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JSON returning [object Object] [duplicate]

Tags:

json

jquery

ajax

I am trying to return the JSON data from the specified URL but when the alert pops up it just shows [object Object] (I realize the object object is not in fact an error). I would like to spit out the position name and other fields in the alert. How do I do this?

Here is an example of the JSON I am looking at (the full file has about 30 postings)

[
  {
    "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"
  }
]

And here is the script I am using.

 $.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);
}
});

Any Help is much appreciated.

like image 801
gschervish Avatar asked Nov 19 '12 15:11

gschervish


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2 Answers

You could always turn the object into a string and alert that.

alert(JSON.stringify(data));
like image 176
William Byrne Avatar answered Oct 03 '22 18:10

William Byrne


Try this:

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);
  }
}
like image 28
eseceve Avatar answered Oct 03 '22 19:10

eseceve