Php mysql错误信息

Question

我在努力使事情变得简单。当我输入一个现有的电子邮件和错误的密码都显示错误。当我输入错误的电子邮件地址和正确的密码格式时，显示电子邮件的错误消息，但是当我输入错误的密码和正确的电子邮件时，两个错误消息都会显示出来，而且应该只输入错误的密码。我只是一个初学者，我想知道这是如何工作的，但我在这里找不到解决方案。我应该为阵列使用不同的名称吗？ 这里是我的代码：

<?php 
session_start(); 
$errmsg_arr = array(); 
$errflag = false; 
//$errmsg_arr2 = array(); 
//$errflag2 = false; 
include('config.php'); 


$firstname=$_POST['firstname']; 
$lastname=$_POST['lastname']; 
$email=$_POST['email']; 
$pword=$_POST['pword']; 
$number=$_POST['number']; 
$house=$_POST['house']; 
$street=$_POST['street']; 
$city=$_POST['city']; 

$min_length = 6; 
    // you can set minimum length of the query if you want 
$result = mysql_query("select 1 from athan_members where email='" 
         . mysql_real_escape_string($email) . "'"); 
$userExists = (mysql_fetch_array($result) !== FALSE); 
mysql_free_result($result); 

if ($userExists = true){ 
$errmsg_arr[] = 'email address is already used'; 
$errflag = true; 
} 
if(strlen($pword)< $min_length){ 
$errmsg_arr[] = 'password must contain not less than 6 characters'; 
$errflag = true; 
} 
else{ 
mysql_query("INSERT INTO athan_members (firstname, lastname, email, number, house1,  street1, city, password) VALUES ('$firstname', '$lastname', '$email', '$number', '$house',  '$street', '$city', '$pword')"); 
header("location: loginuser.php"); 
} 

/*if(strlen($pword) >= $min_length){ 
//this one will not feed in the database if there's a duplicate but still a problem ohmaygawd:3 
//mysql_query("INSERT INTO athan_members (firstname, lastname, email, number, house1,  street1, city, password) VALUES ('$firstname', '$lastname', '$email', '$number', '$house',  '$street', '$city', '$pword') ON DUPLICATE KEY UPDATE") 
mysql_query("INSERT INTO athan_members (firstname, lastname, email, number, house1,  street1, city, password) VALUES ('$firstname', '$lastname', '$email', '$number', '$house',  '$street', '$city', '$pword')"); 
header("location: loginuser.php"); 
} 
else 
{ 
$errmsg_arr[] = 'password must contain not less than 6 characters'; 
$errflag = true; 
}*/ 


if ($errflag) { 
     $_SESSION['ERRMSG_ARR'] = $errmsg_arr; 
     session_write_close(); 
     header("location: new.php"); 
     exit(); 
} 
mysql_close($con); 
?> 

Answer 1

您有：

if ($userExists = true)... 

将分配值$ userExists没有比较和该条款将永远是真（往那个部分）。

您应该比较的是，而不是像这样：

if ($userExists == true) { 

    // exists 
} else { 

    // not 
} 

和更改这个密码校验码：

if(strlen($pword)< $min_length){ 
$errmsg_arr[] = 'password must contain not less than 6 characters'; 
$errflag = true; 
} 
else{ 
mysql_query("INSERT INTO athan_members (firstname, lastname, email, number, house1,  street1, city, password) VALUES ('$firstname', '$lastname', '$email', '$number', '$house',  '$street', '$city', '$pword')"); 
header("location: loginuser.php"); 
} 

这样：

​​

你只检查是否密码太短。不是如果电子邮件已经存在这会检查两者。