Question

我是php的初学者。代码似乎是正确的，但是当我尝试将表单中的数据插入数据库时，它不会进入数据库。我的代码出了什么问题？

这是我的代码

addemployee.php

echo '<form action="addemployee.php" method="POST">';
echo '<table align ="center">';

    echo '<tr><td>First Name: </td> <td><input type="text" name="first" pattern="[A-Za-z]{1,}" title="A-Z only" required></td> </tr>';
    echo '<tr><td>Middle Initial: </td><td><input type="text" name="middle"></td></tr>';
    echo '<tr><td>Last Name:</td><td> <input type="text" name="last"></td></tr>';
    echo '<tr><td>Contact Number:</td><td> <input type="tel" name="contact" maxlength="11"></td></tr>';
    echo '<tr> <td>Province: </td><td> <input type="text" name="provincee"></td></tr>';
    echo '<tr> <td>City: </td><td> <input type="text" name="cityy"></td></tr>';
    echo '<tr> <td>Username:</td><td> <input type="email" name="usernamee"></td></tr>';
    echo '<tr><td>Password:</td><td> <input type="password" name="pass"></td></tr>';

echo '</table>';
echo '<br> <input type="Submit" name="submitt">';

    if(isset($_POST['submitt'])){
    $firstname=$_POST['first'];
    $middleinitial=$_POST['middle'];
    $lastname=$_POST['last'];
    $contactnumber=$_POST['contact'];
    $province= $_POST['provincee'];
    $city =$_POST['cityy'];
    $username=$_POST['usernamee'];
    $password=$_POST['pass']; 
    $type= 'employee';

 $query=("INSERT INTO usertbl (fname,middeinitial,lname,contactnum,province,city,username,password,type) VALUES ('$firstname','$middleinitial','$lastname','$contactnumber','$province',
  '$city','$username','$password','$type')");
  mysqli_query($db, $query);
}
echo '</form>';

Answer 1

请检查更改： -

addemployee.php

<!-- use html form not php generated form -->

<form action="addemployee.php" method="POST">
    <table align ="center">
        <tr><td>First Name: </td> <td><input type="text" name="first" pattern="[A-Za-z]{1,}" title="A-Z only" required></td> </tr>
        <tr><td>Middle Initial: </td><td><input type="text" name="middle"></td></tr>
        <tr><td>Last Name:</td><td> <input type="text" name="last"></td></tr>
        <tr><td>Contact Number:</td><td> <input type="tel" name="contact" maxlength="11"></td></tr>
        <tr> <td>Province: </td><td> <input type="text" name="provincee"></td></tr>
        <tr> <td>City: </td><td> <input type="text" name="cityy"></td></tr>
        <tr> <td>Username:</td><td> <input type="email" name="usernamee"></td></tr>
        <tr><td>Password:</td><td> <input type="password" name="pass"></td></tr>
    </table>
    <br>
    <input type="Submit" name="submitt">
</form>
 <?php
    error_reporting(E_ALL); // check all type of error
    ini_set('display_errors',1); // display those errors
    if(isset($_POST['submitt'])){
        $firstname=$_POST['first'];
        $middleinitial=$_POST['middle'];
        $lastname=$_POST['last'];
        $contactnumber=$_POST['contact'];
        $province= $_POST['provincee'];
        $city =$_POST['cityy'];
        $username=$_POST['usernamee'];
        $password=$_POST['pass']; 
        $type= 'employee';

        $db = mysqli_connect ('hostname','username','password','dbname') or die(mysqli_connect_error()); // provide your details here

        $query=("INSERT INTO usertbl (fname,middeinitial,lname,contactnum,province,city,username,password,type) VALUES ('$firstname','$middleinitial','$lastname','$contactnumber','$province',
  '$city','$username','$password','$type')");
        if(mysqli_query($db, $query)){
           echo "Inserted successfully";
        }else{
           echo "error:-".mysqli_error($db);
        }
    }

?>

注意： - 您的代码向SQLInjection开放。请阅读prepared statements或mysqli_*的{{1}}并使用它们。

代码是正确的，但我无法将数据输入数据库

1 个答案: