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php json jquery和选定盒子
原标题:php json jquery and select box

我有这本《任择议定书》的法典。

$jsonArray = array();
$sql = "SELECT ID,CLIENT FROM PLD_SERVERS";
$result = mysql_query($sql);
while($row = mysql_fetch_array($result)) {
    $jsonArray[] = array( id =>$row[ ID ], client =>$row[ CLIENT ]);
}
echo json_encode($jsonArray);

而这种 j

function autosearchLoadServers()
{

  $.post("php/autosearch-load-servers.php",function(data){
      var toAppend = "";
      for(var i = 0; i < data.length; i++){
          toAppend +=  <option value = "  + data[i].id +  ">  + data[i].client +  </option> ;
      }
      $("#serverSelect").empty();
      $("#serverSelect").html(toAppend);
  });

}

问题是,我只得到未界定的价值。 如何做到这一点? 这些数值在JSON,即用大鼠的火力检查,因此,必须有一些数据变量,但可以理解什么。 我尝试了不同的方法,没有结果。

最佳回答

具体列明员额要求中的数据类型如下:

 $.post("php/autosearch-load-servers.php",function(data){
  var toAppend = "";
  for(var i = 0; i < data.length; i++){
      toAppend +=  <option value = "  + data[i].id +  ">  + data[i].client +  </option> ;
  }
  $("#serverSelect").empty();
  $("#serverSelect").html(toAppend);
}, "json");
问题回答

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