Use jQuery to Output JSON from PHP Page

Question

I have a PHP page that outputs some JSON. After the page loads, I would like to output the JSON using jQuery AJAX. However, nothing is happening. The alert() dialog isn't appearing and it doesn't work with the JavaScript console, either. What is wrong with my jQuery AJAX call?

<script type="text/javascript">
jQuery.ajax({ url: 'http://www.domain.com/page.php',
    dataType: 'json',
    success: function(data) {
        var obj = jQuery.parseJSON(data);
        alert(obj.json_array[0]);
        alert(obj.json_array[1]);
        alert(obj.json_array[2]);
    }
});
</script>

The PHP is:

<?php
$array = [
    'foo' => 'bar',
    'bar' => 'foo',
    'int' => 4
];

$json_array = json_encode($array);
echo $json_array;
?>

Howzieky · Accepted Answer · 2016-09-18 22:53:13Z

1

In PHP, you can have arrays where each element has a key and a value, but once you parse it in javascript, that becomes an object. So using [0], [1], or [2] is invalid. You must use foo, bar, or int. Also, when you echo a string, the name of the variable that holds the string is not saved in any way. The only thing saved (echoed) is, well, the string. In your example, echo $json_array; echoes {"foo":"bar","bar":"foo","int":4}, and when it is parsed, you get an object. So you don't need to say obj.json_array, you just say obj. So to fix your issue,

Replace

alert(obj.json_array[0]);
alert(obj.json_array[1]);
alert(obj.json_array[2]);

with

alert(obj.foo);
alert(obj.bar);
alert(obj.int);

answered Sep 18, 2016 at 22:53

Howzieky

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10 Comments

GTS Joe Over a year ago

Hi Howzieky, I changed it to alert(obj.foo); but still, nothing is happening. No alert and it doesn't work in the JavaScript console either.

Howzieky Over a year ago

Can you put in console.log(obj) right after you set obj? What does it say?

GTS Joe Over a year ago

I get: Uncaught ReferenceError: obj is not defined

Howzieky Over a year ago

You are certain this is after var obj = jQuery.parseJSON(data);? Try saying console.log(data) instead.

Howzieky Over a year ago

No problem, thanks for the update! Good luck with your project!

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