Question

PHP 表单验证，，为什么总是提示验证码错误，而不是先判定用户名和密码

session_start();
if(isset($_POST['denglu']))
{
echo "<script>alert('".$_SESSION['randcode']."');</script>";
$username=$_POST['username'];
$password=$_POST['password'];
$randcode=$_POST['yanzheng'];
if($username!="账号/密码"&&$password!="")
{
if($_SESSION['randcode']==$randcode)

{$conn=mysql_connect("localhost","root","");
mysql_select_db("phpwork",$conn);
mysql_query("set names gb2312");
$sql="select * from 'userinfo' where username='$username'";
$checkuser=mysql_query($sql,$conn);
$rows=mysql_num_rows($checkuser);
$res=mysql_fetch_array($checkuser);
if(intval($rows)==0){echo "<script>function checktext(){alert('不存在的账号');return false;}</script>";}
else if($password!=$res[userpwd]){echo "<script>function checktext(){alert('密码错误');return false;}</script>";}
else {$_SESSION['username']=$username;
$_SESSION['password']=$password; echo "<script>function checktext(){return true;}</script>";}
}

else {echo "<script>function checktext() {alert('验证码错误');return false;}</script>";}
}
else {echo "<script>function checktext()
{
if(login.username.value==''||login.username.value=='账号/密码'){
alert('请输入账号');
return false;
}
else if(login.password.value==''){
alert('请输入密码');
return false;}
}</script>";
}
}

Answer 1

"<script>function checktext(){alert('不存在的账号');return false;}</script>你打印的是一个js自定义函数，没有定义；所以一直到校检验证码时前面即使出错也不会提示，并且继续往下执行，直到运行到检查验证码那一行时，如果验证码不对，他才报错，然后依次检查账号，密码

Answer 2

判断语句结构问题，把结构理清就好了