PHP下拉菜单-我的JavaScript错误 [英] PHP Dropdown menu - i have error in javascript

问题描述

<tr>
                    <td>Menu Name</td>
					<td><select name="menu_name"  önChange="showEmail(this.value)" >
                       
                        <?php require 'connection.php';<br mode="hold" /??>						$sql=mysql_query("select * from managing_menu where type='content'");
						
						?> <option value="">---Select Menu---</option>
								  
								<?php while($row=mysql_fetch_array($sql))<br mode="hold" /??>								{
								$menuId=$row['menuId'];
								$menu_name=$row['menu_name'];
								?>
                        <option value="<?php  echo $menuId;?>"??><?php  echo $menu_name;???></option>
                        <?php } ???>
                      </select> </td>                               
					</tr>
            <tr>
              <td colspan="2">
			  			  		$menuId=@$_REQUEST['menuId'];
					$sql="SELECT * FROM managing_menu WHERE menuId=$menuId";
					//echo "SELECT * FROM managing_menu WHERE menuId=$menuId";
					$r = mysql_query($sql);
					$row = @mysql_fetch_array($r);
					?>
				<input type="text" name="description" value="<?php echo $row['description'];?>"??></td></tr>
<script type="text/javascript" language="javascript">
				      function showEmail(menuId)
						{
						var url = "aa.php";
						if(menuId=='')
						{
						alert('Please Select the Menu Name');
						return false;
						}
						else
							{
							url = url + "?menuId=" + menuId;
							window.location = url;
							} 
						}
</script>

我在Java脚本中出现错误

i have error in java script

推荐答案

sql = mysql_query(" )； ?> < option value = " > ---选择菜单--- </ > <?php while (

sql=mysql_query("select * from managing_menu where type='content'"); ?> <option value="">---Select Menu---</option> <?php while(

row = mysql_fetch_array(

row=mysql_fetch_array(

sql))< ; br mode = " /??> {

sql))<br mode="hold" /??> {

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