douge3492
2016-01-05 14:30
浏览 419
已采纳

未捕获的错误:调用未定义的函数mysql_escape_string()

Fatal error: Uncaught Error: Call to undefined function mysql_escape_string() in C:\xampp\htdocs\phoenixprojectegister.php:16 Stack trace: #0 {main} thrown in C:\xampp\htdocs\phoenixprojectegister.php on line 16

How to fix this?

<?php
require("config.php");
?>
<?php
if(isset($_POST['submit'])){

$email1 = $_POST['email1'];
$email2 = $_POST['email2'];
$pass1 = $_POST['pass1'];
$pass2 = $_POST['pass2'];

if($email1 == $email2) {
    if($pass1 == $pass2) {
//All good. Nastavi broo.

$name = mysql_escape_string($_POST['name']);
$lname = mysql_escape_string($_POST['lname']);
$uname = mysql_escape_string($_POST['uname']);
$email1 = mysql_escape_string($email1);
$email2 = mysql_escape_string($email2);
$pass1 = mysql_escape_string($pass1);
$pass2 = mysql_escape_string($pass2);

mysql_query("INSERT INTO `users` (`id`, `name`, `lname`, `uname`, `email`, `pass`) VALUES (NULL, '$name', '$lname', '$uname', '$email1', '$pass1')") or die (mysql_error());



}else{
  echo "Sorry, your password is not corrext.";
  exit();
}
}else{
  echo "Sorry!";
}

} // brace for submit conditional

$form = <<<EOT
<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" />
</form>
EOT;
echo $form;

?>

Well I know that I was try to mix mysql and mysqli....

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4条回答 默认 最新

  • dsour68888 2016-01-05 14:53
    已采纳

    To help you out here... (too long for a comment)

    Your require("config.php"); should contain the following:

    Sidenote: Use the proper settings for your host.

    $link = mysqli_connect("localhost", "username", "mpassword", "database") or die($link);
    

    Then changing your escape functions to use the mysqli_ version of it and passing the connection parameter to it:

    $name = mysqli_real_escape_string($link, $_POST['name']);
    $lname = mysqli_real_escape_string($link, $_POST['lname']);
    $uname = mysqli_real_escape_string($link, $_POST['uname']);
    $email1 = mysqli_real_escape_string($link, $email1);
    $email2 = mysqli_real_escape_string($link, $email2);
    $pass1 = mysqli_real_escape_string($link, $pass1);
    $pass2 = mysqli_real_escape_string($link, $pass2);
    

    Again, same thing for the query. Using the i version and passing connection to it as the first parameter.

    mysqli_query($link, "INSERT INTO ...
    

    Check for errors on your query using mysqli_error($link);

    So you could modify the query to read as

    $query = mysqli_query($link, "INSERT INTO ...
    

    and doing

    if(!$query){
       echo "Error: " . mysqli_error($link);
       }
    

    Also read the following on Stack in regards to API mixing:

    • Can I mix MySQL APIs in PHP?
    • You can't. mysql_ with mysqli_ or PDO etc. do NOT intermix together. You must use the same one from connecting to querying.

    Footnotes.

    Passwords

    I also noticed that you may be storing passwords in plain text. This is not recommended. If you intend on going LIVE with this at some point, do NOT store passwords as plain text in your database.

    Consult the following.

    Other links:

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  • duan19805 2017-03-12 15:39

    Change your PHP version back to PHP 5.6 and it will working fine.

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  • douyuefu1372 2017-04-29 15:17

    It seems I don`t know which version of PHP you are Using!

    If Your PHP is 7.x.x then only one thing to do is just rename

    $uname = mysqli_real_escape_string($link, $_POST['uname']);

    to

    $uname = $_POST['uname'];

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  • drsfgwuw61488 2017-10-20 06:23

    Best is to downgrade the PHP version to 5.6 and all will set. Error will 100% removed then.

    Hope this helps!

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