为什么我输入正确用户名跟密码后都不会跳转的呢?<?php
include("conn.php");
include("head.php");
if($_post['submit']){
$sql = "select * from login";
// where username = $_post[username]";
$squery = mysql_query($sql);
$row = mysql_fetch_array($squery);
if($_post[username]==$row[username]){
if($_post[password]==$row[password]){
echo "<script language=\"javascript\">location.href='list.php';</script>";
}
}
}
?><script language=javascript>
function CheckLogin(){
if(myform.username.value==""){
alert("请填写id");
return false;
}
if(myform.password.value==""){
alert("请填写密码");
return false;
}else{return;}
}
</script>
<form action="" name="myform" method="post" onsubmit="CheckLogin()">
ID:<input type="text" name="username"/><br/>
PW:<input type="password" name="password"/><br>
<input type="submit" name="submit" value="登 陆"/>
</form>
include("conn.php");
include("head.php");
if($_post['submit']){
$sql = "select * from login";
// where username = $_post[username]";
$squery = mysql_query($sql);
$row = mysql_fetch_array($squery);
if($_post[username]==$row[username]){
if($_post[password]==$row[password]){
echo "<script language=\"javascript\">location.href='list.php';</script>";
}
}
}
?><script language=javascript>
function CheckLogin(){
if(myform.username.value==""){
alert("请填写id");
return false;
}
if(myform.password.value==""){
alert("请填写密码");
return false;
}else{return;}
}
</script>
<form action="" name="myform" method="post" onsubmit="CheckLogin()">
ID:<input type="text" name="username"/><br/>
PW:<input type="password" name="password"/><br>
<input type="submit" name="submit" value="登 陆"/>
</form>
拿到数据库中运行一下看看。
if($_post['submit']){
==》
if(isset($_POST['submit'])){这行下面加上: print_r($_POST); //看能取到值吗
if($_post['submit']){
之后添加了print_r($_post);语句,打印不出什么来,没有报错!最后我把if($_post['submit']){
去掉了,一刷新直接跳转了,不用验证的!
正确的验证方法是select * from login where name = $_POST['name'];然后检验这个密码与输入的密码是否一致,如果一致则登录成功。