Question

    <script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jquery/1.7.2/jquery.min.js"></script>     
<script type="text/javascript" src="js/ddslick.js"></script>     

<div id="myDropdown"></div>     

 <?  
     $req = "SELECT firstname,id "  ."FROM user "   ."WHERE firstname LIKE '%".$_REQUEST['term']."%' ";  
     $query = mysql_query($req); 
     while($row = mysql_fetch_array($query)) {  
    $results[] = array('text' => $row['firstname'],'value' => $row['id'],'selected' => 
    'false','description' => $row['firstname']); }   

     $r=json_encode($results);

    ?> 

<script>

     $('#myDropdown').ddslick({    
     data:<? echo $r?>,    
     width:300,    
     selectText: "Select your preferred social network",   
      imagePosition:"right",    
     onSelected: function(selectedData){         

     }   
     });     
</script>

Answer 1

与上面提到的user007一样，您忘记了结束“＆gt;”为最后一个脚本块。

另外：

onSelected: function(selectedData){         //callback function: do something with selectedData;     }    });

将抛出错误，因为注释“// calback”在结束}}之前;）;标记，导致解析器抛出错误，因为您的代码格式不正确，使其无法运行。一个有用的工具是监控浏览器控制台。

此代码在ajax加载函数中不起作用，但它以普通的php形式工作

1 个答案: