比较电子邮件和密码是否为真时，在else条件中出现语法错误[重复]

Question

//Help me to solve the error in below code:

/*解析错误：语法错误，C:\xampp\htdocs\myfiles\register.php中第30*/行出现意外的“else”(T_ELSE)

<?php
require('config.php');

if(isset($_POST['submit']))
{       // first curly brace
        $email1=$_POST['email1'];
        $email2=$_POST['email2'];
        $pass1=$_POST['pass1'];
        $pass2=$_POST['pass2'];

        if($email1 == $email2)
        {//second curly brace
            if($pass1 == $pass2)
            {//third curly brace
        //I have to add this function here so i can't take else condition here
            $name = mysql_escape_string($_POST['$name']);
            $lname = mysql_escape_string($_POST['$lname']);
            $uname = mysql_escape_string($_POST['$uname']);
            $email1 = mysql_escape_string($_POST['$email1']);
            $email2 = mysql_escape_string($_POST['$email2']);
            $pass1 = mysql_escape_string($_POST['$pass1']);
            $pass2 = mysql_escape_string($_POST['$pass2']);

            }//third closed
            else
            {//fourth curly brace
                echo "Your passwords do not match<br/>";
                exit();//Working properly upto here
            }//fourth closed
            else //It show me the error in this line
            {   //fifth curly brace
                echo "Your email do not match<br/>";
                exit();
            }//fifth closed
        }//second closed

$form = '       
<form action="register.php" method="POST">

First Name:<input type="text" name="name"/><br/>
Last Name:<input type="text" name="lname"/><br/>
Username:<input type="text" name="uname"/><br/>
Email:<input type="text" name="email1"/><br/>
Confirm Email:<input type="text" name="email2"/><br/>
Password:<input type="password" name"pass1"/><br/>
Confirm Password:<input type="password" name="pass2"/><br/>
<input type="submit" value="Register" name="submit"/><br/>
</form>
';//The problem of EOT is solved    
echo $form;
}//sixth closed
?>

Answer 1

看起来您添加了一个额外的else语句，但没有关闭第一个else语句。虽然，因为你要退出，你有很多不必要的额外杂乱。这应该与您的代码具有相同的功能。下面是一个更容易处理此表单的方法：

if( isset($_POST['submit']) )
{       
    $email1 = $_POST['email1'];
    $email2 = $_POST['email2'];
    $pass1  = $_POST['pass1'];
    $pass2  = $_POST['pass2'];

    if ( $email1 != $email2 )
    {
        echo "Your email does not match<br/>";
        exit;
    }

    if ( $pass1 != $pass2 )
    {
        echo "Your passwords do not match<br/>";
        exit;
    }
}
else
{
    echo '      
    <form action="register.php" method="POST">

    First Name:<input type="text" name="name"/><br/>
    Last Name:<input type="text" name="lname"/><br/>
    Username:<input type="text" name="uname"/><br/>
    Email:<input type="text" name="email1"/><br/>
    Confirm Email:<input type="text" name="email2"/><br/>
    Password:<input type="password" name"pass1"/><br/>
    Confirm Password:<input type="password" name="pass2"/><br/>
    <input type="submit" value="Register" name="submit"/><br/>
    </form>
    ';
}
?>

从可读性的观点来看，这是更好的遵循，并将使事情更容易理解。