问题描述
为什么会这样:
foreach( $store as $key => $value){
$value = $value.".txt.gz";
}
unset($value);
print_r ($store);
Array
(
[1] => 101Phones - Product Catalog TXT
[2] => 1-800-FLORALS - Product Catalog 1
)
我正在尝试获取 101Phones - 产品目录 TXT.txt.gz
I am trying to get 101Phones - Product Catalog TXT.txt.gz
对正在发生的事情的想法?
Thoughts on whats going on?
好吧，我找到了解决方案……我的数组中的变量有我看不到的值……正在做
Alright I found the solution...my variables in my array had values I couldn't see...doing
$output = preg_replace('/[^(x20-x7F)]*/','', $output);
echo($output);
清理并使其正常工作
推荐答案
文档 http://php.net/manual/en/control-structures.foreach.php 清楚地说明了您遇到问题的原因:
The doc http://php.net/manual/en/control-structures.foreach.php clearly states why you have a problem:
为了能够在循环中直接修改数组元素，在 $value 前面加上 &.在这种情况下，值将通过引用赋值."
"In order to be able to directly modify array elements within the loop precede $value with &. In that case the value will be assigned by reference."
<?php
$arr = array(1, 2, 3, 4);
foreach ($arr as &$value) {
$value = $value * 2;
}
// $arr is now array(2, 4, 6, 8)
unset($value); // break the reference with the last element
?>
只有当迭代数组可以被引用(即如果它是一个变量)时，才可能引用 $value.以下代码不起作用:
Referencing $value is only possible if the iterated array can be referenced (i.e. if it is a variable). The following code won't work:
<?php
/** this won't work **/
foreach (array(1, 2, 3, 4) as &$value) {
$value = $value * 2;
}
?>
这篇关于更改 foreach 循环内的值不会更改正在迭代的数组中的值的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持跟版网！