Question

由于某些原因，我的代码没有更新mySQL数据库，但它没有报告任何错误。

register.php （表格）

        <form class="register_form" action="action.php?do=register" method="post">
            Team Name*: <input type="text" name="teamname" required />
            Team Region*: <input type="text" name="teamregion" maxlength="4" required />
            Team Leader*: <input type="text" name="teamleader" maxlength="16" required />
            Team Members: <input type="text" name="teammembers"  />
            <input name="register_submit" type="submit" value="Register" />
        </form>

action.php的

    <?php
        $con=mysqli_connect("192.185.#.###","########_reg","#######","#########");
        if (mysqli_connect_errno()) {echo "Failed to connect to MySQL: " . mysqli_connect_error();}
            $action = $_GET['do'];
                if($action=="register") {
                    $teamname = $_POST["teamname"];
                    $teamregion = $_POST["teamregion"];
                    $teamleader = $_POST["teamleader"];
                    $teammembers = $_POST["teammembers"];
                        $result = mysqli_query($con, "INSERT INTO teams (teamname, region, teamleader, teammembers, wins, loses)
                        VALUES (" . $teamname . "," . $teamregion . "," . $teamleader . "," . $teammembers . ",0,0);");
                }
    ?>

为什么这不能正常工作？

Answer 1

这是一个包含预习语句的工作示例，通常使用“更好”代替query

<强> action.php的

$con = new mysqli('localhost', 'root', '', 'dachi');

if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
}

if (isset($_GET['do']) && $_GET['do'] === "register") {
    $teamname = $_POST["teamname"];
    $teamregion = $_POST["teamregion"];
    $teamleader = $_POST["teamleader"];
    $teammembers = $_POST["teammembers"];
    $wins = 0;
    $loses = 0;

    $stmt = $con->prepare("INSERT INTO `teams` (`teamname`,`region`,`teamleader`,`teammembers`,`wins`,`loses`) VALUES (?,?,?,?,?,?)");
    $stmt->bind_param('ssssii', $teamname, $teamregion, $teamleader, $teammembers, $wins, $loses);
    $stmt->execute();
    $stmt->close();
}

<强> register.php

<form class="register_form" action="action.php?do=register" method="post">
    Team Name*: <input type="text" name="teamname" required />
    Team Region*: <input type="text" name="teamregion" maxlength="4" required />
    Team Leader*: <input type="text" name="teamleader" maxlength="16" required />
    Team Members: <input type="text" name="teammembers"  />
    <input name="register_submit" type="submit" value="Register" />
</form>

Answer 2

你应该改变：

if($action=="register") {

到

if($action=="register_submit") {

因为您的输入的标记名称设置为值 register_submit 而不是 register 。

并将$action = $_GET['do'];更改为$action = $_POST['register_submit'];

Answer 3

register.php（表格）

<form class="register_form" action="action.php?do=register" method="post">

<强> action.php的

$action = $_GET['do'];

不更新SQL数据库

3 个答案: