Php 如何为用户插入登录数据？_Php_Mysql_Post

Php 如何为用户插入登录数据？

php mysql post

Php 如何为用户插入登录数据？,php,mysql,post,Php,Mysql,Post,我创建了一个注册页面，但出现了一个错误注意：未定义变量：中的语句第35行的D:\xampp\htdocs\123\signup.php 它插入到数据库中，但php文件显示错误。如何将其插入数据库但仍显示错误注册页面 <!doctype html> <html> <head><title>ABU E-ADVISING</title> <link rel="stylesheet" type="text/css" href

我创建了一个注册页面，但出现了一个错误

注意：未定义变量：中的语句第35行的D:\xampp\htdocs\123\signup.php

它插入到数据库中，但php文件显示错误。如何将其插入数据库但仍显示错误

注册页面

<!doctype html>
<html>
  <head><title>ABU E-ADVISING</title>
  <link rel="stylesheet" type="text/css" href="style.css" \>
</head>
<body>
  <?php

    $Username1 = $_POST['usr'];
   $Pass1 = $_POST['pass'];
   $Surname = $_POST['sur'];
   $Othernames = $_POST['oname'];
   $DOB = $_POST['dob'];
   $email = $_POST['email'];
   $Tel = $_POST['tel'];
   $Add = $_POST['add'];
   $Postcode = $_POST['postcode'];



   $db = mysql_connect('localhost','root','ROOT');
   if (!$db)
   {
     print "<h1>Unable to Connect to MySQL</h1>";
   }

   $dbname = 'adv';

   $btest = mysql_select_db($dbname);

   if (!$btest)
   {
     print "<h1>Unable to Select the Database</h1>";
   }
   $statement .= "insert into user (userName, Password, FName, LName, DOB,Email, Tele, Address, PCode) ";
   $statement .= "values (";
   $statement .= "'".$Username1."', '".$Pass1."',  
    '".$Surname."','".$Othernames."','".$DOB."','".$Tel."', '".$Add."', 
    '".$Postcode."','".$email."'";
   $statement .= ")";
   $result = mysql_query($statement);

   if ($result)
   {
     echo "<br>Author Added: ".$Surname.", ".$Othernames;
   } else {
      $errno = mysql_errno($db);

      if ($errno == '1062') {
        echo "<br>Author is already in Table: <br />".$mylastname.", 
  ".$myfirstname;
      } else {
        echo("<h4>MySQL No: ".mysql_errno($result)."</h4>");
        echo("<h4>MySQL Error: ".mysql_error($result)."</h4>");
        echo("<h4>SQL: ".$statement."</h4>");
        echo("<h4>MySQL Affected Rows: ".mysql_affected_rows($result)."  
   </h4>");
      }

      print 'NotAdded';
   }

 ?>
  </body>

有人能帮忙吗？

这里是您遇到错误的地方：

$statement .= "insert into user (userName, Password, FName, LName, DOB,Email, Tele, Address, PCode) ";

您正在连接以前未声明的字符串变量

应改为：

 $statement = "insert into user (userName, Password, FName, LName, DOB,Email, Tele, Address, PCode) ";

更新以下行

$statement .= "insert into user (userName, Password, FName, LName, DOB,Email, Tele, Address, PCode) ";

到

$x.=$y

表示

$x=$x$y

因此，如果在此语句之前未定义

$x

，您将得到第二个

$x

（在

之后）的通知

不要使用
mysql.*
。它已被弃用且不安全。

您试图在第35行的$statement变量之后追加一些内容

$statement未定义，您正在尝试附加到它

试试看

<!doctype html>
<html>
    <head>
        <title>ABU E-ADVISING</title>
        <link rel="stylesheet" type="text/css" href="style.css" \>
    </head>
    <body>
        <?php
        $Username1 = $_POST['usr'];
        $Pass1 = $_POST['pass'];
        $Surname = $_POST['sur'];
        $Othernames = $_POST['oname'];
        $DOB = $_POST['dob'];
        $email = $_POST['email'];
        $Tel = $_POST['tel'];
        $Add = $_POST['add'];
        $Postcode = $_POST['postcode'];

