Question

这是我正在使用的PHP块：

$reg = @$_POST['reg'];

$fn = "";
$ln = "";
$un = "";
$em = "";
$em2 = "";
$pswd = "";
$pswd2 = "";
$d = "";
$u_check = "";

$fn = strip_tags(@$_POST['fname']);
$ln = strip_tags(@$_POST['lname']);
$un = strip_tags(@$_POST['username']);
$em = strip_tags(@$_POST['email']);
$em2 = strip_tags(@$_POST['email2']);
$pswd = strip_tags(@$_POST['password']);
$pswd2 = strip_tags(@$_POST['password2']);
$d = date("Y-m-d"); //Year - Month - Day

if ($reg) {}
if ($em==$em2){}
$u_check = mysql_query("SELECT username FROM users WHERE username='$un'");
$check = mysql_num_rows ($u_check);
if ($check == 0){}
if ($fn&&$ln&&$un&&$em&&$em2&&$pswd&&$pswd2){}
if ($pswd==$pswd2){}
if (strlen($un)>25||strlen($fn)>25||strlen($ln)>25){
echo "The maximum amount of character is 25! Please try again";
}
else
{
if (strlen($pswd)>30||strlen($pswd)<5){}
echo "Your password be between 5 and 30 characters long!";
}
else
{
$pswd = md5($pswd);
$pswd2 = md5($pswd2);
$query = mysql_query("INSERT INTO users VALUES (' ', '$un', '$fn', '$ln',   
'$em','$pswd','$d','0')");
}

以下是其他两个陈述：

else
{
if (strlen($pswd)>30||strlen($pswd)<5){}
echo "Your password be between 5 and 30 characters long!";
}
else
{
$pswd = md5($pswd);
$pswd2 = md5($pswd2);
$query = mysql_query("INSERT INTO users VALUES (' ', '$un', '$fn', '$ln',   
'$em','$pswd','$d','0')");
}

当我只有：

else
{
if (strlen($pswd)>30||strlen($pswd)<5){}
echo "Your password be between 5 and 30 characters long!";
}

我没有收到错误消息，但是当我添加：

else
{
$pswd = md5($pswd);
$pswd2 = md5($pswd2);
$query = mysql_query("INSERT INTO users VALUES (' ', '$un', '$fn', '$ln', '$em','$pswd','$d','0')");
}

刷新时收到此错误消息：

解析错误：第39行的C：\ xampp \ htdocs \ Socially \ index.php中的语法错误，意外的'else'（T_ELSE）

我正在使用YouTube教程，这是他输入的内容，他没有收到错误消息。这是链接：https://www.youtube.com/watch?v=EgqVNMTnmDQ&list=PLA7F9875BD031DC16&index=36

该视频于2013年完成。

如果有人可以帮助我，我们将不胜感激。

Answer 1

if错误{打开控制块，}关闭它。例如：

if (strlen($pswd)>30||strlen($pswd)<5){}

当密码超过30个字符或小于5时，您什么都不做。（为什么还要将密码限制为30个字符？）

然后，无论条件如何，您都会回复该消息：

echo "Your password be between 5 and 30 characters long!";

附加说明：

您的控制块应缩进。

您不应将用户数据直接放入SQL查询中。这就是注射的发生方式。 strip_tags无法阻止SQL注入。

尝试：

<?php
$reg = @$_POST['reg'];
$fn = "";
$ln = "";
$un = "";
$em = "";
$em2 = "";
$pswd = "";
$pswd2 = "";
$d = "";
$u_check = "";
$fn = strip_tags(@$_POST['fname']);//dont use @, no need for error supression, resolve the errors.
$ln = strip_tags(@$_POST['lname']);
$un = strip_tags(@$_POST['username']);
$em = strip_tags(@$_POST['email']);
$em2 = strip_tags(@$_POST['email2']);
$pswd = strip_tags(@$_POST['password']);
$pswd2 = strip_tags(@$_POST['password2']);
$d = date("Y-m-d"); //Year - Month - Day
if ($reg) {}//does nothing
if ($em==$em2){}//does nothing
$u_check = mysql_query("SELECT username FROM users WHERE username='$un'");
$check = mysql_num_rows ($u_check);
if ($check == 0){}//does nothing
if ($fn&&$ln&&$un&&$em&&$em2&&$pswd&&$pswd2){}//does nothing
if ($pswd==$pswd2){}//does nothing
if (strlen($un)>25||strlen($fn)>25||strlen($ln)>25){
    echo "The maximum amount of character is 25! Please try again";
} else {
    if (strlen($pswd)>30||strlen($pswd)<5){
        echo "Your password be between 5 and 30 characters long!";
    } else {
        $pswd = md5($pswd); //should upgrade hashing algorithm
        $pswd2 = md5($pswd2);//not used
        $query = mysql_query("INSERT INTO users VALUES (' ', '$un', '$fn', '$ln',   
        '$em','$pswd','$d','0')");//open to SQL injections
    }
}

我一直收到错误消息，我在PHP中使用了else函数

1 个答案: