Develop Reference

Display error on ajax response from php function

I have some ajax call like this
function ExportData() {
var data = {
action: "export_database", // the name of your PHP function!
};
jQuery.ajax({
type: "POST",
url: ajaxurl,
data: data,
beforeSend: function () {},
success: function (data) {
alert(data);
},
});
}
And php function like this
function export_database(){
return $response;
}
The problem is in that response I have something like this
{
"type": "https://tools.ietf.org/html/rfc7231#section-6.5.1",
"title": "One or more validation errors occurred.",
"status": 400,
"traceId": "|3fa58ee1-48bf0cb9f60bfa25."
}
I want to alert only title, but when i try data.title , i got undefine
Do I must encode or decode something, thanks?

This is what you need. Just access the object by data.title and it will show in the alert()
You need to define dataType as json in your request.
If its does not work then use JSON.parse(data) like this:
var response = JSON.parse(data)
alert(response.title)
Try below:
function ExportData() {
var data = {
action: "export_database", // the name of your PHP function!
};
jQuery.ajax({
type: "POST",
url: ajaxurl,
dataType: 'json'
data: data,
beforeSend: function () {},
success: function (data) {
alert(data.title);
},
error: function(error){
//Error
alert(error.title)
}
});
}
Hope this helps.

Try Below:
function ExportData() {
var data = {
action: "export_database", // the name of your PHP function!
};
jQuery.ajax({
type: "POST",
url: ajaxurl,
data: data,
beforeSend: function () {},
success: function (data) {
var parsedData = jQuery.parseJSON(data)
alert(parsedData.title);
},
});
}

You have to use JSON.parse() for accessing data objects Like this:
function ExportData() {
var data = {
action: "export_database", // the name of your PHP function!
};
jQuery.ajax({
type: "POST",
url: ajaxurl,
data: data,
beforeSend: function () {},
success: function (data) {
var res = JSON.parse(data)
alert(res.title);
},
});
}

php sending values to ajax faild error

I Am trying to send value from ajax to php and retrieve it just to test that everything is work, when i click in a button to test i got error and alert('Failed') Appears , how can i fix it in order to get success? thanks
Ajax :
var a = "test";
$.ajax({
url: "search.php",
dataType: "json",
data: a ,
success: function(data) {
alert('Successfully');
},
error: function(data) {
alert('Failed');
}
})
PHP :
<?php
$pictures = "img1";
echo json_encode($pictures);
?>

I refined your code slightly and it works.
var a = "test";
$.ajax({
type: 'POST',
url: 'search.php',
data: 'a=' + a,
dataType: 'json',
cache: false,
success: function (result) {
alert('Successful');
},
error: function (result) {
alert('Failed');
}
});

If you're requesting a JSON, use the $.getJSON from jQuery, it's aready parse the JSON into a JSON object for you.
Seems that you're not return an actual JSON from server, maybe this is what is causing the error.
If you're seeing the 'Failed' message probably the problem is a 500 error which is a server error.
Try this code above.
Javascript:
var a = "test";
$.getJSON("search.php", {
a: a
}, function (json) {
console.log(json);
});
PHP:
<?php
$pictures = ["img1"];
echo json_encode($pictures);
The only way to this not work, is if you have a huge mistake on you webserver configuration.

Your ajax is wrong, it should be:
var a = "test";
$.ajax({
type: "POST",
url: "search.php",
dataType: "json",
data: {a:a},
success: function(data) {
alert('Successfully');
},
error: function(data) {
alert('Failed');
}
});

How to call the php function using jquery ajax

I am using the following script
<script>
$(document).ready(function(){
$("#view").click(function(){
var requestId = $("#hdnRequestId").val();
$.ajax({
type: "POST",
url: "enquiryProcess.php",
data: requestId,
cache: false,
success: function(data){
console.log(data);
}
});
return false;
});
});
My controller function is
<?php
include('enquiry_function.php');
$functionObj=new Enquiry();
if(isset($_POST['requestId']))
{
$qt_request_id=$_POST['requestId'];
$responce=$functionObj->view_enquiry_request($qt_request_id);
echo json_encode($responce);
}
?>
And my model function is
class Enquiry
{
public function view_enquiry_request($qt_request_id)
{
$query=mysql_query("SELECT * FROM quote_request WHERE qt_request_id='$qt_request_id'");
$result=mysql_fetch_assoc($query);
return $result;
}
}
I did not get any error.But result in console message is empty.How to get the result from php in jquery ajax.please help me.

