Question

我想将选择项目的用户名添加到我的数据库中。例如，如果三个用户选择该项目，我希望它像user1，user2，user3 ..我得到不同的用户进入数据库，但我不能让逗号分开..我使用爆炸但仍然无法正常工作

<?php

session_start();
$date = date("Y-m-d H:i:s");

include("php/connect.php");

if (isset($_SESSION['username'])){

           $query1 = mysql_query("SELECT * FROM user WHERE username='$username'");

    $username = $_SESSION['username'];

    $submit = $_POST["submit"];
    $tests = $_POST["test"];

    // If the user submitted the form.
    // Do the updating on the database.
    if (!empty($submit))
    {
        if (count($tests) > 0)
        {
            foreach ($tests as $test_id => $test_value)
            {
                switch ($test_value)
                {

                    case 1:
                        mysql_query("UPDATE test SET win = win + 1 WHERE id = '$test_id'");
                        mysql_query("UPDATE test SET date = '$date'  WHERE id = '$test_id'");
                        $username = explode(",", $username);
                        $cnt = count($username);
                        $slice = array_slice($username, 0,10);
                        mysql_query("UPDATE test SET users = CONCAT(users, '$username')  WHERE id = '$test_id'");
                    break;

                    case 'X':
                        mysql_query("UPDATE test SET draw = draw + 1 WHERE id = '$test_id'");
                        mysql_query("UPDATE test SET date = '$date'  WHERE id = '$test_id'");
                    break;

                    case 2:
                        mysql_query("UPDATE test SET lose = lose + 1 WHERE id = '$test_id'");
                        mysql_query("UPDATE test SET date = '$date'  WHERE id = '$test_id'");
                    break;

                    default:
                        // DO NO THING.
                }
            }
        }

    }



    // Whenever this wiil be fetched it will be updated.
    $query = "SELECT * FROM test ORDER BY `id` ASC LIMIT 3";
    $result = mysql_query($query);

    echo "<h2>Seria A</h2><hr/>
    <br/>Welcome, ".$username."! <a href='php/logout.php'><b>LogOut</b></a><br/>";

    while($row = mysql_fetch_array($result)){

        $id = $row['id'];
        $home = $row['home'];
        $away = $row['away'];
        $win = $row['win'];
        $draw = $row['draw'];
        $lose = $row['lose'];


        echo "<br/>",$id,") " ,$home, " - ", $away;

        echo "

        <form action='seria.php' method='post'>

        <select name='test[$id]'>        
            <option value=\"\">Parashiko</option>
            <option value='1'>1</option>
            <option value='X'>X</option>
            <option value='2'>2</option>
       </select>

       <input type='submit' name='submit' value='Submit'/>

        <br/>

        </form><br/>";        

        echo "Totali ", $sum = $win+$lose+$draw, "<br/><hr/>"; 

    } 

    }else{

        $error = "<div id='hello'>Duhet te besh Log In qe te vendosesh parashikime ndeshjesh<br/><a href='php/login.php'>Kycu Ketu</a></div>";

    }

 ?>

Answer 1

修改

好的，这应该对你有用....我还清理了你的多个执行查询......

这将检索测试行，获取用户名，将当前用户名附加到列表中，然后重新插入....

case 1: $sql = mysql_query("SELECT * FROM test WHERE id = '$test_id'"); $users = mysql_fetch_row($sql); $usernames = $users['users'].",".$username; mysql_query("UPDATE test SET users = '$usernames', win = win + 1, date = '$date' WHERE id = '$test_id'"); break; case 'X': mysql_query("UPDATE test SET date = '$date', draw = draw + 1 WHERE id = '$test_id'"); break; case 2: mysql_query("UPDATE test SET date = '$date', lose = lose + 1 WHERE id = '$test_id'"); break; default:

只需将其插入您的switch语句
即可
另外......出于安全考虑，你想要逃避用户输入的数据......

$submit = mysql_real_escape_string($_POST["submit"]); $tests = mysql_real_escape_string($_POST["test"]);

或者更好的是，切换到使用mysqli_ *或PDO，mysql is deprecated，很快就会从PHP中删除。

Answer 2

我不知道当你拿到它时用户名是什么。

如果$username是 ARRAY ，请在执行此行之前：$username = explode(",", $username);然后您应该执行$username = implode(",",$username);

所以如果是这样的话：

INPUT: $username = array('fred','frank','jill');

$username = implode(",",$username);

OUTPUT: fred,frank,jill

爆炸功能不起作用

2 个答案: