使用PHP将数据传递到数据库

debugcn 发表于 Dev

塔纳戈尔

我是编程新手。

今天，我开始学习一些mysqli，以便可以为“练习站点”制作一个登录表单，但是我遇到了问题。如果您检查代码，您将理解我想要做什么。

有什么帮助吗？因为这行不通。它不会传递我在表单中输入的数据。而且，这是正确的方法吗？如果没有，那是更专业的方法吗？提前致谢。

<form action = "<?php $_PHP_SELF ?>" method ="POST">
 Username: <input type ="text" name = "username"/> </br>
 Password: <input type ="password" name = "password"/> </br>
 Email: <input type = "text" name = "email"/> </br>
 <input type = "button" value = "Submit"/> 
</form>

<?php

ini_set('display_errors', '1');

   $dbhost = 'localhost';
   $dbuser = 'root';
   $dbpass = ''; 
   $dbname = 'dbtesting';
   $username = $_POST["username"];
   $password = $_POST['passsword'];
   $email = $_POST['email'];

   $query = "INSERT INTO mywebpageusers (username, password, email)
             VALUES ('$username', '$password', '$email')";

   $conn = new mysqli($dbhost, $dbuser, $dbpass, $dbname);

   if($conn->connect_error > 0)
   {
    die ('Could not connect to database [' . $conn->connect_error . ']');
   }
   echo 'Connected succesfully!!';

  $conn->query($query);

?>

阮南

您有一些错误：

忘记了功能“回声”
$ _PHP_SELF不是PHP变量，应该是$ _SERVER ['PHP_SELF']，最好使用$ _SERVER ['SCRIPT_NAME']。
提交按钮的类型错误：type =“ submit”

“ $ conn-> connect_error”是一个字符串，当与0比较时，它将变为0 =>（0> 0）将返回false，因此函数“ die”永远不会执行

 <form action = "<?php echo $_SERVER['PHP_SELF']; ?>" method ="POST">
 Username: <input type ="text" name = "username"/> </br>
 Password: <input type ="password" name = "password"/> </br>
 Email: <input type = "text" name = "email"/> </br>
 <input type = "submit" value = "Submit"/> 
</form>

<?php

ini_set('display_errors', '1');

   $dbhost = 'localhost';
   $dbuser = 'root';
   $dbpass = '';
   $dbname = 'dbtesting';
   $username = $_POST['username'];
   $password = $_POST['passsword'];
   $email = $_POST['email'];

   $query = "INSERT INTO mywebpageusers (username, password, email)
             VALUES ('$username', '$password', '$email')";

   $conn = new mysqli($dbhost, $dbuser, $dbpass, $dbname);
   var_dump($conn->connect_error); //You can debug variables by var_dump function
   if ($conn->connect_error) {
       die('Could not connect to database ['.$conn->connect_error.']');
   }
   echo 'Connected succesfully!!';

  $conn->query($query);

?>

希望对您有所帮助。此代码足以满足实际需要。

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