javascript exception when using foreach function

Question

I am trying to parse an associated array returned from a PHP functoin like this:

$.ajax({
        type: "GET",
        url: "../controllers/Checkpoint.php/getAllCheckpoints"
    }).done(function( result ) {
        result.foreach(function(index, value){
            alert(index + "  "+ value);
        });
    }).error(function(){

    });

i return the array like this in php

foreach($result as $row) {
            $checkpoints[$row['ID']] = $row['Name'];
        }
        return json_encode($checkpoints);

the result of that array reading from the chrome F12 console is

{"1":"check point 1","2":"check point 2"}

I got this error in the javascript code:

ReferenceError: forEach is not defined

could you help me please

Your code has foreach (lower e), but your error message has forEach (upper E). Which did you actually use? — ajp15243
– ajp15243, Commented Dec 11, 2014 at 17:07
by the way, {"1":"check point 1","2":"check point 2"} is not an array. — Igwe Kalu
– Igwe Kalu, Commented Dec 11, 2014 at 17:15
@SultanBaby It's not, and OP does call it that, but OP also called it an associative array, which is correct. This is an unfortunate clashing of PHP and JS terminology (personally I hate the term associative array, since it confuses people with terms for objects and normal arrays). — ajp15243
– ajp15243, Commented Dec 11, 2014 at 17:18
AFAIK, 'associated' is different from 'associative'. But most importantly, OP's recount of the issue is inconsistent as you already noted - there's no way one would use foreach but then get a mention of forEach in the error message. — Igwe Kalu
– Igwe Kalu, Commented Dec 11, 2014 at 17:23
@SultanBaby Good catch, I read "associated" as "associative". That is clearly incorrect. — ajp15243
– ajp15243, Commented Dec 11, 2014 at 17:25

JoeKincognito · Accepted Answer · 2014-12-11 19:08:05Z

2

I would suggest using the jQuery.each() generic iterator function in combination with the javascript JSON.parse() to accomplish what you are attempting to do.

jQuery.each()

JSON.parse

.done(function( result ) {
    checkpoints = JSON.parse(result);
    $.each( checkpoints, function( index, value) {
         alert(index + "  "+ value);
    });
});

edited Dec 11, 2014 at 19:08

answered Dec 11, 2014 at 17:31

JoeKincognito

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Comments

Girish · Accepted Answer · 2014-12-11 17:30:29Z

1

On client side

you should use object.forEach() function e capital letter or jQuery function $.each(object)

    result.forEach(function(index, value){
        alert(index + "  "+ value);
    });

If you have not set response header Content-type: application/json then you will need to parse response content as JSON object use JSON.parse() function

.done(function( result ) {

    JSON.parse(result).forEach(function(index, value){
        alert(index + "  "+ value);
    });
})

On Server side

return mean function that return and assign value into any variable, AJAX response need output buffer so echo content on server side and terminate script after output buffer

    foreach($result as $row) {
        $checkpoints[$row['ID']] = $row['Name'];
    }
    //return json_encode($checkpoints);
    header('Content-type: application/json');
    echo json_encode($checkpoints);
    exit();

header('Content-type: application/json'); means response content type as json object

edited Dec 11, 2014 at 17:30

answered Dec 11, 2014 at 17:22

Girish

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2 Comments

ajp15243 Over a year ago

For older browsers that don't have JSON, there's also jQuery.parseJSON.

Girish Over a year ago

yes @ajp15243 may be required, thanks for suggestion

ray · Accepted Answer · 2014-12-11 17:33:44Z

1

If you have the data on client, on browser then Php is no is problem ok. When you is on client. Javascript native isnt have function foreach, You can use $.each function

$.ajax({
    type: "GET",
    url: "../controllers/Checkpoint.php/getAllCheckpoints"
}).done(function( result ) {
    $.each(result, function(i, data){
        console.log(i, data); // this print on console the data that you is run
        // for this you should have installed firebug on firefox or chrome already installed some similar.
    });
}).error(function(){

});

answered Dec 11, 2014 at 17:33

ray

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Comments

Tobi Klostermair · Accepted Answer · 2014-12-11 17:10:30Z

0

At first: You have to return the php result with "echo" and not with "return".

answered Dec 11, 2014 at 17:10

Tobi Klostermair

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