php foreach as $var 'Undefined variable'

Question

The undefined variable is $new_string.

I wrote this little script because I needed it, (don't ask why).

    $string  = "abcdefghijklmnopqrstuvwxyz"; // just used as an example

    // $string becomes "badcfehgjilknmporqtsvuxwzy"

    $now_in_pairs = str_split($string, 2);

    $reverse = array_map('strrev', $now_in_pairs);

    foreach($reverse as $r)   { 

    $new_string .= $r;  

    }

    echo $new_string;

I know I could simply say, $new_string = NULL at the start to avoid the Undefined variable but that doesn't help me to understand why it isn't defined.

In very layman's terms, $r equals the value of each pairs in the array?

How can $new_string be undefined when it equals $r?

Jenz · Accepted Answer · 2014-09-26 05:18:50Z

2

foreach($reverse as $r)   { 
    $new_string .= $r;  
    }

$new_string .= $r; means $new_string = $new_string . $r; Here you are appending $r to the value of variable $new_string which is not declared or defined before this code ie, which doesn't exist. You can solve this issue by declaring the variable above foreach like :

$new_string="";
foreach($reverse as $r)   { 
        $new_string .= $r;  
        }

edited Sep 26, 2014 at 5:18

answered Sep 26, 2014 at 5:13

Jenz

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1 Comment

Tom Over a year ago

I understand. Thank you. I assumed the first loop would be considered the equals and the following loops be the appended. My assumption was wrong.

Todor · Accepted Answer · 2014-09-26 06:36:36Z

1

The OP is looking for solution where $new_string doesnt need to be initialized beforehand.

The solution is to use the implode function of PHP

$new_string = implode($reverse);
//another usage with glue parameter:
//$new_string = implode('', $reverse);

edited Sep 26, 2014 at 6:36

answered Sep 26, 2014 at 5:45

Todor

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1 Comment

Tom Over a year ago

Nice. Thank you. Although I didn't know the name, Implode was the function I was actually looking for when I stumbled upon that I could use .= $r to combine the output of the pairs.

abhinsit · Accepted Answer · 2014-09-26 05:22:59Z

0

There is no error in the output of your program:-

http://codepad.org/9FDqrh3D

Its giving gthe correct output for $new_string

answered Sep 26, 2014 at 5:22

abhinsit

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2 Comments

Tom Over a year ago

@abhinsit Sorry I forgot to mention that the code works as planned. I've been using it. I opened the error_log and saw the page full of undefined variable notices that I didn't expect to find.

Christian Over a year ago

Sure, when suppressing error_reporting, the notice won't be shown, but the problem still exists.

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