小编典典

JSON返回[object Object]

ajax

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。


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2020-07-26

共1个答案

小编典典

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);
  }
}
2020-07-26