php 更改 foreach 循环内的值不会更改正在迭代的数组中的值
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Changing value inside foreach loop doesn't change value in the array being iterated over
提问by user1179295
Why does this yield this:
为什么会产生这个：
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
)
I am trying to get 101Phones - Product Catalog TXT.txt.gz
我正在尝试获取 101Phones - 产品目录 TXT.txt.gz
Thoughts on whats going on?
关于发生了什么的想法？
EDIT: Alright I found the solution...my variables in my array had values I couldn't see...doing
编辑：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......做
$output = preg_replace('/[^(\x20-\x7F)]*/','', $output);
echo($output);
Cleaned it up and made it work properly
清理它并使其正常工作
回答by Michel
The doc http://php.net/manual/en/control-structures.foreach.phpclearly states why you have a problem:
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：
"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."
“为了能够直接修改循环中的数组元素，在 $value 前面加上 &。在这种情况下，值将通过引用分配。”
<?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
?>
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:
仅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 $value。以下代码将不起作用：
<?php
/** this won't work **/
foreach (array(1, 2, 3, 4) as &$value) {
$value = $value * 2;
}
?>
回答by Andy
Try
尝试
foreach( $store as $key => $value){
$store[$key] = $value.".txt.gz";
}
回答by deceze
The $valuevariable in the array is temporary, it does not refer to the entry in the array.
If you want to change the original array entry, use a reference:
$value数组中的变量是临时的，它不引用数组中的条目。
如果要更改原始数组条目，请使用引用：
foreach ($store as $key => &$value) {
// ^ reference
$value .= '.txt.gz';
}
回答by Daniel
You are rewriting the value within the loop, and not the key reference in your array.
您正在重写循环中的值，而不是数组中的键引用。
Try
尝试
$store[$key] = $value.".txt.gz";
回答by k102
pass $valueas a reference:
$value作为参考传递：
foreach( $store as $key => &$value){
$value = $value.".txt.gz";
}
回答by Baba
Try
尝试
$catalog = array();
foreach( $store as $key => $value){
$catalog[] = $value.".txt.gz";
}
print_r ($catalog);
OR
或者
foreach( $store as $key => $value){
$store[$key] = $value.".txt.gz";
}
print_r ($store);
Depends on what you want to achieve
取决于你想要达到的目标
Thanks :)
谢谢 ：）
回答by Chibueze Opata
I believe this is what you want to do:
我相信这是你想要做的：
foreach( $store as $key => $value){
$store[$key] = $value.".txt.gz";
}
unset($value);
print_r ($store);
回答by Ankush Jetly
foreach(array_container as & array_value)
Is the way to modify array element value inside foreach loop.
是在 foreach 循环中修改数组元素值的方法。
回答by Jaros?aw Gomu?ka
How about array map:
数组映射如何：
$func = function($value) { return $value . ".txt.gz"; };
print_r(array_map($func, $store));
