PHP foreach 引用

<?php

$a = array(
	"str1" => "val1",
	"str2" => "val2",
	"str3" => "val3",
	"str4" => "val4",
);

$b = array(
	"str1", 
	"str2",
	"str3",
	"str4",

);

foreach ($b as &$val) {
	$val = $a[$val];
}

unset($val);

foreach ($b as $val) {
	echo "-----$b[3]";
	echo $val."
";
}




?>

<?php$a = array("str1" => "val1","str2" => "val2","str3" => "val3","str4" => "val4",);$b = array("str1", "str2","str3","str4",);foreach ($b as &$val) {$val = $a[$val];}foreach ($b as $val) {echo "-----$b[3]";echo $val." ";}?>

输出：

-----val1val1
-----val2val2
-----val3val3
-----val3val3

 

 

可以看到，因为第一个foreach 的引用最后的$[2] 引用没有关闭，所以，在第二个foreach 中已然在不断的对地址进行写操作；

 

通过打印$b[2] 的值，可以看出。随着foreach 的运行，$b[2] 的值在不断的改变。这就导致当foreach 运行到$b[1] 的时候，$b[2] 的值

 

就等于$b[1],  那么在最后一次运行时，$b[2] = $b[2]  就是的结果与$b[1] 一样了

 

 

解决方案：

使用unset () 方法将引用变量释放后再进行第二次foreach 循环。

 

<?php

$a = array(
	"str1" => "val1",
	"str2" => "val2",
	"str3" => "val3",
	"str4" => "val4",
);

$b = array(
	"str1", 
	"str2",
	"str3",
	"str4",

);

foreach ($b as &$val) {
	$val = $a[$val];
}

unset($val);

foreach ($b as $val) {
	echo "-----$b[3]";
	echo $val."
";
}




?>

输出结果：

-----val4val1
-----val4val2
-----val4val3
-----val4val4

 

 

方案二（不怎么好）：

 

 

<?php

$a = array(
	"str1" => "val1",
	"str2" => "val2",
	"str3" => "val3",
	"str4" => "val4",
);

$b = array(
	"str1", 
	"str2",
	"str3",
	"str4",

);

foreach ($b as &$val) {
	$val = $a[$val];
}

// unset($val);

foreach ($b as $item) {
	echo "-----$b[3]";
	echo $item."
";
}




?>

 

换掉第二次foreach 中的as  后的变量，不使用被引用的变量，那么不会出现这样的情况。

输出：

-----val4val1
-----val4val2
-----val4val3
-----val4val4

 

总结： 建议在引用之后使用unset() 对其释放。