Question

这是我的sample.json文件

  "general": {
    "option_name" : "option_name",
    "filter_name" : "filter name"
}

我想用密钥

更新值

$string = "general.filter_name";

$updateContent = "new filtername";

$langArray = explode('.',$string);

print_r($langArray);

/*Array
(
[0] => general
[1] => filter_name
) */


$file='assets/sample.json';

$jsonString = file_get_contents($file);
$data = json_decode($jsonString, true);

**$data['general']['filter_name'] = $updateContent; **


$newJsonString = json_encode($data);
file_put_contents($file, $newJsonString);

这里我想分配数组

 [0] => general
 [1] => filter_name

应该是

$data['general']['filter_name']

如何定义这样的数组？感谢

Answer 1

你可以使用引用来做到这一点：

$json = file_get_contents('assets/sample.json');
$json = json_decode($json, true);

$path = "general.filter_name";
$path = explode('.', $path);

$ref = &$json;

foreach ($path as $key) {
    $ref = &$ref[$key];
}

$ref = "new filtername";
unset($ref);

工作原理：

为您的阵列创建参考：$ref = &$json;
遍历$path项以获取$json['general']['filter_name']的值：
1. 在第一次迭代$ref将引用$json['general']
2. 在第二次迭代$ref上将引用$json['general']['filter_name'] - 这正是我们想要的
分配$ref = "new filtername";时，您需要分配$json['general']['filter_name'] = "new filtername";
不要忘记删除引用unset($ref);，如果不这样做，就有可能更改$ref，从而更改$json['general']['filter_name']

如何在PHP中使用explode值分配数组变量？

1 个答案: