java - 使用 PHP 使用 JSONP 返回对象而不是结果

Question

来自远程域的我的 JSONP 响应是:

jQuery183012824459988766945_1354016515353([{"StudentID":"BA1122","LastName":"BAG","FirstName":"RON"},{"StudentID":"B770","LastName":"BAEN","FirstName":"AI"},{"StudentID":"B994","LastName":"BALD","FirstName":"AARON"},{"StudentID":"B580","LastName":"Balzstin","FirstName":"Manda"},{"StudentID":"B932","LastName":"BAR","FirstName":"ABBAS"},{"StudentID":"B139","LastName":"BANES","FirstName":"ALAN"},{"StudentID":"B718","LastName":"Baen","FirstName":"Alex"},{"StudentID":"B524","LastName":"BAER","FirstName":"ANA"}])

我正在尝试使用数据:

$.getJSON('http://www.remote_domain.com/json.php?callback=?',function(res){
    ParseJson(res);
    function ParseJson(data){
        for (var key in data) {
            if (data.hasOwnProperty(key)) {
                alert(key + " -> " + data[key]);
            }
        }
    }
});

我似乎只获取了对象的数据？

更新:

 $num_columns = $rs->Fields->Count ();
 $arrColumns = array();

 for ($i=0; $i < $num_columns; $i++) {
         $arrColumns[] = $rs->Fields($i);
         $newArr[] = $rs->Fields($i)->name; 
 }

 $arrResult = array();

 while (!$rs->EOF) {
     $arrRow = array();
     for ($i=0; $i < $num_columns; $i++) {
        $arrRow[$newArr[$i]] = $arrColumns[$i]->value;
     }
     $arrResult[] = $arrRow;

     $rs->MoveNext();
 }

 echo $_GET['callback'] . '(' . json_encode($arrResult) . ')';

我的 JSON 返回如下: [{"First":"John"},{"Last":"Doe"}] 包装在对象中的数组。我需要返回一个对象或数组，以便我可以在客户端处理，例如: {"First":"John"},{"Last":"Doe"}

<script>
    $.getJSON('http://remote.domain.com/json.php?callback=?',function(res){
        alert('Results: '+res.Last);
    });
</script>

我认为这可能是我的方式 json_encode($Array);并且不使用类？

Answer 1

试试这个

$.ajax({
    url: 'http://www.remote_domain.com/json.php',
    dataType: 'jsonp',
    success: function(data){

        $.each(data, function(k, v){
            console.log('key: ' + k + ' val: ' + v);
        });
    }
});