CODEWITHSUNDEEP
jqueryjson
user1783933
user1783933

Reputation: 1145

Use of use jQuery.getJSON() for a Url

How i use jQuery.getJSON() to obtain data for this URL?

http://gbrds.gbif.org/registry/organisation/a3c228d0-3110-11db-abb8-b8a03c50a862.json?op=contacts

If i look the result in a browser, i get this result:

[{"position":"","lastName":"","phone":"+39 06 6118286","type":"technical","city":"","country":"","isPrimaryContact":true,"postalCode":"","address":"","email":"[email protected]","description":"","province":"","firstName":"Milko Skofic","salutation":"","key":"48"},{"position":"","lastName":"","phone":"39-06-6118204","type":"administrative","city":"","country":"","isPrimaryContact":true,"postalCode":"","address":"IPGRI, Via Tre Denari, 472/a, 00057, Maccarese, Rome, Italy,","email":"[email protected]","description":"","province":"","firstName":"Ms. Sonia Dias","salutation":"","key":"49"}]

Upvotes: 0

Views: 200

Answers (3)

Felipe Oriani
Felipe Oriani

Reputation: 38626

Take a look at the getJSON method on the jQuery documentation.

The sintaxe:

$.getJSON(url, data, function success);

So, you can try something like this:

$.getJSON("http://gbrds.gbif.org/registry/organisation/a3c228d0-3110-11db-abb8-b8a03c50a862.json?op=contacts", null, function(data) {


      // loop in your result if it is an array
      $.each(data, function(i, item) {

         // use data[i].property to access each property of your array.
         // for sample:

         var p = data[i].position;
         var l = data[i].lastName;

      });​

   });

Upvotes: 1

gen_Eric
gen_Eric

Reputation: 227310

You need to use JSONP because it's not on the same domain.

$.getJSON("http://gbrds.gbif.org/registry/organisation/a3c228d0-3110-11db-abb8-b8a03c50a862.json?op=contacts&callback=?", function(data) {
    console.log(data);
});

Upvotes: 0

Miquel
Miquel

Reputation: 15675

If you are requesting the URL from a different schema/host/port combination (e.g. https://gbrds.gbif.com:8080) your browser will throw a security exception because of a violateion of the same-origin policy

One way around this would be if you could implement jsonp which jquery also supports.

Upvotes: 0

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