0

我在使用 Jquery 插件自动完成检索数据时遇到了问题。这是我的例子：

<!doctype html>
<html lang="en">
<head>
  <meta charset="utf-8">
  <title>jQuery UI Autocomplete - Remote JSONP datasource</title>
  <link rel="stylesheet" href="http://code.jquery.com/ui/1.11.2/themes/smoothness/jquery-ui.css">
  <script src="http://code.jquery.com/jquery-1.10.2.js"></script>
  <script src="http://code.jquery.com/ui/1.11.2/jquery-ui.js"></script>
  <!-- <link rel="stylesheet" href="/resources/demos/style.css"> -->
  <style>
  .ui-autocomplete-loading {
    background: white url("images/ui-anim_basic_16x16.gif") right center no-repeat;
  }
  #city { width: 25em; }
  </style>
  <script>
  $(function() {
    function log( message ) {
      $( "<div>" ).text( message ).prependTo( "#log" );
      $( "#log" ).scrollTop( 0 );
    }

    $( "#city" ).autocomplete({
      source: function( request, response ) {
        //$.getJSON('/http://search-profuturo-iiwjjl6diqb7gluv2inqunahea.us-west-2.cloudsearch.amazonaws.com/2013-01-01/search?', { q: request.term }, function(data){ response(data); });


        $.ajax({
          url: "http://search-DOMAIN-iiwjjl6diqb7gluv2inqunahea.us-west-2.cloudsearch.amazonaws.com/2013-01-01/search?",
          dataType: "jsonp",
          data: {
            q: request.term
          },
          success: function( data ) {
            response( data );
          }
        });

      },
      minLength: 3,
      select: function( event, ui ) {
        log( ui.item ?
          "Selected: " + ui.item.label :
          "Nothing selected, input was " + this.value);
      },
      open: function() {
        $( this ).removeClass( "ui-corner-all" ).addClass( "ui-corner-top" );
      },
      close: function() {
        $( this ).removeClass( "ui-corner-top" ).addClass( "ui-corner-all" );
      }
    });
  });
  </script>
</head>
<body>

<div class="ui-widget">
  <label for="city">Your city: </label>
  <input id="city">
</div>

<div class="ui-widget" style="margin-top:2em; font-family:Arial">
  Result:
  <div id="log" style="height: 200px; width: 300px; overflow: auto;" class="ui-widget-content"></div>
</div>


</body>
</html>

这是回应：

http://search-DOMAIN-iiwjjl6diqb7gluv2inqunahea.us-west-2.cloudsearch.amazonaws.com/2013-01-01/search?&callback=jQuery11020041068303398787975_1421364823901&q=starwars&_=1421364823902

这会附加所有字符串：

&callback=jQuery11020041068303398787975_1421364823901

那是错误，该字符串不是必需的。

我需要这样的网址：

http://search-DOMAIN-iiwjjl6diqb7gluv2inqunahea.us-west-2.cloudsearch.amazonaws.com/2013-01-01/search?q=starwars

我需要一个 JSON 响应。

我能怎么做？

4

1 回答 1

0

Try this

$.ajax({
    url: 'http://search-DOMAIN-iiwjjl6diqb7gluv2inqunahea.us-west-2.cloudsearch.amazonaws.com/2013-01-01/search',
    data: {
      q: 'starwars'
    }
    jsonp : false,
    jsonpCallback: 'jsonCallback',
    cache: 'true',
    dataType: 'jsonp'
});

function jsonCallback(data) {
  console.log(data);
}
于 2015-01-15T23:56:24.143 回答