Accessing outside variable using anonymous function as params

问题

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;

回答1:

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).

回答2:

What about rewriting 'fetch' to call $func only once ?

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

This way you would call $func only once and re-process the array once fetched? Not sure about the performance even tho calling 200 times a function doesn't sound like a good idea.

来源：https://stackoverflow.com/questions/8403908/accessing-outside-variable-using-anonymous-function-as-params

标签

php

closures

scope