Notice: Trying to get property of non-object in C:\xampp\htdocs\ihelploginapi\index.php on line 5

Question

This code was working 2 days before, but now i am getting an error:

Trying to get property of non-object in C:\xampp\htdocs\ihelploginapi\index.php on line 4.

Somebody please help me out.

<?php

    $json = file_get_contents('php://input');
    $obj = json_decode($json,TRUE);
    $tag = $obj->{'tag'};
?>

can you `var_dump($obj)` on the line after json_decode and post the results? — Scott, Jan 22 '16 at 17:33
Error:Parse error: syntax error, unexpected '$tag' (T_VARIABLE) in C:\xampp\htdocs\ihelploginapi\index.php on line .. — Nitish Kumar, Jan 22 '16 at 17:37
code :{'tag'}; if ($tag != '') { // Get tag $tag = $obj->{'tag'}; — Nitish Kumar, Jan 22 '16 at 17:37
Possible duplicate of [Reference - What does this error mean in PHP?](http://stackoverflow.com/questions/12769982/reference-what-does-this-error-mean-in-php) — Abdulla Nilam, Jan 22 '16 at 17:38

score 1 · Answer 1 · answered Jan 22 '16 at 17:36

json_decode does not give you an object. It gives you an array. You want to access it as such:

$tag = $obj['tag'];

or to re-write the var names more accurately

$json = file_get_contents('php://input');
$php_array = json_decode($json,TRUE);
$tag = $php_array['tag'];

score 0 · Answer 2 · answered Jan 22 '16 at 17:37

0

In the related line use :

$tag = $obj['tag'];

answered Jan 22 '16 at 17:37

pritesh

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score 0 · Answer 3 · answered Jan 22 '16 at 17:37

0

The second argument to json_decode() tells it to convert JSON objects to PHP associative arrays, not PHP objects. So you need to use $obj['tag'] instead of $obj->tag. Or change the decode line to

$obj = json_decode($json);

answered Jan 22 '16 at 17:37

Barmar

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500
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Notice: Trying to get property of non-object in C:\xampp\htdocs\ihelploginapi\index.php on line 5

3 Answers3