Question

我如何使用jQuery.getJSON（）获取此URL的数据？

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

如果我在浏览器中查看结果，我会得到以下结果：

[{"position":"","lastName":"","phone":"+39 06 6118286","type":"technical","city":"","country":"","isPrimaryContact":true,"postalCode":"","address":"","email":"m.skofic@cgiar.org","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":"eurisco@cgiar.org","description":"","province":"","firstName":"Ms. Sonia Dias","salutation":"","key":"49"}]

Answer 1

查看jQuery文档中的getJSON方法。

Sintaxe：

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

所以，你可以尝试这样的事情：

$.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;

      });

   });

Answer 2

如果您要求来自其他架构/主机/端口组合的网址（例如https://gbrds.gbif.com:8080），则由于违反了same-origin policy

，您的浏览器会抛出安全例外

如果你能实现jquery也支持的jsonp，那么就可以解决这个问题。

Answer 3

您需要使用JSONP，因为它不在同一个域中。

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

使用jQuery.getJSON（）用于Url

3 个答案: