Question

假设我有以下JSON字符串：

$json = '[{"Name":" Jim", "ID":"23", "Age": "0"},{"Name":" Bob", "ID":"53", "Age": "0"}]';

我如何只在更新的JSON字符串中显示属性'Name'？例如，我希望在更新的变量$json2中将代码转换为此代码：

$json2 = '[{"Name":" Jim"},{"Name":" Bob"}]';

我尝试使用下面的代码执行此操作，但收到以下错误：

注意：未定义的索引：第9行的名称

$json = '[{"Name":" Jim", "ID":"23", "Age": "0"},{"Name":" Bob", "ID":"53", "Age": "0"}]';
$decode = json_decode($json, 'false'); 
$json2 = json_encode($decode['Name']);

echo $json2;

$json2返回'null'。

Answer 1

对于PHP 5.3 +：

<?php
$json = '[{"Name":" Jim", "ID":"23", "Age": "0"},{"Name":" Bob", "ID":"53", "Age": "0"}]';
$decode = json_decode($json, true);

$newArray = array_map(function ($array) {
    return ['Name' => $array['Name']];
}, $decode);

echo json_encode($newArray);

Answer 2

$json = '[{"Name":" Jim", "ID":"23", "Age": "0"},{"Name":" Bob", "ID":"53", "Age": "0"}]';
$decoded = json_decode($json, true); 

$transformed = array_map(function (array $item) {
    return array_intersect_key($item, array_flip(['Name']));
}, $decoded);

$json2 = json_encode($transformed);

array_intersect_key是the easiest method从数组中提取特定键，并且在整个数组中array_map中执行此操作是您正在寻找的。

解码完整数组并对其进行编码

2 个答案: