PHP - Unable to send JSON correctly

Question

I'm not sure why this isn't working, as a lot of the examples on the internet suggest doing it this way. Anyway, I have a SQL result that I've converted to JSON and now I'm trying to use that with Javascript.

json_encode($test, true); ?>

<script type="text/javascript">
var obj = (<?php echo $test; ?>);
alert(obj.toSource());
</script>

This does not work and Chrome gives me an error of "illegal character" and the Javascript variable somehow displays some x-debug HTML from the PHP server: enter image description here

If I simple echo the JSON out to display on the webpage that works fine without any errors. What am I doing wrong?

Are you sure you're assigning $test the JSON encoded value? Also what are the brackets for? — kero
– kero, Commented Apr 22, 2014 at 14:14
it's some sort of stack trace for maybe like undefined variable — Daniel Cheng
– Daniel Cheng, Commented Apr 22, 2014 at 14:15

ElmoVanKielmo · Accepted Answer · 2014-04-22 14:16:31Z

4

Do it like this:

$test = json_encode($test, true);

json_encode doesn't change the variable in place.

answered Apr 22, 2014 at 14:16

ElmoVanKielmo

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Jeff Lambert · Accepted Answer · 2014-04-22 14:18:18Z

1

You're doing a couple of things wrong here..

json_encode($test, true);

I think you're probably thinking of json_decode, but the second parameter to json_encode is supposed to be a bitmask of options. Passing true here is probably wrong.

@ElmoVanKielmo is also correct, the variable doesn't change because you call a function, you must reassign the variable to the return value.

answered Apr 22, 2014 at 14:18

Jeff Lambert

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Rangad · Accepted Answer · 2014-04-22 14:17:23Z

0

You got hmtl that looks line an xdebug error/notice message. Fix that before you proceed! (You cut out the part where the message is put).

Additionally you do not encode $test correctly. json_encode returns the changed value and does not modify it by reference.

answered Apr 22, 2014 at 14:17

Rangad

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