.= syntax php undefined variable

Question

I got this error message

Undefined variable: x in ../../../../.php on line 35

I get the error on this line.

$x .= $y->getContent();

This line of code is in a foreach loop.

How do I get rid of the error message.

If I replace the .= with just = I'm not getting the correct output.

I hope I provided enough information

And what does .= do? Thanks in advance.

perhaps you are trying to append to an uninitialised variable? try adding `$x = '';` above your `foreach () { ... }` code. — Latheesan, Mar 19 '14 at 11:22
first define it as blank value before foreach bcoz u append value in it so must be define — wild, Mar 19 '14 at 11:22

score 3 · Accepted Answer · answered Mar 19 '14 at 11:25

.= is used (in your code) to concatenate the value of $x with result of ->getContent() call on $y and write the result back into $x.

Is like write $x = $x.$y

Of course if $x does not exists (like in your example I suppose; with "not exists" I mean that hasn't a value), regardless how you wrote your expression, this will fail. Moreover, $x and $y will be considerated strings so, please pay attention to your variables type (you can't concatenate two object, for example)

score 1 · Answer 2 · answered Mar 19 '14 at 11:23

1

$x.=$y is a shortcut for $x=$x.$y

so if $x = 'cat' and $y = 'fish' then the result of $x.=$y is 'catfish'

As to your error, you need to create the variable $x 1st, out side the loop:

$x='';
foreach($var as $y){
    $x.=$y;
}

answered Mar 19 '14 at 11:23

Steve

20,703
5
41
67

score 1 · Answer 3 · answered Mar 19 '14 at 11:25

1

$x = '';
foreach()
{
   $x .= $y->getContent();
}

*it is must to define $x becouse you are append value in $x *

answered Mar 19 '14 at 11:25

wild

340
1
3
14

score 0 · Answer 4 · answered Mar 19 '14 at 11:25

0

Define x before your loop : $x = '';

answered Mar 19 '14 at 11:25

Sadok SFAR

309
2
6

score 0 · Answer 5 · answered Mar 19 '14 at 11:27

The .= operator is a string operator to concatenate strings.

$x .= $yis the same as $x = $x.$y

You could read about string operator s in the official PHP reference

If $x don't exist you can't concatenate it with $y, this is the reason of the error message.

.= syntax php undefined variable

5 Answers5