未將數據輸入數據庫的PHP代碼

Question

我無法將表單發送到我的數據庫。我試了一個小時來解決問題;我用「echo mysqli_error（）;」解決了我的代碼但沒有收到任何錯誤，只是缺少新的數據到我的數據庫（看起來好像提交表單不重新加載頁面發送信息）。該數據庫適用於我的其餘頁面使用config.php，connectDB.php和header.php，所以問題不在數據庫設置。我相信這個錯誤是在「isset（$ _ POST ['submit']）」內，因爲它不會使用header（）重新加載頁面，但是我提供了其餘的代碼以防萬一。 （我知道這個代碼不針對SQL注入的保護，表單驗證是下一步我解決這個問題之後）：執行

<?php 
require_once ("Includes/config.php"); 
require_once ("Includes/connectDB.php"); 
include("Includes/header.php"); 
if (isset($_POST['submit'])){ 
       $name = $_POST['name']; 
       $email = $_POST['email']; 
       $content = $_POST['content']; 
       $query = "INSERT INTO requests (name, email, content) VALUES (?, ?, ?)"; 

       $statement = $databaseConnection->prepare($query); 
       $statement->bind_param('sss', $name, $email, $content); 
       header('Location: /index.php'); 

       $statement->execute() 
       $statement->store_result(); 
       setcookie("nameErr"," ", time()+3600); 
       setcookie("emailErr"," ", time()+3600); 
       setcookie("contentErr"," ", time()+3600); 
       setcookie("contentSucc","Announcement Request Successful", time()+3600); 
       header('Location: /request.php'); 

} 
?> 
<div id="main"> 
     <ol> 
      <li> 
       <label for="name">Name:</label> 
       <input type="text" name="name" value="" id="name" /> 
       <span class="error">* <?php if (isset($_COOKIE["nameErr"])){echo $_COOKIE["nameErr"];}?></span> 
      </li> 
      <li> 
       <label for="email">Email:</label> 
       <input type="text" name="email" value="" id="email" style = "position: relative; left: 3px;"/> 
       <span class="error" style = "position: relative; left: 3px;">* <?php if (isset($_COOKIE["emailErr"])){echo $_COOKIE["emailErr"];}?></span> 
      </li> 
      <li> 
        <label for="content">Requested Announcement:</label><br> 
        <textarea rows="18" cols="140" name="content" id="content"></textarea> 
        <br> 
        <span class="error"> <?php if (isset($_COOKIE["contentErr"])){echo $_COOKIE["contentErr"];}?></span> 
        <span class="error"> <?php if (isset($_COOKIE["contentSucc"])){echo $_COOKIE["contentSucc"];}?></span> 
      </li> 
     </ol> 
     <input type="submit" name="submit" value="Submit " style = "position: relative; left: 40px;" /> 
     <!-- onclick="window.location='request.php';" --> 
     <p> 
      <a class="cancel" href="index.php" style = "position: relative; left: 40px;">Cancel</a> 
     </p> 
</div> 

Answer 1

1. 您重定向到SQL之前，另一個網頁。
2. 您在行尾沒有分號。
3. 您沒有<form>圍繞<input>標籤。
4. 你不應該做非常多的事來阻止SQL注入。只要你正確使用MySQLi，你的罰款。

更改此

  $statement->bind_param('sss', $name, $email, $content); 
      header('Location: /index.php'); 

      $statement->execute() 

這個

  $statement->bind_param('sss', $name, $email, $content); 
      $statement->execute(); 
      $statement->close(); 

      header('Location: /index.php');