Question

我正在尝试使用一种形式将数据插入两个单独的mysql表中。

在下面的PHP代码中，我有两个INSERT语句，一个用于Table Games，一个用于Table Results。如果我仅使用“ INSERT INTO Games ...”或“ INSERT INTO Results ...”语句运行代码（将表头的形式修改为仅包括相应的一个或两个相应字段），则会插入数据正确地输入数据库。当我尝试同时进行操作时，我无法弄清楚自己在做什么错。只是要注意，当我在phpmyadmin中同时运行两个INSERT语句时，数据正确插入了两个表中。

我已经阅读了有关mysqli_multi_query的信息，并尝试了各种选项来使用它，但没有任何结果，因为我不确定如何将其与正在使用的以下准备语句一起使用。

if($stmt = mysqli_prepare($link, $sql))

这是php部分：

<?php

// Define variables and initialize with empty values
$tourn_id = "";
$game_id = "";
$player_id = "";
$tourn_name_err = "";

// Processing form data when form is submitted
if($_SERVER["REQUEST_METHOD"] == "POST"){

    // Check input errors before inserting in database
    if(empty($tourn_name_err)){

        // Prepare an insert statement
        $sql = "INSERT INTO Games (idTournaments) VALUES (?);
        INSERT INTO Results (idGames, idPlayers) VALUES (?, ?);";

        if($stmt = mysqli_prepare($link, $sql)){
            // Bind variables to the prepared statement as parameters
            mysqli_stmt_bind_param($stmt, "iii", $param_tourn_id, $param_game_id, $param_player_id);

            // Set parameters
            $param_tourn_id = trim($_POST["tourn_id"]);
            $param_game_id = trim($_POST["game_id"]);
            $param_player_id = trim($_POST["player_id"]);

            // Attempt to execute the prepared statement
            if(mysqli_stmt_execute($stmt)){
                // Redirect to login page
                header("location: welcome.php");
            } else{
                echo "Something went wrong. Please try again later.";
            }
        }

        // Close statement
        mysqli_stmt_close($stmt);
    }

    // Close connection
    mysqli_close($link);
}
?>

这是HTML表单部分：

<div class="wrapper">
    <form class="form-signin" action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]); ?>" method="post">
        <h1 class="form-signin-heading">New Match</h1>

        <div class="form-group <?php echo (!empty($tourn_name_err)) ? 'has-error' : ''; ?>">
            <label>Enter Tournament ID here ...:</label>
            <input type="number" name="tourn_id"class="form-control" value="<?php echo $tourn_id; ?>">
            <span class="help-block"><?php echo $tourn_name_err; ?></span>
        </div>

        <div class="form-group <?php echo (!empty($tourn_name_err)) ? 'has-error' : ''; ?>">
            <label>Enter Game ID here ...:</label>
            <input type="number" name="game_id"class="form-control" value="<?php echo $game_id; ?>">
            <span class="help-block"><?php echo $tourn_name_err; ?></span>
        </div>

        <div class="form-group <?php echo (!empty($tourn_name_err)) ? 'has-error' : ''; ?>">
            <label>Enter Player ID here ...:</label>
            <input type="number" name="player_id"class="form-control" value="<?php echo $player_id; ?>">
            <span class="help-block"><?php echo $tourn_name_err; ?></span>
        </div>

        <div class="form-group">
            <input type="submit" class="btn btn-primary" value="Submit">
            <input type="reset" class="btn btn-default" value="Reset">
        </div>
    </form>
</div>

Answer 1

好的，经过多次尝试错误的尝试并摆脱了原始php代码中的“准备语句”，它才能正常工作。我还尽可能遵循手册中的mysqli_multi_query示例。

我不确定是否需要像下面这样设置参数。

$param_tourn_id = trim($_POST["tourn_id"]);

这是有效的完整php代码：

<?php

// Define variables and initialize with empty values
$tourn_id = "";
$game_id = "";
$player1_id = "";
$tourn_name_err = "";


// Processing form data when form is submitted
if($_SERVER["REQUEST_METHOD"] == "POST"){


        // Set parameters
        $param_tourn_id = trim($_POST["tourn_id"]);
        $param_game_id = trim($_POST["game_id"]);
        $param_player1_id = trim($_POST["player1_id"]);


        // Prepare an insert statement
        $sql = "INSERT INTO Games (idTournaments) VALUES ($param_tourn_id);";
        $sql .= "INSERT INTO Results (idGames, idPlayers) VALUES ($param_game_id, $param_player1_id)";


        /* execute multi query */
        if (mysqli_multi_query($link, $sql)) {
            do {
                /* store first result set */
                if ($result = mysqli_store_result($link)) {
                    while ($row = mysqli_fetch_row($result));
                    mysqli_free_result($result);
                }
                /* print divider */
                if (mysqli_more_results($link));
            } while (mysqli_next_result($link));
        }

        /* close connection */
        mysqli_close($link);

    }

?>

使用一种HTML格式插入两个SQL表

1 个答案: