Question

我有一个jQuery Post到PHP页面，生成一个JSON响应。它成功生成了JSON，因为我显示了整个响应。但就我而言，我无法提醒它。任何想法都赞赏。谢谢。 JSON看起来像：

[{"Part_Field":"Part_Note:3","Part_Value":"ValueofPart"},{"Part_Field":"Ft_In:3","Part_Value":"12"}, ...

jQuery的：

 $.post('/scripts/update_detail.php' , field_id + "=" + value, function(data){
              var d = document.getElementById("displayjson");
                d.innerHTML = data;

               $.each( data, function( key, value ) {
                  alert( value.Part_Field );
                });


   });

Answer 1

你最后忘记了字符串},"json");。

更正后的代码应为：

 $.post('/scripts/update_detail.php' , field_id + "=" + value, function(data){
              var d = document.getElementById("displayjson");
                d.innerHTML = data;

               $.each( data, function( key, value ) {
                  alert( value.Part_Field );
                });


   },"json"); // Here, you need to tell jQuery that whatever response
// we are getting from server is a JSON string, in your case, 
//it was considering it as a string.

Jquery Post与JSON响应工作

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