“if”正在工作或“其他”正在工作

Question

我想制作一个用户验证脚本。在这里，如果用户想要注册时，他/她必须填写登记表并点击“注册”

PHP

<?php 
$usrnm=$_POST["userName"]; 
$email=$_POST["mailID"]; 
$pwd=$_POST["Password"]; 
$firstName=$_POST["firstName"]; 
$lastName=$_POST["lastName"]; 
$confpwd=$_POST["ConfirmPassword"]; 
if ($pwd == $confpwd) 
{ 
if (!$con = @mysql_connect("localhost", "root","","login")) 
{ 
echo "connection unsuccessful\n"; 
} 
if (!$selectdb = mysql_select_db("login",$con)) 
{ 
echo "database selection unsuccessful\n"; 
} 
$sql = "SELECT userName FROM userdetails WHERE userName='$usrnm'"; 
$sql2 = "INSERT INTO userdetails (userName, Password,mailID, firstName, 
lastName) VALUES ('$usrnm','$pwd','$email','$firstName','$lastName')"; 
$retval = mysql_query($sql, $con); 
while($row = mysql_fetch_row($retval)) 
{ 

如果字段数大于0，这意味着用户名已经存在于数据库中，并且ELSE将信息添加到数据库中。我的问题是，ELSE条件不起作用，IF正在工作。我甚至尝试过使用ISSET，但仍然没有运气。

$fields=mysql_num_fields($retval); 
if ($fields>0) 
{echo "Username already exists";} 
else 
{$retval2 = mysql_query($sql2, $con); 
echo "Information added"; 
} 
} 
} 
} 
else 
{ 
echo "Opps..."; 
} 
mysql_close($con); 
?> 

Answer 1

您有1个额外的大括号接近您的else声明。

这是你的代码的清理版本：

if ($pwd == $confpwd){ 
    if (!$con = @mysql_connect("localhost", "root","","login")){ 
     echo "connection unsuccessful\n"; 
    } 
    if (!$selectdb = mysql_select_db("login",$con)){ 
     echo "database selection unsuccessful\n"; 
    } 
    $sql = "SELECT userName FROM userdetails WHERE userName='$usrnm'"; 
    $sql2 = "INSERT INTO userdetails (userName, Password,mailID, firstName, lastName) VALUES ('$usrnm','$pwd','$email','$firstName','$lastName')"; 
    $retval = mysql_query($sql, $con); 
    while($row = mysql_fetch_row($retval)){ 
     $fields=mysql_num_fields($retval); 

     if ($fields>0){ 
      echo "Username already exists"; 
     } else { 
      $retval2 = mysql_query($sql2, $con); 
      echo "Information added"; 
     } 
    } 
} else { 
    echo "Opps..."; 
} 
mysql_close($con); 

我没有测试它也不想这样做，因为你真的需要为使用mysqli()或PDO()，你绝对应该验证用户输入。

Answer 2

我猜你编写查询的方式也是错误的。您正在使用字符串变量，因此请按照以下格式编写查询。

$sql = "SELECT userName FROM userdetails WHERE userName='".$usrnm."'"; 

$sql2 = "INSERT INTO userdetails (userName, Password,mailID, firstName, lastName) VALUES('".$usrnm."','".$pwd."','".$email."','".$firstName."','".$lastName."')"; 

并且还在else语句后使用了额外的右括号。请删除。