What is unexpected T_VARIABLE in PHP?

Question

I get this PHP error:

Parse error: syntax error, unexpected T_VARIABLE

From this line:

$list[$i][$docinfo['attrs']['@groupby']] = $docinfo['attrs']['@count'];

Is there anything wrong with this line?

X-Ref: [PHP Parse/Syntax Errors; and How to solve them? - Unexpected T_VARIABLE](http://stackoverflow.com/a/18092267) for the other common causes and context-specific fixes. — mario, Aug 18 '15 at 14:56

score 198 · Accepted Answer · edited Apr 23 '20 at 10:33

198

There might be a semicolon or bracket missing a line before your pasted line.

It seems fine to me; every string is allowed as an array index.

edited Apr 23 '20 at 10:33

Peter Mortensen

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answered Sep 23 '09 at 09:30

knittl

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Oh you reminded me,it's a bracket missed.Why it's called T_VARIABLE? – omg Sep 23 '09 at 09:32
2

The error message tells you what's there (unexpected), not what's missing. And there's a variable, so it tells you exactly that. – OregonGhost Sep 23 '09 at 09:34
2

Because the unexpected `$list` is a variable. – Michael Krelin - hacker Sep 23 '09 at 09:34
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because php was expecting a bracket and found a variable, that's why it tells you ‘unexpected variable’ – knittl Sep 23 '09 at 09:37
Ohh! it's a semicolon missed :-\ – Jayani Sumudini May 12 '21 at 16:35

score 26 · Answer 2 · edited Apr 23 '20 at 10:34

It could be some other line as well. PHP is not always that exact.

Probably you are just missing a semicolon on previous line.

How to reproduce this error, put this in a file called a.php:

<?php
  $a = 5
  $b = 7;        // Error happens here.
  print $b;
?>

Run it:

eric@dev ~ $ php a.php

PHP Parse error:  syntax error, unexpected T_VARIABLE in
/home/el/code/a.php on line 3

Explanation:

The PHP parser converts your program to a series of tokens. A T_VARIABLE is a Token of type VARIABLE. When the parser processes tokens, it tries to make sense of them, and throws errors if it receives a variable where none is allowed.

In the simple case above with variable $b, the parser tried to process this:

$a = 5 $b = 7;

The PHP parser looks at the $b after the 5 and says "that is unexpected".

score 2 · Answer 3 · edited Apr 23 '20 at 10:35

2

In my case it was an issue of the PHP version.

The .phar file I was using was not compatible with PHP 5.3.9. Switching interpreter to PHP 7 did fix it.

edited Apr 23 '20 at 10:35

Peter Mortensen

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answered Jul 18 '18 at 09:23

Zank

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What is unexpected T_VARIABLE in PHP?

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