JSON返回

我试图从指定的URL返回JSON数据但是当弹出警报时它只显示[object Object]（我意识到对象对象实际上不是错误）。 我想在警报中吐出位置名称和其他字段 。 我该怎么做呢？

这是我正在看的JSON的一个例子（完整文件有大约30个post）

[ { "m_id": 473644, "m_positionName": "Application Monitoring Software Engineer", "m_positionLocations": [ {} ], "m_active": true, "m_description": "Job Responsibilities:\r\n\r\n-Create world class application monitoring tools and dashboards for our health care applications\r\n\r\n-Develop business rules to pro actively identify and re-mediate system-level issues before they occur.\r\n\r\n-Create business intelligence reports for internal and external use as a supplement to software products.\r\n\r\n\r\n\r\nJob Requirements:\r\n\r\n-BS or MS Degree in computer science or any engineering discipline.\r\n-4+ years of experience with Java (or other object-oriented programming language).\r\n-Experience in SQL, Struts, Hibernate, Spring, Eclipse, JSP, JavaScript.\r\n-Highly motivated and self-driven personality.\r\n-Excellent interpersonal and leadership skills.\r\nA vision for the future and a desire to make a difference.\r\n-Experience with Maven, Tomcat, PostgreSql, Jasper Reports,", "m_postedDate": "Jun 29, 2012 9:17:19 AM", "m_closingDate": "Jun 29, 2013 12:00:00 AM" } ] 

这是我正在使用的脚本。

  $.ajax({ type: "GET", url: '/wp-content/themes/twentyeleven/js/jobopenings.json', async: false, beforeSend: function(x) { if(x && x.overrideMimeType) { x.overrideMimeType("application/j-son;charset=UTF-8"); } }, dataType: "json", success: function(data){ alert(data); } }); 

任何帮助深表感谢。

Answer 1

试试这个：

 success: function(data) { var _len = data.length; , post, i; for (i = 0; i < _len; i++) { //debugger post = data[i]; alert("m_positionName is "+ post. m_positionName); } } 

Answer 2

您始终可以将对象转换为字符串并提醒它。

 alert(JSON.stringify(data)); 

Answer 3

当jQuery收到json时，jQuery会自动将其转换为javascript对象。 因此， data只包含您准备使用的对象。 如果要访问响应的原始文本，可以执行以下操作：

 success: function(data, textStatus, jqXHR){ alert(jqXHR.responseText); }