Passing variables from PHP to AJAX success function

Question

My aim is to update a WordPress post using AJAX. My code so far:

Script:

$.ajax({
    type: 'POST',
    url: ajax_url,
    data: {
        'action': 'wp_post',
        'ID': post_id,
        'post_title': post_title
},
success: function( data ) {
        $( '.message' )
        .addClass( 'success' )
        .html( data );
},
error: function() {
        $( '.message' )
        .addClass( 'error' )
        .html( data );
    }       
});

PHP:

function wp_post() {

    $post['ID'] = $_POST['ID'];
    $post['post_title'] = $_POST['post_title'];
    $post['post_status'] = 'publish';

    $id = wp_update_post( $post, true );

    if ( $id == 0 ) {
        $error = 'true';
        $response = 'This failed';
        echo $response;
    } else {
        $error = 'false';
        $response = 'This was successful';
        echo $response;
    }

}

As you can see the $response variable in my PHP function is being passed to the success function in my script and the value of $response is displayed on the page.

I want to modify my success function to do something like this:

success: function( data ) {
    if( $error == 'true' ) {
        // do something
    } else {
        // do something else
    }
},

The problem is, I am having trouble passing both the $response and $error variables in my PHP function to the success function in my scipt.

Can anyone let me know how to pass $response and $error to my script's success function?
Is there a better approach I should be taking?

I'm newish to AJAX so forgive me if the question is very basic.

why don't you use standard http result code instead of reinventing a protocol on top of a protocol? — njzk2
– njzk2, Commented Jan 7, 2014 at 15:50
Thanks njzk2 - as I said I'm newish to AJAX. Any chance you could provide an example of how to do that? — henrywright
– henrywright, Commented Jan 7, 2014 at 15:51
Encode the response as JSON object. Check out json_encode in PHP and JSON.parse in JavaScript. — elclanrs
– elclanrs, Commented Jan 7, 2014 at 15:52
use php.net/manual/en/function.http-response-code.php to set a failure code, which will get you into the ajax error function. — njzk2
– njzk2, Commented Jan 7, 2014 at 15:57

taxicala · Accepted Answer · 2014-01-07 15:58:15Z

13

You shoud encode the response of the php script as json, as follows:

function wp_post() {

    $post['ID'] = $_POST['ID'];
    $post['post_title'] = $_POST['post_title'];
    $post['post_status'] = 'publish';

    $id = wp_update_post( $post, true );

    $response = array();

    if ( $id == 0 ) {
        $response['status'] = 'error';
        $response['message'] = 'This failed';
    } else {
        $response['status'] = 'success';
        $response['message'] = 'This was successful';
    }

    echo json_encode($response);

}

And then, in your javascript code:

success: function( data ) {
    if( data.status == 'error' ) {
        // error handling, show data.message or what you want.
    } else {
        // same as above but with success
    }
},

answered Jan 7, 2014 at 15:58

taxicala

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3 Comments

henrywright Over a year ago

Thanks taxicala - Do I need to add a data type to my $.ajax()?

taxicala Over a year ago

It's better if you do :) Glad I could help you!

Kurt Van den Branden Over a year ago

If you don't add a datatype, jquery will not know if it's json. You can also do this at the top of your php script with header('content-type: text/json');

Jonast92 · Accepted Answer · 2014-01-07 15:55:40Z

2

You can create a JSON array like this one, in the backend:

$arr = array('error' => true, 'something' => 'foo');
echo json_encode($arr);

And then parse the json array to fetch the returned values, like this:

success: function( data ) {
    var error = '';
    var something = '';
    for(var i = 0; i < data.length ; i++)
    {
        error = data[i].error;
        something = data[i].something;
    }
    if( error ) {
        // do something
    } else {
        // do something else
    }
},

Wherea you echoed the array from the backend to the frontend, you can't simply access PHP variables within the JavaScript.

Note that there might be a syntax error, since I'm not testing it.

answered Jan 7, 2014 at 15:55

Jonast92

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6 Comments

pathfinder Over a year ago

If you are passing a single string back in your php, it is absolutely unnecessary to go through the whole json process as long as your javascript ajax function asks for data returned in html.

henrywright Over a year ago

pathfinder - I'm trying to pass 2 strings though. $response and $error

pathfinder Over a year ago

he only needs the one though, it's good or it's bad. why bloat code? But you guys can have the points. I will go back to making money on my compact code now. I really hate what stack overflow has become. It's like a video game now.

henrywright Over a year ago

pathfinder - No I need 2. I need 'it's good or bad' AND 'the message'

Jonast92 Over a year ago

You can send as many variables over as you want, what is the problem, henry?

|

Community · Accepted Answer · 2023-11-17 20:33:26Z

1

What you are looking for is json_encode()

This converts a PHP array to JSON.

For example:

$dataArray = array( 'message' => 'Error', 'data' => data);
echo json_encode($dataArray);

edited Nov 17, 2023 at 20:33

CommunityBot

11 silver badge

answered Jan 7, 2014 at 15:53

Phil

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4 Comments

pathfinder Over a year ago

yours does not even answer his question. why bother critiquing. It makes stackoverflow a total annoyance in my life.

pathfinder Over a year ago

the question is: "I want to modify my success function to do something like this:"

Jonast92 Over a year ago

Stack over flow is not about giving 100% working code, even though it's nice when possible, and this is what OP is looking for, it's no biggy if he has to do some research on what has been tipped to him.

pathfinder Over a year ago

you and phil are telling him to do unnecessary steps.

Cristián Dagnino · Accepted Answer · 2015-01-28 06:13:08Z

if you use:

echo json_encode($response);

on your .php function, remember to use:

var data = $.parseJSON(data);

on your ajax success.

Example:

function php_ajax_function() {

    // whatever you want to do...

    $response = array();

    if ( $id == 0 ) {
        $response['status'] = 'error';
        $response['message'] = 'This failed';
    } else {
        $response['status'] = 'success';
        $response['message'] = 'This was successful';
    }

    echo json_encode($response);
}

And then, in your javascript code:

success: function( data ) {

    console.log(data);

    var data = $.parseJSON(data);

    if( data.status == 'error' ) {
        // do something
    } else {
        // do other thing
    }
}

Ryan · Accepted Answer · 2014-01-07 15:53:56Z

0

You cannot directly use PHP variables within JavaScript. The best you can do is manipulate PHP output in your JavaScript code.

For example, you could print $response as either 'error' or 'no_error' - and in your AJAX callback, check that var data does not equal 'error' (For example).

answered Jan 7, 2014 at 15:53

Ryan

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Comments

pathfinder · Accepted Answer · 2014-01-07 15:53:29Z

-2

the javascript variable data contains your echoed $response variable. So using your example, it would be something like this. Be sure you are also asking for html as your return data in the ajax function. here is the docs on that: http://api.jquery.com/jquery.ajax/

success: function( data ) {
    if( data == 'This failed' ) {
        // do something
    } else {
        // do something else
    }
},

answered Jan 7, 2014 at 15:53

pathfinder

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4 Comments

pathfinder Over a year ago

works great for me, I use this method all the time. Then since I have it in html, I don't have to monkey with json or anything, I just pass a single statement from php and it works great.

henrywright Over a year ago

Hi pathfinder - thanks for your help with this. I'll keep your approach in mind for when I need just one variable. Nice and simple.

Jonast92 Over a year ago

data is an object, not a string.

pathfinder Over a year ago

the data is a string if you want it to be. If you look at original php he was passing one simple string. My function works and you all click without thinking. Yours works too, but it bloats the code. By saying this answer is incorrect you limit possibilities in an effort to be right.

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