Question

我相当确定我只犯了一个简单的错误，但我遇到了麻烦。基本上，每当有人填写正常表单时，他们输入的数据将提交到db表，稍后将发布在站点的另一页上。但是，这似乎不起作用。

如果可能的话，我只是想在正确的方向上帮助我实际做错了什么。

在此先感谢，我希望这篇文章不会太烦人。我是新来的：，）

PHP：

<?php 
    session_start();
    include_once("connection.php");

    if(isset($_POST['post'])) {
        $match_name = strip_tags($_POST['match_name']);
            $team1 = strip_tags($_POST['team1']);
                $player1 = strip_tags($_POST['player1']);
                $player2 = strip_tags($_POST['player2']);
                $player3 = strip_tags($_POST['player3']);
                $player4 = strip_tags($_POST['player4']);
                $player5 = strip_tags($_POST['player5']);

            $team2 = strip_tags($_POST['team2']);
                $player6 = strip_tags($_POST['player6']);
                $player7 = strip_tags($_POST['player7']);
                $player8 = strip_tags($_POST['player8']);
                $player9 = strip_tags($_POST['player9']);
                $player10 = strip_tags($_POST['player10']);

        $match_name = mysqli_real_escape_string($dbCon, $match_name);
            $team1 = mysqli_real_escape_string($dbCon, $team1);
                $player1 = mysqli_real_escape_string($dbCon, $player1);
                $player2 = mysqli_real_escape_string($dbCon, $player2);
                $player3 = mysqli_real_escape_string($dbCon, $player3);
                $player4 = mysqli_real_escape_string($dbCon, $player4);
                $player5 = mysqli_real_escape_string($dbCon, $player5);
            $team2 = mysqli_real_escape_string($dbCon, $team2);
                $player6 = mysqli_real_escape_string($dbCon, $player6);
                $player7 = mysqli_real_escape_string($dbCon, $player7);
                $player8 = mysqli_real_escape_string($dbCon, $player8);
                $player9 = mysqli_real_escape_string($dbCon, $player9);
                $player10 = mysqli_real_escape_string($dbCon, $player10);

        $sql = "INSERT INTO `match` (match_name, team1, player1, player2, player3, player4, player5, team2, player6, player7, player8, player9, player10) VALUES ('$team1, '$player1', '$player2', '$player3', '$player4', '$player5', '$team2, '$player6', '$player7', '$player8', '$player9', '$player10')";              

        if($match_name == "") {
            echo "You're missing a title for varible <strong>Match Title</strong> | <a href='post.php'>Go back</a>";
            return;
        }

        if($team1 == "") {
            echo "You're missing a title for varible <strong>Team 1 Name</strong> | <a href='post.php'>Go back</a>";
            return;
        }

        if($team2 == "") {
            echo "You're missing a title for varible <strong>Team 2 Name</strong>";
            return;
        }

        mysqli_query($dbCon, $sql);

        header("Location: index.php");
    }
?>

Answer 1

从@arkascha和@Jay Blanchard已经提到的所有事情向前迈进。

您在' VALUES和'$team1 '$team2附近的SQL查询中遗漏了单引号$sql = "INSERT INTO `match` (match_name, team1, player1, player2, player3, player4, player5, team2, player6, player7, player8, player9, player10) VALUES ('$team1, '$player1', '$player2', '$player3', '$player4', '$player5', '$team2, '$player6', '$player7', '$player8', '$player9', '$player10')";：

而不是：

$sql = "INSERT INTO `match` (match_name, team1, player1, player2, player3, player4, player5, 
team2, player6, player7, player8, player9, player10) 
VALUES ('$team1', '$player1', '$player2', '$player3', '$player4', '$player5', 
'$team2', '$player6', '$player7', '$player8', '$player9', '$player10')";

试试这个：

/controllers/front/CmsController.php

PHP不将提交的数据上传到数据库

1 个答案: