我想从mysql表中填充下拉值。我使用了加载事件
 在html页面中,但我不知道此代码不起作用。

HTML页面

<!DOCTYPE html>
    <html>
    <head>
    <script src="jquery.js"></script>
    <script>
    $(document).ready(function(){
      $("button").click(function(){
        $("#select1").load("se2.php");
      });
    });
    </script>
    </head>
    <body>

    <select id="select1"></select>
    <button>Get External Content</button>

    </body>
    </html>


se2.php

 <?php
    $dbhandle = mysql_connect("localhost","root","")
     or die("Unable to connect to MySQL");
    $selected = mysql_select_db("student",$dbhandle)
      or die("Could not select examples");
    $result1 = mysql_query("SELECT column1 FROM information");
     while($row=mysql_fetch_array($result1))
    {
   if($row['column1']!=NULL)
    {
    echo "<option value='$row[column1]'>$row[column1]</option>";
    }
    }
    ?>

最佳答案

改变这个

 <select id="select1"></select>


对此

<div id="select1"></div>


并将select标签添加到php

 $result1 = mysql_query("SELECT column1 FROM information");
echo "<select>";
     while($row=mysql_fetch_array($result1))
    {
   if($row['column1']!=NULL)
    {
    echo "<option value='$row[column1]'>$row[column1]</option>";
    }
    }
echo "</select>";


并给按钮一个ID

<button id="button">Get External Content</button>

09-10 08:52
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