我正在尝试从指定的URL返回JSON数据,但是当警报弹出时,它仅显示[object Object](我意识到object对象实际上不是错误)。 我想在警报中吐出职位名称和其他字段 。我该怎么做呢?
这是我正在查看的JSON示例(完整文件包含大约30个帖子)
[ { "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\n-A 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。
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); } }