Please change
var requestId = $("#hdnRequestId").val();
$.ajax({
type: "POST"
, url: "enquiryProcess.php"
, data: {"requestId":requestId}
, cache: false
, success: function (data) {
console.log(data);
}
});
Pass data as PlainObject or String or Array. See jQuery documentation here http://api.jquery.com/jquery.ajax/

How do i parse json data retrieved from php script in javascript or jquery

Please somebody help me, am new to ajax, i have been trying to read json data from php script but just no success.
when i console.log the data i get this,
{"name":"JOHN","surname":"FIGO","sex":"M","Records_id":"1","student":""}.
and when i do this console.log(data[2]); i simply get n character. what i want is to get the values, for example, console.log(data['name']); should give JOHNor console.log(data[0]); should give JOHN. when i use either javascript or jquery parse methods, i get an error, i dont understand. Here is are the codes;
<?php
$conn= new mysqli('localhost', 'root', '', 'Goldfinger');
$query= 'SELECT * FROM records';
$result= $conn->query($query);
while($row = $result->fetch_assoc()) {
echo json_encode($row);
}
?>
and jquery;
$('#one').click(function () {
$.ajax({
url: 'ajaxtesting.php',
type: 'POST',
success: function (data) {
if (data) {
console.log(data['name']);
};
},
error: function () {
$('div:not(#one)').text('error dude!');
}
})
})
pardon my code if it's very poor. Thank you in advance.

Put dataType : 'json', inside ajax setting like so :
$.ajax({
url: 'ajaxtesting.php',
type: 'POST',
dataType : 'json', //<------------- here
success: function (data) {
if (data) {
console.log(data['name']);
};
},
error: function () {
$('div:not(#one)').text('error dude!');
}
})
or simply parse inside success callback :
$.ajax({
url: 'ajaxtesting.php',
type: 'POST',
success: function (data) {
var myData = $.parseJSON( data ); //<------- here
if ( myData ) {
console.log(myData['name']);
};
},
error: function () {
$('div:not(#one)').text('error dude!');
}
})

How to get the PHP var after Jquery $.ajax

I use $.ajax to send some data to testajax.phpenter code here where data are processing. Result of this proces is link. How to get access to this? I think about sessions or cookies. Set in session $test and get access in another files.
$.ajax({
url: 'testajax.php',
type: 'post',
dataType: 'json',
success: console.log('success');
data: { json : jsonObj }
});
In testajax.php I have someting like this
echo "PDF: <a href=images/test.pdf>$dir.pdf</a>";
$test = "PDF: <a href=images/test.pdf>$dir.pdf</a>";
How to get the $test or output the echo after call $.ajax

You can use success' function to get request from PHP.
$.ajax({
url: 'testajax.php',
type: 'post',
dataType: 'html',
data: { json : jsonObj }
success: function(data) {
var request = data;
console.log(data);
$('#link_container').html(data);
}
});
Your console now holds PDF: <a href=images/test.pdf>($dir content).pdf</a>.
Also the variable request holds the same result that you can use anywhere in the script.
Result is appended to #link_container.

Use the success() method for display the result.
$.ajax({
url: 'testajax.php',
type: 'post',
dataType: 'json',
success: function (data){ // <= define handler for manupulate the response
$('#result').html(data);
}
data: { json : jsonObj }
});

Use below code It may help you.
$.ajax({
url: 'testajax.php',
type: 'post',
dataType: 'json',
data: { json : jsonObj },
onSuccess: successFunc,
onFailure: failureFunc
});
function successFunc(response){
alert(response);
}
function failureFunc(response){
alert("Call is failed" );
alert(response);
}