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PHP/MySQL检查登录详细信息_Php_Mysql_Html_Database_Forms - Fatal编程技术网

PHP/MySQL检查登录详细信息

PHP/MySQL检查登录详细信息,php,mysql,html,database,forms,Php,Mysql,Html,Database,Forms,我有下面的代码,在提交表单时应检查登录详细信息: <html><body> <?php $con = mysql_connect("localhost","will","blahblah"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("blahblah", $con); $password = $_POST['password']; $use

我有下面的代码,在提交表单时应检查登录详细信息:

<html><body>
<?php
$con = mysql_connect("localhost","will","blahblah");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
  mysql_select_db("blahblah", $con);
$password = $_POST['password'];
$username = $_POST['username'];
$query2=mysql_query("SELECT * FROM users  WHERE  pass =md5('$password') AND user = $username)or die(mysql_error());
if(mysql_num_rows($query2)==1) {
$sql="INSERT INTO links (link, notes, username)
VALUES
('". mysql_escape_string($_POST['link']) ."','". mysql_escape_string($_POST['notes']) ."','". mysql_escape_string($_POST['username']) ."')";

if (!mysql_query($sql,$con))
  {
  die('Error: ' . mysql_error());
  }
echo "1 link added";

mysql_close($con)
}
else {
echo "Username or password was not correct, please try again";
}
?>
</body></html>

我真的不知道发生了什么。谢谢

您缺少的结尾引号

$query2=mysql_query("SELECT * FROM users  WHERE  pass =md5('$password') AND user = $username)or die(mysql_error());
应该是

$query2=mysql_query("SELECT * FROM users  WHERE  pass =md5('$password') AND user = $username")or die(mysql_error());

请看上面的几行:

$query2=mysql\u查询(“

没有结尾“



Will Evans,请停止滥发问题。为什么不买一本关于PHP的好书读一读。错误是你忘了一个'or'"在某个地方终止字符串。你应该使用带有语法突出显示的编辑器。不需要因为这是一个新手问题而进行向下投票?呵呵,即使我们可能会将语法突出显示为正确回答了问题。@Znarkus:请撤销你的向上投票。他今天已经发布了11个质量相当低的问题。这是鼓励他不会让情况好转。伙计们,我才刚刚开始,我需要帮助……如果你愿意帮助,那么我会在其中得到错误未知列“john”,其中clauseI get unknown column“john”,clauseYou在查询中没有引用大部分字符串。
user='$username'
我已经引用了它们,我得到了一个inc当我在几秒钟前创建密码或用户名时,请更正密码或用户名…即使密码正确,也会出现错误…设置密码的注册表单如下:function addNewUser($username,$password){global$connection;$password1=md5($password);$username=mysql\u real\u escape\u string($username);$q=“INSERT INTO”.TBL\u USERS。“值('$username','$password1')”;返回mysql_查询($q,$connection);似乎输入的内容不正确。可能希望检查数据库中是否有条目。有针对用户的条目和md5密码。它们位于正确的列中。链接表中没有条目,因为它说密码或用户名错误,所以无法访问该条目……a有问题如果我使用:$query2=mysql\u query(“从pass=md5(“$password”)和user=username的用户中选择*)或die(mysql\u error()),那么你发布的实际查询是有效的。请使用你的安全方法发布一个固定的查询。你必须确保你的mysql\u real\u escape\u字符串是用户名,否则它会有bug。如果你有
$username=$\u post['username']
,使其成为
$username=mysql\u real\u escape\u字符串($\u POST['username']);
$query2=mysql_query("SELECT * FROM users  WHERE  pass =md5('$password') AND user = $username")or die(mysql_error());
<html><body>
<?php
$con = mysql_connect("localhost","will","blahblah");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
  mysql_select_db("blahblah", $con);
$password = $_POST['password'];
$username = $_POST['username'];
$query2=mysql_query("SELECT * FROM users  WHERE  pass = '".md5('$password')."' AND user = '".mysql_real_escape_string($username)."'") or die (mysql_error());
if(mysql_num_rows($query2)==1) {
$sql="INSERT INTO links (link, notes, username)
VALUES
('". mysql_real_escape_string($_POST['link']) ."','". mysql_real_escape_string($_POST['notes']) ."','". mysql_real_escape_string($_POST['username']) ."')";

if (!mysql_query($sql,$con))
  {
  die('Error: ' . mysql_error());
  }
echo "1 link added";

mysql_close($con)
}
else {
echo "Username or password was not correct, please try again";
}
?>
</body></html>