Passing PHP array to js with ajax

Question

I already looked up questions about this topic but couldn't figure my problem out. I have a php file which contains the array:

$data = ['logged' => $_SESSION['loggedin'], 'sessName' => $_SESSION['name']];
            echo json_encode($data);

Here's my AJAX code, but I have no idea what should I put in "data". Basically my goal is to use the $data array in my Javascript code. (So i can manipulate DOM with conditions).

<script>
    $.ajax({
    type: 'POST',
    dataType: "json",
    url:'sign-in.php',
    data: 
    success: function(data)
    {
    try {
        data = JSON.parse(data);
       
    }catch(e) {}
    console.log(data);
}
 });
</script>

data in your ajax success method contains the array. Try, for example, var logged = data.logged; — GrumpyCrouton
– GrumpyCrouton, Commented Aug 18, 2020 at 19:14
Just use data variable as it is. When you set dataType of the AJAX call, the response value is handled as JSON, and it is parsed for you by jQuery. "dataType "json" : Evaluates the response as JSON and returns a JavaScript object." — Teemu
– Teemu, Commented Aug 18, 2020 at 19:16
"but I have no idea what should I put in "data"" - data is already being supplied to the success function. Have you checked what's already in data before trying to modify it? — David
– David, Commented Aug 18, 2020 at 19:17
What did you expect to get from your javascript variable data? What did your console.log(data) show? — Koala Yeung
– Koala Yeung, Commented Aug 18, 2020 at 19:23
Use header('Content-type: application/json'); echo json_encode($data); — Sobir
– Sobir, Commented Aug 18, 2020 at 19:23

Koala Yeung · Accepted Answer · 2020-08-19 02:48:35Z

2

By specifying dataType: "json" in your $.ajax call, jQuery will automatically parse your JSON data into javascript object / array for you. You can probably remove the JSON.parse form you code.

Also there is an extra data: line, which would be a javascript syntax error.

<script>
$.ajax({
    type: 'POST',
    dataType: "json",
    url:'sign-in.php',
    success: function (data) {
        console.log(data);
    },
});
</script>

One more thing. Your PHP code, expects both 'loggedin' and 'name' to be set in your $_SESSION. If not, your PHP (depends on settings) might generate warning message in between and cause JSON parsing error.

You can use the null coalescing operator (introduced since PHP 7.0) to assign some value if either or both values are not set:

$data = [
    'logged' => $_SESSION['loggedin'] ?? FALSE,
    'sessName' => $_SESSION['name'] ?? '',
];
echo json_encode($data);

Updated: Add proper handling to potential invalid key issue.

edited Aug 19, 2020 at 2:48

answered Aug 18, 2020 at 19:28

Koala Yeung

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4 Comments

Dániel Mozsár Over a year ago

I tried this, but I can't see anything in the console

Dániel Mozsár Over a year ago

I tried to validate JSON and that's what I've got: Error: Parse error on line 4: ..."sessName": "asd1"}; ----------------------^ Expecting 'EOF', '}', ',', ']', got 'undefined'

Koala Yeung Over a year ago

You can open the "Network" tab in your browser's debugging console before your load the page. There should be a "Response" sub-tab somewhere in it. You can see the actual output there when you load the page and trigger the JSON request.

Teemu Over a year ago

@DánielMozsár Make sure there's no white-space or any other characters outside of the PHP tags in sign-in.php file. The response is not only the output of echo, everything outside of the PHP tags is collected to the output buffer with the output of the script commands, and the content of the buffer is sent as the response when the file ends.

AmraniCh · Accepted Answer · 2020-08-18 23:19:02Z

0

In your example you haven't intiliaze the session with session_start(), also it recomanded to indicate the response content type, and i fix also your ajax request :

PHP :

session_start();

$data = [
    'logged' => $_SESSION['loggedin'], 
    'sessName' => $_SESSION['name'],
];

header("Content-type: application/json");
echo json_encode($data);
exit();

Jquery :

$.ajax({
    type: 'GET',
    dataType: "json",
    url:'sign-in.php',
    success: function(data)
    {
        try {
            data = JSON.parse(data);
        } catch(e) {}
            console.log(e);
        }
    }
 });

answered Aug 18, 2020 at 23:19

AmraniCh

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1 Comment

Dániel Mozsár Over a year ago

So, i think there's something wrong with my JSON. SyntaxError: JSON.parse: unexpected character at line 1 column 5 of the JSON data. Here's my JSON (php generates this with encode): {"logged":true,"sessName":"asd1"};

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