简单的PHP注册表-工作了几个月，没有任何更改，现在mysql插入失败
这张登记表有效地使用了好几个月。我什么也没变。现在，它通过了重复名称、电子邮件和密码检查的所有条件，然后无法插入mysql并返回简单的PHP注册表-工作了几个月，没有任何更改，现在mysql插入失败,php,registration,Php,Registration,这张登记表有效地使用了好几个月。我什么也没变。现在，它通过了重复名称、电子邮件和密码检查的所有条件，然后无法插入mysql并返回“发生了错误。您的帐户没有创建。”我不明白为什么。语法改变了还是怎么了 <div id="backdrop"></div> <div id="register"> <img src="http://www.staketheclaim.com/wp-content/themes/retlehs-ro
“发生了错误。您的帐户没有创建。” <div id="backdrop"></div>
<div id="register">
<img src="http://www.staketheclaim.com/wp-content/themes/retlehs-roots-c526a84/dropbox/2012/rotate/header<?php echo(rand(1,4)); ?>.png" style="margin-left: -25px;margin-top: -20px;" />
<div id="regi" style="width:400px;float: right;">
<?php
if ($username && $userid) {
echo "<div id='log-re' style='margin-left: 6px;width: 413px;'><h2>You must logout to register a new account. Not your Account?</h2>" . "<br /><br /><div id='cta'><a href='http://www.staketheclaim.com/logout/' class='button' style='padding-left: 36px;font-size: 24px;top: 2px;right: -160px;'>Logout Now</a></div></div>";
}
else {
if ($_POST['registerbtn']) {
$getuser = $_POST['user'];
$getemail = $_POST['email'];
$password = $_POST['pass'];
$getretypepass = $_POST['retypepass'];
if ($getuser) {
if(strpos($getuser, ' ') > 0 == false ){
if ($getemail) {
if ($password) {
if ($getretypepass) {
if ( $password === $getretypepass) {
if ( (strlen($getemail) >= 7) && (strstr($getemail, "@")) && (strstr($getemail, "."))){
require("base.php");
$query = mysql_query("SELECT * FROM users WHERE username='$getuser'");
$numrows = mysql_num_rows($query);
if ($numrows == 0) {
$query = mysql_query("SELECT * FROM users WHERE email='$getemail'");
$numrows = mysql_num_rows($query);
if ($numrows == 0) {
$password = md5(md5("ss3verds4g".$password."ss357rd5sg"));
$date = date("F d Y");
$code = md5(rand());
$bio = "Bio";
$location = "Location";
mysql_query("INSERT INTO users VALUES (
'','$getuser', '$password', '$getemail', '0', '$code', '$date', '$bio', '$location', '1'
)");
$query = mysql_query("SELECT * FROM users WHERE username='$getuser'");
$numrows = mysql_num_rows($query);
if ($numrows == 1){
$site = "http://www.staketheclaim.com";
$webmaster = "noreply <noreply@staketheclaim.com>";
$headers = "From: $webmaster";
$subject = "Activate Your Account";
$message = "Thanks for registering. Click the link below to activate your account.\n";
$message .= "$site/activate/?user=$getuser&code=$code\n";
$message .= "You must activate your account to login.";
if (mail($getemail, $subject, $message, $headers )) {
$errormsg = "You have been registered. You must activate your account from the activition link send to <b>$getemail</b>.";
$getuser = "";
$getemail = "";
}
else
else
$errormsg ="An error has occured. Your account was not created.";
}
else
$errormsg ="Their is already a user with that email.";
}
else
$errormsg ="Their is already a user with that username.";
mysql_close;
}
else
$errormsg = "You must enter a valid email address to register.";
}
else
$errormsg = "Your passwords did not match.";
} else
$errormsg = "You must retype you password to register.";
} else
$errormsg = "You must enter a password to register.";
} else
$errormsg = "You must enter you email to register.";
} else
$errormsg = "Your username cannot have any spaces.<br />";
} else
$errormsg = "You must enter a username to register.<br />";
} $form = "<form action='' method='post' style='margin-top:-20px;'>
<h2>Sign up for StakeTheClaim™.<br /> It's free!</h2>
<br />
<font color='red'>$errormsg</font>
<br />
<br />
Username:
<br />
<input type='text' name='user' value='$getuser' style='' />
<br />
<br />
Email:
<br />
<input type='text' name='email' value='$getemail' />
<br />
<br />
Password:
<br />
<input type='password' name='pass' value='' />
<br />
<br />
Re-Password:
<br />
<input type='password' name='retypepass' value='' />
<br />
<input type='submit' name='registerbtn' value='Register' />
</form>";
echo $form;
}
?></div></div>
.png“style=”左边距：-25px；利润上限：-20px；" />
这可能是mysql的数据类型问题（嗯，实际上是限制）。您说它在update
上失败，但我只看到insert
语句
运行insert的表users
上的列类型是什么
基本上，如果达到列类型的限制，插入将失败。这与您所说的一致
编辑：另外，PHP正在贬低mysql\u*
调用。换成mysqli\u*
或另一种类似PDO
的替代方法。换成mysqli
这段代码很容易，但是你的代码需要全面检查，它充满了问题。你运行的是什么版本的PHP？你的代码容易受到sql injection和xss…是的，这是一个非常基本的功能。有更好的代码/插件，我可以轻松实现吗？或者有什么东西可以在修复后保护此代码？我想首先要做的是了解代码中发生了什么。您对代码目前如何工作有任何疑问吗？您可能希望执行某种f更改为密码的双md5编码方法（实际上是非常蹩脚的方法）。首先查看crypt（）
。用户的输入列类型为VARCHAR（50）。列类型的“限制”是什么？比如VARCHAR（50）您输入了70个字符或其他您所指的内容？哎呀，我又多了一列，所以最后忘记了一个“”。至于加密，在这种情况下我如何使用crypt（）。所有列都是VARCHAR（50）？是否有主键（通常是id
列）。如果所有列都是VARCHAR（50），应该没有问题。对于crypt（）
，请尝试：