\n"; }
echo "{\"galaksi\": ".$galaksi.",";
echo $lb."\"hari\": ".$hari.",";
echo $lb."\"planets\": [";
//etc
//etc
}
So I replaced all the individual echoes with $JSONstr like below.
if ($_GET["t"]=="json") {
$viewID=$_GET["viewID"];
$galaksi=$_GET["galaksi"];
$con=mysqli_connect($server, $user, $password, $database);
$sql="SELECT Hari FROM berita WHERE Galaksi=".$galaksi;
$hari=1;
$result = mysqli_query($con,$sql); while(($row = mysqli_fetch_array($result)) ){$hari=$row['Hari']; }
$lb="";
$JSONstr="";
// if ($_GET["t"]!="json") { $JSONstr="<PRE>\n"; $lb="\n"; }
$JSONstr=$JSONstr."{\"galaksi\": ".$galaksi.",";
$JSONstr=$JSONstr.$lb."\"hari\": ".$hari.",";
$JSONstr=$JSONstr.$lb."\"planets\": [";
//etc
//etc
//and at the end:
echo $JSONstr;
}
Then I added the echo $JSONstr; at the end. Originally I did that so that I can do :
echo json_encode($JSONstr);
but this creates {\"Galaksi\" : 1} at the JSON output instead of the intended { "Galaksi": 1 }
So I removed the json_encode and just output the string.
Also I had to remove the
if ($_GET["t"]!="json"){ $JSONstr="<PRE>\n"; $lb="\n"; }
I also used a different JSON tester this time.
Originally I used http://www.freeformatter.com/json-validator.html which says JSON Valid for my initial JSON output. Then I used this one, which said that my JSON output url was invalid, although if I copy+paste the output string it returned valid. http://jsonformatter.curiousconcept.com/
So after making those changes and removing the "<PRE>", the curiousconcept validator gave me a valid status.
Then I used this in the javascript, and I am now able to retrieve expected values.
Thank you all, hope this helps someone else too.

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