Question

我正在利用<代码>json_encode(>>)从一个阵列中生成json,该阵列运转正常,但使用钥匙:按惯例从阵列中的价值。但是,我只想改变json产出中的关键名称。这样做是可能的? 或者,我是否准备了json的钥匙:人工估价?

例:

$arr = array ( a =>1, b =>2, c =>3, d =>4, e =>5);

echo json_encode($arr);

O/P

{"a":1,"b":2,"c":3,"d":4,"e":5}

我想。

{"foo":1,"something":2,"bar":3,"foo":4,"baz":5}

编辑: 我不能篡改原来的阵列。

Answer 1

只有当你重写时。您可以使用:

$rewriteKeys = array( a  =>  foo ,  b  =>  something ,  c  =>  bar ,  d  =>  foo ,  e  =>  baz );

$newArr = array();
foreach($arr as $key => $value) {
  $newArr[ $rewriteKeys[ $key ] ] = $value;
}

echo json_encode($newArr);

不知道这符合你的目标。

Answer 2

You can always json_decode it and then re-encode it. But it s going to be easiest if you simply prepare your keys before you encode it.

Answer 3

There is another option I described here. The main idea is to consider JSON as a string and then use str_replace or preg_replace(str_replace for regexp).

There is a code for your case.

$mapping_array  = array( a  =>  foo ,  b  =>  something ,  c  =>  bar ,  d  =>  foo ,  e  =>  baz );

$tmp_arr =  array_map(function($k){ return  / .$k. /u ; }, array_keys($mapping_array));

$new_json =  preg_replace($tmp_arr, array_values($mapping_array), $old_json);

Answer 4

This may be a short answer:

foreach ($list as $key => $val) {
    $newList[$key] = [
         foo    =>  $val[ a ],
         something  =>    $val[ b ],
         bar  => $val[ c ],
         foo  => $val[ d ],
         baz  => $val[ e ],
    ];
}

友情链接