使用PHP從表單插入數據到SQLite3數據庫

Question

我想創建一個註冊並將數據發佈到SQlite數據庫文件。

我的形式如下：

<form action="registerprocess_test.php" class="form-horizontal" id= 
"register_form" method="post" name="register_form" role="form"> 
    <h2>Registration Form</h2> 
    <div class="form-group"> 
     <label class="col-sm-3 control-label" for="firstname">First 
     Name</label> 
     <div class="col-sm-6"> 
      <input autofocus="" class="form-control" id="firstname" name= 
      "firstname" placeholder="First Name" type="text"> 
     </div> 
    </div> 
    <div class="form-group"> 
     <label class="col-sm-3 control-label" for="lastname">Last 
     Name</label> 
     <div class="col-sm-6"> 
      <input autofocus="" class="form-control" id="lastname" name= 
      "lastname" placeholder="Last Name" type="text"> 
     </div> 
    </div> 
    <div class="form-group"> 
     <label class="col-sm-3 control-label" for="email">Username</label> 
     <div class="col-sm-6"> 
      <input class="form-control" id="username" name="username" 
      placeholder="Username" type="text"> 
     </div> 
    </div> 
    <div class="form-group"> 
     <label class="col-sm-3 control-label" for= 
     "password">Password</label> 
     <div class="col-sm-6"> 
      <input class="form-control" id="password" name="password" 
      placeholder="Password" type="password"> 
     </div> 
    </div> 
    <div class="form-group"> 
     <label class="col-sm-3 control-label" for="country">Country</label> 
     <div class="col-sm-6"> 
      <select class="form-control" id="country" name="country"> 
       <option> 
        United Kingdom 
       </option> 
       <option> 
        United States 
       </option> 
      </select> 
     </div> 
    </div><!-- /.form-group --> 
    <div class="form-group"> 
     <label class="control-label col-sm-3">Gender</label> 
     <div class="col-sm-6"> 
      <div class="row"> 
       <div class="col-sm-4"> 
        <label class="radio-inline"><input id="femaleRadio" 
        type="radio" value="Female">Female</label> 
       </div> 
       <div class="col-sm-4"> 
        <label class="radio-inline"><input id="maleRadio" name= 
        "gender" type="radio" value="Male">Male</label> 
       </div> 
      </div> 
     </div> 
    </div><!-- /.form-group --> 
    <div class="form-group"> 
     <div class="col-sm-6"> 
      <div class="checkbox"></div> 
     </div><!-- /.form-group --> 
     <div class="form-group"> 
      <div class="col-sm-6 col-sm-offset-3"> 
       <div class="checkbox"> 
        <label><input type="checkbox">I accept <a href= 
        "#">Terms & Conditions</a></label> 
       </div> 
      </div> 
     </div><!-- /.form-group --> 
     <div class="form-group"> 
      <div class="col-sm-6 col-sm-offset-3"> 
       <button class="btn btn-primary btn-block" type= 
       "submit">Register</button> 
      </div> 
     </div> 
    </div> 
</form> 

然後我的PHP是這樣的：

<?php 
    try 
{ 
//open the database 
$db = new PDO('sqlite:users.db'); 


$firstname = $_POST["firstname"]; 
$lastname = $_POST["lastname"]; 
$username = $_POST["username"]; 
$password = $_POST["password"]; 
$gender = $_POST["gender"]; 
$country = $_POST["country"]; 


//Insert record 

$db->exec("INSERT INTO registered_users (firstname, lastname, username, password, gender, country) VALUES ('$firstname', '$lastname', '$username', '$password', '$gender', $country);"); 

//now output the data to a simple html table... 
print "<table border=1>"; 
print "<tr><td>firstname</td><td>lastname</td><td>username</td><td>password</td><td>gender</td><td>country</td></tr>"; 
$result = $db->query('SELECT * FROM registered_users'); 
foreach($result as $row) 
{ 
    print "<tr><td>".$row['firstname']."</td>"; 
    print "<td>".$row['lastname']."</td>"; 
    print "<td>".$row['username']."</td>"; 
    print "<td>".$row['password']."</td>"; 
    print "<td>".$row['gender']."</td>"; 
    print "<td>".$row['country']."</td>"; 
} 
print "</table>"; 

$db = NULL; 
} 
catch(PDOException $e) 
{ 
print 'Exception : ' .$e->getMessage(); 
} 

?> 

我知道我，因爲它連接到數據庫顯示的當前數據在表中的數據庫中，按點擊su bmit。但是，不會從登記表中插入數據。我在我的PHP代碼中丟失了重要的東西嗎？

Answer 1

我做了一些錯誤檢查，這是因爲我的表中有另一列，我沒有插入。所以我從數據庫中刪除了這一列，現在它工作正常。