        $db = mysql_connect('localhost','root','ROOT');
        if (!$db)
        {
            print "<h1>Unable to Connect to MySQL</h1>";
        }

        $dbname = 'adv';
        $btest = mysql_select_db($dbname);

        if (!$btest)
        {
            print "<h1>Unable to Select the Database</h1>";
        }

        $statement = "insert into user (userName, Password, FName, LName, DOB,Email, Tele, Address, PCode) ";
        $statement .= "values (";
        $statement .= "'".$Username1."', '".$Pass1."', '".$Surname."','".$Othernames."','".$DOB."','".$Tel."', '".$Add."', '".$Postcode."','".$email."'";
        $statement .= ")";
        $result = mysql_query($statement);

        if ($result)
        {
            echo "<br>Author Added: ".$Surname.", ".$Othernames;
        } else {
            $errno = mysql_errno($db);

            if ($errno == '1062') {
                echo "<br>Author is already in Table: <br />".$mylastname.", ".$myfirstname;
            } else {
                echo("<h4>MySQL No: ".mysql_errno($result)."</h4>");
                echo("<h4>MySQL Error: ".mysql_error($result)."</h4>");
                echo("<h4>SQL: ".$statement."</h4>");
                echo("<h4>MySQL Affected Rows: ".mysql_affected_rows($result)."</h4>");
            }

        print 'NotAdded';
        }
        ?>
    </body>


阿布电子咨询公司
请mysql.*函数已经过时，而且不安全。使用或代替。另外，你完全可以使用mysql。停止使用mysql，我知道这与你的问题无关，但这对未来会更好，避免使用mysql，使用mysqli、pdo或任何其他。
$statement = "insert into user (userName, Password, FName, LName, DOB,Email, Tele, Address, PCode) ";

<!doctype html>
<html>
    <head>
        <title>ABU E-ADVISING</title>
        <link rel="stylesheet" type="text/css" href="style.css" \>
    </head>
    <body>
        <?php
        $Username1 = $_POST['usr'];
        $Pass1 = $_POST['pass'];
        $Surname = $_POST['sur'];
        $Othernames = $_POST['oname'];
        $DOB = $_POST['dob'];
        $email = $_POST['email'];
        $Tel = $_POST['tel'];
        $Add = $_POST['add'];
        $Postcode = $_POST['postcode'];

        $db = mysql_connect('localhost','root','ROOT');
        if (!$db)
        {
            print "<h1>Unable to Connect to MySQL</h1>";
        }

        $dbname = 'adv';
        $btest = mysql_select_db($dbname);

        if (!$btest)
        {
            print "<h1>Unable to Select the Database</h1>";
        }

        $statement = "insert into user (userName, Password, FName, LName, DOB,Email, Tele, Address, PCode) ";
        $statement .= "values (";
        $statement .= "'".$Username1."', '".$Pass1."', '".$Surname."','".$Othernames."','".$DOB."','".$Tel."', '".$Add."', '".$Postcode."','".$email."'";
        $statement .= ")";
        $result = mysql_query($statement);

        if ($result)
        {
            echo "<br>Author Added: ".$Surname.", ".$Othernames;
        } else {
            $errno = mysql_errno($db);

            if ($errno == '1062') {
                echo "<br>Author is already in Table: <br />".$mylastname.", ".$myfirstname;
            } else {
                echo("<h4>MySQL No: ".mysql_errno($result)."</h4>");
                echo("<h4>MySQL Error: ".mysql_error($result)."</h4>");
                echo("<h4>SQL: ".$statement."</h4>");
                echo("<h4>MySQL Affected Rows: ".mysql_affected_rows($result)."</h4>");
            }

        print 'NotAdded';
        }
        ?>
    </body>

<html>
    <head>
    </head>
    <body>
        <?php
        if($bla == $bla) {
            echo "bla = bla";
        }
        ?>
    </body>
</html>