情景:
用户输入参考编号,根据参考编号,我应该显示相当于它的位置。
SQL:
require_once('conn.php');
$refnum = (isset($_POST['refOff'])) ; //Get filename set in form
$query = mysql_query("SELECT * FROM pilot WHERE geo=$refnum");
// display query results
while($row = mysql_fetch_array($query))
{
$rname =$row['rname'];
$pname =$row['pname'];
$mname =$row['mname'];
}
HTML:
<tr>
<td width="283" height="32">Region:</span> </td>
<td width="407"> <input type="text"value="<?php echo $rname;?>"/></td>
</tr>
<tr>
<td width="283" height="32">Province:</span> </td>
<td width="407"> <input type="text"value="<?php echo $pname;?>"/></td>
</tr>
<tr>
<td width="283" height="32">City:</span> </td>
<td width="407"> <input type="text"value="<?php echo $mname;?>"/></td>
</tr>
问题:
正在显示错误,指出rname,pname和mname未定义。有什么问题?再次感谢
答案 0 :(得分:0)
首先,我相信你的html中有这种输入元素:
<label for='refOff'>Reference Number: </label>
<input type='text' id='refOff' name='refOff'/>
代码中的这一行:
$refnum = (isset($_POST['refOff']))
只返回一个布尔值(即true或false),并且永远不会返回用户输入的实际值#refffff&#39; html输入元素。这应该使用三元运算符很好地工作:
$refnum = (isset($_POST['refOff']))? $_POST['refOff'] : null;
if($refnum){
$query = mysql_query("SELECT * FROM pilot WHERE geo=$refnum");
// display query results
while($row = mysql_fetch_array($query))
{
$rname =$row['rname'];
$pname =$row['pname'];
$mname =$row['mname'];
}
}
古德勒克!