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      更改 foreach 循环内的值不会更改正在迭代的数组

      时间:2023-09-23

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                本文介绍了更改 foreach 循环内的值不会更改正在迭代的数组中的值的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

                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html5模板网!

                上一篇:foreach 是否保证在 php 中按数组顺序迭代? 下一篇:如何在php中的foreach语句的每三个结果上输出一个

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