将表单数据传递给 php 变量。变量似乎是空的

Question

我是 PHP/MySQL 的菜鸟。我一直在环顾四周，但我不知道出了什么问题。脚本的目的：更新数据库中 item 1 和 item 2 的值，用户 ID 在 txtUser.

中输入

当我在查询中使用 userId 附近的 "fixed" 值时，我的脚本开始工作。但是，当我尝试使用变量（$player）时，它不起作用。好像我的变量是空的...

HTML:

<body>
   <form id="form1" action="http://www.something.com/TestScript1.php" method="post" enctype="application/x-www-form-urlencoded">

    <div>
    <button type="submit" id="submit" value="Submit" title="SAVE">SAVE</button>
    </div>
    <div>
        <input id="txtUser" name="txtUser" type="text" />
        <input id="txtItem1" name="txtItem1" type="text" />
        <input id="txtItem2" name="txtItem2" type="text" />
    </div>

</form>
</body>

PHP :

$host = "localhost";
$user = "username";
$password = "password";
$database = "database";
$player = mysqli_real_escape_string($connection,$_POST['txtUser']);
$connection = mysqli_connect($host,$user,$password,$database) or die ("connection to server failed");
mysqli_select_db($connection,$database) or die ("couldn’t select database");

$item1 = mysqli_real_escape_string($connection,$_POST['txtItem1']);
$item2 = mysqli_real_escape_string($connection,$_POST['txtItem2']);

$query = "UPDATE table SET item1=$item1, item2=$item2 WHERE userId=$player";
$result = mysqli_query($connection,$query)
or die ("couldn’t execute update query: ".mysqli_error($connection));
echo "<h4>Data saved in the database</h4>";
mysqli_close($connection);

?>

Answer 1

请把这行放在

$player = mysqli_real_escape_string($connection,$_POST['txtUser']);

$连接后

$connection = mysqli_connect($host,$user,$password,$database) or die ("connection to server failed");

喜欢

$connection = mysqli_connect($host,$user,$password,$database) or die ("connection to server failed");
$player = mysqli_real_escape_string($connection,$_POST['txtUser']);

因为你在 mysqli_real_escape_string 中使用 $connection 变量，甚至在它初始化之前

Answer 2

未为行

定义 $connection

$player = mysqli_real_escape_string($connection,$_POST['txtUser']);

请在创建 $connection 后移动此行

$connection = mysqli_connect($host,$user,$password,$database) or die ("connection to server failed");

Answer 3

<?php

    $host = "localhost";
    $user = "username";
    $password = "password";
    $database = "database";



    $conn = mysqli_connect($host,$user,$password,$database) or die ("connection to server failed");
    mysqli_select_db($conn,$database) or die ("couldn't select database");

    /* The $conn object must be declared before first / any calls to mysqli_real_escape_string */
    $player = mysqli_real_escape_string($conn,$_POST['txtUser']);
    $item1 = mysqli_real_escape_string($conn,$_POST['txtItem1']);
    $item2 = mysqli_real_escape_string($conn,$_POST['txtItem2']);

    /* Values in query should be encapsulated in quotes if they are strings.. are they? */
    $query = "UPDATE `table` SET `item1`='$item1', `item2`='$item2' WHERE userId='$player';";

    /* Try not to reveal too many details in the event of an error, reduce the attack surface if possible! */
    $result = mysqli_query($conn,$query) or die ("update failed: ");
    echo "<h4>Data saved in the database</h4>";
    mysqli_close($conn);

?>

将表单数据传递给 php 变量。变量似乎是空的

Issue passing form data to php variable. Variable seems empty

php

mysql

post

sql-update