php mysql错误信息

Question

我试图让事情变得简单。当我输入现有的电子邮件和错误的密码时，两个错误都会出现。当我输入错误的电子邮件地址和正确的密码格式时，会显示电子邮件的错误消息，但是当我输入错误的密码并更正电子邮件时，两个错误消息都会显示，这应该只是错误的密码。我只是一个初学者，我想知道这些东西是如何工作的，但我在这里找不到解决方案。我应该为数组使用不同的名称吗？ 这是我的代码:

<?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

你有:

if ($userExists = true)...

这将为不比较的 $userExists 赋值，并且该子句将始终为真(转到该部分)。

你应该这样比较:

if ($userExists == true) {

  // exists 
} else {

  // not
}

并更改此密码校验码:

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");
}

为此:

if(strlen($pword)< $min_length){
 $errmsg_arr[] = 'password must contain not less than 6 characters';
 $errflag = true;
}

if (!$errflag) {

// No errors.

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");
}

您只检查密码是否太短。如果电子邮件已经存在，则不会。这检查两者。