时间戳

首页 > 解决方案 > 使用php将表单中的数据填充到数据库中

php - 使用php将表单中的数据填充到数据库中

问题描述

我想将表单中的数据插入数据库。

我已经搜索了很长时间，但无法弄清楚我做错了什么。任何帮助将不胜感激

HTML

<!DOCTYPE html>
<html>
  <head>
    <title>PHP insertion</title> 
    <link href="css/insert.css" rel="stylesheet">
  </head> 
<body>
  <div class="maindiv">
  <!--HTML Form -->
    <div class="form_div">
      <div class="title"> 
        <h2>Book Information</h2>
      </div>
      <form action="new 12.php" method="post">
      <!-- Method can be set as POST for hiding values in URL--> 
        <h3>Enter the Details</h3> 
        <label>Access number</label>
        <input class="input" name="access" type="text" value=""><br> 
        <label>Title</label>
        <input class="input" name="title" type="text" value=""><br> 
        <label>Author</label> 
        <input class="input" name="author" type="text" value=""><br> 
        <label>Edition</label> 
        <input class="input" name="edition" type="text" value=""><br> 
        <label>Publisher</label> 
        <input class="input" name="publisher" type="text" value=""><br> 
        <input class="submit" name="submit" type="submit" value="Submit"><br> 
      </form> 
    </div> 
  </div> 
</body> 
</html>

PHP

  <?php
    $link = mysqli_connect("localhost", "root", "admin", "library");
    if($link === false){
        die("ERROR: Could not connect. " . mysqli_connect_error());
    }


    $sql = "INSERT INTO library (access,title,author,edition,publisher) VALUES ('$access','$title','$author','$edition','$publisher')";
    if(mysqli_query($link, $sql)){
        echo "Records inserted successfully.";
    } else{
        echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
    }

    mysqli_close($link);
  ?>

标签： phphtmlmysqlmysqli

解决方案

首先，我强烈建议进行 PDO 连接，例如：

<?php
$dbh = new PDO('mysql:host=localhost;dbname=test', $user, $pass);
?>

第二：我会将 id 放入所有输入 第三：你错过了$_POST["value"]所以你没有发送信息 html 应该如下所示：

 <!DOCTYPE html>
 <html> 
 <head> 
 <title>PHP insertion
 </title> 
<link href="css/insert.css" rel="stylesheet"> 
</head> 

<body> <div class="maindiv"> 
<!--HTML Form --> 
<div class="form_div"> 
<div class="title"> 
<h2>Book Information</h2> 
</div> 
<form action="new 12.php" method="post"> 
<!-- Method can be set as POST for hiding values in URL--> 
<h3>Enter the Details</h3> 
<label>Access number</label> <input class="input" name="access" id="access" type="text" value=""><br> 
<label>Title</label> <input class="input" name="title" id="title" type="text" value=""><br> 
<label>Author</label> <input class="input" name="author" id="author" type="text" value=""><br> 
<label>Edition</label> <input class="input" name="edition" id="edition" type="text" value=""><br> 
<label>Publisher</label> <input class="input" name="publisher" id="publiser" type="text" value=""><br> 
<input class="submit" name="submit" type="submit" value="Submit"><br> 
</form> 
</div> 
</div> 
</body>
</html>

像这样的php：

<?php
try {
    $conn = new PDO("mysqli:server = yoursever; Database = yourdatabase", "user", "pass");
    $conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
}
catch (PDOException $e) {
    print("Error connecting to Server.");
    die(print_r($e));
}

$access= $_POST['access'];
$title= $_POST['title'];
$author= $_POST['author'];
$edition= $_POST['edition'];
$publisher= $_POST['publisher'];

$sql = "INSERT INTO library (access, title, author, edition, publisher) VALUES (?,?,?,?,?)";
$stmti= $conn->prepare($sql);
$stmti->execute([$access, $title, $author, $edition, $publisher]);

if ($stmti->error){
      echo "ERROR";
    }
    else{
        echo "Records inserted successfully.";
    }
$conn->close();
?>

这是在服务器中插入信息以防止 SQL 注入的一种安全方法，请阅读“Bobby Tables”XKCD 漫画中的 SQL 注入如何工作？更好地了解如何防止感染到您的服务器

推荐阅读