PHP 请求未获得价值

编程问答 2022-05-18

问题描述

我有一个滑块，其中一个位于标签中。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();
?>

网址：“http://localhost/insert.PHP?myrange=10”

它正在运行，并生成一行，但该行完全是空的......所以我不知道发生了什么......

解决方法

正确的代码：

<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？

  • 我更改了方法 method="POST" 而不是 method="$_POST"
  • 将变量 GET 上的 POST 更改为 $getRangeValue
  • 使用准备语句代替简单且不安全的查询

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