问题可能出在: $sql="select manager_name from managers where manager_name='$manager_name'"; 你 不仿 用 echo $sql;看看,结果! 其实我想可以这么写的: $sql="select manager_name from managers where manager_name='".$manager_name."'";试试看,肯定对的!
不行,我早试过了,echo $sql执行后的结果是 select manager_name from managers where manager_name=''没有找到该用户!但是我数据库里面确实有啊,我用另外的一个程序试过了ceshi.php:<?php $link=mysql_connect("localhost","root",""); mysql_select_db("student",$link); $query="select * from managers where manager_name='kingover'"; $result=mysql_query($query,$link); $num=mysql_fetch_array($result); if($num==0) { echo "没有找到该用户!"; } else { for($i=0;$i<=num;$i++) { $manager_id=mysql_result($result,$i,"manager_id"); $manager_name=mysql_result($result,$i,"manager_name"); $password=mysql_result($result,$i,"password"); $comment=mysql_result($result,$i,"comment"); echo"<br>编号:$manager_id"; echo"<br>帐号:$manager_name"; echo"<br>密码:$password"; echo"<br>备注:$comment"; }} mysql_close($link); ?>
如果你的php>4.2,请用$_POST[]得到传递的值。。
select manager_name from managers where manager_name=''证明你的$manager_name传过来的值是空的,请根据楼上的建议检查
"select manager_name from managers where manager_name=''"证明你的 $sql="select manager_name from managers where manager_name='$manager_name'";这一条语句出了问题,原因有二 其一:$sql 语句中的 $manager_name与你 form 传过来的变量不是一个,检查 一下是否出错了; 其二:在$result=mysql_query($sql,$db_id);语句后加上下面的语句 $error=mysql_error(); echo $error; 检查一下是否报错?
$sql="select manager_name from managers where manager_name='$manager_name'";
你 不仿 用 echo $sql;看看,结果!
其实我想可以这么写的:
$sql="select manager_name from managers where manager_name='".$manager_name."'";试试看,肯定对的!
select manager_name from managers where manager_name=''没有找到该用户!但是我数据库里面确实有啊,我用另外的一个程序试过了ceshi.php:<?php
$link=mysql_connect("localhost","root","");
mysql_select_db("student",$link);
$query="select * from managers where manager_name='kingover'";
$result=mysql_query($query,$link);
$num=mysql_fetch_array($result);
if($num==0)
{
echo "没有找到该用户!";
}
else
{
for($i=0;$i<=num;$i++)
{
$manager_id=mysql_result($result,$i,"manager_id");
$manager_name=mysql_result($result,$i,"manager_name");
$password=mysql_result($result,$i,"password");
$comment=mysql_result($result,$i,"comment");
echo"<br>编号:$manager_id";
echo"<br>帐号:$manager_name";
echo"<br>密码:$password";
echo"<br>备注:$comment";
}}
mysql_close($link);
?>
测试程序如下:
所传递的内容如下:<br><br>
用户名: <? echo($HTTP_POST_VARS['manager_name']['name']); ?><BR>
密 码: <? echo($HTTP_POST_VARS['password']['password']); ?><BR>我输入用户名字和密码都为kingover的时候执行后的结果是:用户名:k
密 码:k这是怎么回事?
$sql="select manager_name from managers where manager_name='$manager_name'";这一条语句出了问题,原因有二
其一:$sql 语句中的 $manager_name与你 form 传过来的变量不是一个,检查
一下是否出错了;
其二:在$result=mysql_query($sql,$db_id);语句后加上下面的语句
$error=mysql_error();
echo $error;
检查一下是否报错?