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如何将json数据从php解析为jQuery

[英]How to parse json data from php to jQuery

I have a curl statement that pulls json formatted array to php. 我有一个curl语句,将json格式的数组拉到php。 I then want to transfer this array to jQuery so that the client side will hold the array. 然后我想将此数组传输到jQuery,以便客户端将保存该数组。 I'm currently using the below method: 我目前正在使用以下方法:

<script>var obj = jQuery.parseJSON( <?php echo var_dump($json_short); ?> );</script>

The client sees something like: 客户看到类似的东西:

<script>var obj = jQuery.parseJSON( array(1) {
  ["search"]=>
  array(50) {
    [0]=>
    array(6) {
      ["id"]=>
      string(6) "641279"
      ["description"]=>
      string(36) "Instyle - Responsive Portfolio Theme"
      ["url"]=>
      string(69) "http://themeforest.net/item/instyle-responsive-portfolio-theme/641279"
      ["type"]=>
      string(9) "wordpress"
      ["sales"]=>
      string(3) "135"
      ["rating"]=>
      string(3) "4.5"
    }
    ....
  }
}
 );</script>

Will obj now hold the array? obj现在会持有阵列吗? is this the right way becuase I'm getting an error: 这是正确的方法,因为我得到一个错误:

Uncaught SyntaxError: Unexpected token { 

PHP has json_encode function already, you should use that. PHP已经有了json_encode函数,你应该使用它。

Would look like: 看起来像:

<script>
var a = <?php echo json_encode($json_short); ?>;

</script>

你不能使用直接转储,你需要先json_encode:

<script>var obj = <?php echo json_encode($json_short) ?>;</script>

I don't understand what you're trying with <script>var obj = jQuery.parseJSON( <?php echo var_dump($json_short); ?> );</script> 我不明白你在尝试使用<script>var obj = jQuery.parseJSON( <?php echo var_dump($json_short); ?> );</script>

in PHP try echo json_encode($json_short); 在PHP中尝试echo json_encode($json_short);

the var_dump function doesn't dump it as a json object. var_dump函数不会将其转储为json对象。

use json_encode instead of var_dump . 使用json_encode而不是var_dump

Don't use var_dump first off. 首先不要使用var_dump。 Next make sure you have converted your variable to a json array you have a normal array. 接下来确保已将变量转换为具有普通数组的json数组。

<script>var obj = jQuery.parseJSON( <?php echo json_encode($json_short); ?> );</script>

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