PHP 请求未获得价值

Question

我有一个滑块，其中一个在标签中。我想将滑块放入数据库。在“index.php”中：

<form class="" action="insert.php" method="$_POST">
<input type="range" min="1" max="10" value="5" class="slider" id="myRange" name="myrange">
  <p>Value: <span id="demo"></span></p>
  <button id="btn1">Click Here</button>
</form>

并在 insert.php 中：

    <?php
$getRangeValue = $_GET['myRange'];
$mysqli = new mysqli("localhost","root","passwd","table_name");
//
// Check connection
if ($mysqli -> connect_errno) {
  echo "Failed to connect to MySQL: " . $mysqli -> connect_error;
  exit();
}

$mysqli -> query("INSERT INTO ertekek (RangeValue, anotherValue) VALUES ($getRangeValue, 11)");
echo "New record has id: " . $mysqli -> insert_id;
$mysqli -> close();
?>

url: "http:///localhost/insert.php?myrange=10"

它运行，它生成一行，但该行完全是空的...所以我不知道发生了什么...

Answer 1

正确代码：

<form class="" action="insert.php" method="POST">
   <input type="range" min="1" max="10" value="5" class="slider" id="myRange" name="myrange">
  <p>Value: <span id="demo"></span></p>
  <button id="btn1">Click Here</button>
</form>

并在 insert.php 中：

<?php
$getRangeValue = $_POST['myrange'];
$mysqli = new mysqli("localhost","root","passwd","table_name");
//
// Check connection
if ($mysqli -> connect_errno) {
  echo "Failed to connect to MySQL: " . $mysqli -> connect_error;
  exit();
}

$query = $mysqli->prepare("INSERT INTO ertekek (RangeValue, anotherValue) VALUES (?, 11)");
$query->bind_param('i',$getRangeValue);
$query->execute();
echo "New record has id: " . $mysqli -> insert_id;
$mysqli -> close();
?>

我改变了什么？

我改变了方法 method="POST" 而不是 method="$_POST"
将变量 $getRangeValue

GET

POST

使用准备语句而不是简单且不安全的查询

Link 我推荐你阅读

What is the difference between POST and GET?
How can I prevent SQL injection in PHP?

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