PHP变量范围：通过引用将默认值的参数传递给匿名函数？

Question

I'm trying to loop through a mysql result and run an anonymous function on each record, but I'm having some trouble with variable scope. The problem is part of a much bigger flow but I've created a simplified example here which demonstrates the issue simply:

function runLoop($some_mysql_resource){
    $res = 'RESULTS: ';
    sqlEach($some_mysql_resource, function($sql){
        $res .= $sql['id'].',';
    })
    return $res;
);

Is there any way to make the runLoop function above output the following without changing the logical flow?

RESULTS: 1,2,3,4,etc...

Answer 1

Pass $res as an argument "by reference" to your callback using use

function runLoop($some_mysql_resource){
    $res = 'RESULTS: ';
    sqlEach($some_mysql_resource, function($sql) use (&$res) {
        $res .= $sql['id'].',';
    })
    return $res;
);

I assume that $sql is defined somewhere in your real code, and so will have scope