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Home/ Questions/Q 6908087
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Editorial Team
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Editorial Team
Asked: May 27, 20262026-05-27T08:31:03+00:00 2026-05-27T08:31:03+00:00

Basically I use this handy function to processing db rows (close an eye on

  • 0

Basically I use this handy function to processing db rows (close an eye on PDO and/or other stuff)

function fetch($query,$func) {
    $query = mysql_query($query);   
    while($r = mysql_fetch_assoc($query)) {
        $func($r);
    }
}

With this function I can simply do:

fetch("SELECT title FROM tbl", function($r){
   //> $r['title'] contains the title
});

Let’s say now I need to concatenate all $r['title'] in a var (this is just an example).

How could I do that? I was thinking something like this, but it’s not very elegant:

$result = '';
fetch("SELECT title FROM tbl", function($r){
   global $result;
   $result .= $r['title'];
});

echo $result;
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-27T08:31:04+00:00Added an answer on May 27, 2026 at 8:31 am

    You have to use use as described in docs:

    Closures may also inherit variables from the parent scope. Any such
    variables must be declared in the function header. Inheriting
    variables from the parent scope is not the same as using global
    variables. Global variables exist in the global scope, which is the
    same no matter what function is executing.

    Code:

    $result = '';
    fetch("SELECT title FROM tbl", function($r) use (&$result) {
       $result .= $r['title'];
    });
    

    But beware (taken from one of comments in previous link):

    use() parameters are early binding – they use the variable’s value at
    the point where the lambda function is declared, rather than the point
    where the lambda function is called (late binding).

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