PHP-MySQLi处理两次查询或错误

我有一个非常奇怪的错误,我根本不理解.
我有以下以两种方式运行的PHP脚本.

该脚本外壳将新用户添加到mysql数据库.我要检查的是数据库中是否已存在用户名.如果是这样,请设置一个用于决定如何进行的变量.
如果我执行此代码,则此行

if(strcmp($row["username"], $addUser_name) == 0)

被执行(或最好跳入),但我100％确信在执行此行之前,该名称在数据库中不存在(或不应该存在).现在,即使将$errorName设置为1并打印测试,我想要添加到数据库中的数据集也会出现在我的数据库中,但是最后一部分开始于

if(!isset($errorName))

不执行.我一遍又一遍地用简单的echo语句检查了它.

为了查看我的数据集,我删除了这一部分的注释

while ($row = mysqli_fetch_assoc($allUserNames))
{
    echo $row["username"]."\n";
}

这部分是正确执行的

if(!isset($errorName))

但是我在数据库中两次找到了数据集.

我根本不了解,为什么Scipt的行为方式如此.我已经尝试了许多不同的方法,但是我无法弄清楚我在做什么错.

<?php
    include "auth/auth1.php";
    include "functions/connectToDB.php";
    include "functions/test_input_XSS.php";

    if(isset($_GET["startCheck"])) //TODO Mache auch GET noch POST
    {
        //Sollte niemals true sein! Passiert nur, wenn man Unsinn macht
        if(strcmp($_GET["addUser_pw"], $_GET["addUser_pwRepeat"]) !== 0) { die; }

        $servername     = "localhost";
        $databasename   = "X";
        $mysqluser      = "X";
        $mysqlpass      = "X";
        $addUser_name   = $_GET["addUser_name"];

        $connection = connectToDB($servername, $mysqluser, $mysqlpass, $databasename);

        if(mysqli_connect_errno())
        {
            printf("Connect failed!");
            die();
        }

        $query_getAllUserNames = "SELECT username FROM user;";
        $allUserNames = mysqli_query($connection, $query_getAllUserNames);

        /*while ($row = mysqli_fetch_assoc($allUserNames))
        {
            echo $row["username"]."\n";
        }*/

        while ($row = mysqli_fetch_assoc($allUserNames))
        {
            if(strcmp($row["username"], $addUser_name) == 0)
            {
                $errorName = 1;
                echo "test";
            }
        }

        if(!isset($errorName))
        {
            $username = test_input_for_XSS($_GET["addUser_name"]);
            $password = hash("sha256", $_GET["addUser_pw"]);
            $permission = test_input_for_XSS($_GET["addUser_permission"]);

            $query_addUser = "INSERT INTO user (username, passwordhash, permissionlevel) VALUES ('".$username."', '".$password."', '".$permission."');";

            $addUserSuccess = mysqli_query($connection, $query_addUser);

            if($addUserSuccess !== 1)
            {
                $hostname = $_SERVER['HTTP_HOST'];
                $path = dirname($_SERVER['PHP_SELF']);
                echo"Success";
                //header("Location: http://".$hostname.($path == "/" ? "" : $path)."/userManagment.php?added=".$username."");
            }
            else
            {
                echo "ANNOYING_ERROR";
            }
        }

        //Tidy up
        mysqli_free_result($allUserNames);
        mysqli_close($connection);
    }
?>

这是相应的HTML代码,此后紧跟在同一文件中：

        <?php
            include "home.php";
        ?>

        <section>
            <h3>Einen neuen Benutzer hinzufügen</h3>

            <?php
                if(isset($errorName))
                {
                    echo '<p class="warningMessage">Der Nutzername <b>'.$_GET["addUser_name"].'</b> ist bereits vergeben.<br />Bitte wählen Sie einen anderen aus!</p>';
                }
            ?>

            <form method="GET" action="addUser.php">
                <table>
                    <tr>
                        <td>Nutzername:</td>
                        <td><input type="text" name="addUser_name" required pattern="\w+" /></td>
                        <td></td>
                        <td class="annotation">z.B.: Vorname</td>
                    </tr>

                    <tr>
                        <td>Passwort:</td>
                        <td><input type="password" name="addUser_pw" required pattern="(?=.*\d)(?=.*[a-z])(?=.*[A-Z]).{6,}" title="Passwort muss mindestens 6 Zeichen, Groß- und Kleinbuchstaben, sowie Zahlen enthalten"
                                    onChange="this.setCustomValidity(this.validity.patternMismatch ? '' : ''); if(this.checkValidity()){ form.addUser_pwRepeat.pattern = this.value; }" /></td>
                        <td></td>
                        <td class="annotation">Muss Groß- und Kleinbuchstaben, Zahlen und mindestens 6 Zeichen enthalten</td>
                    </tr>
                    <tr>
                        <td>Passwort wiederholen:</td>
                        <td><input type="password" name="addUser_pwRepeat" required pattern="(?=.*\d)(?=.*[a-z])(?=.*[A-Z]).{6,}" /></td>
                        <td></td>
                        <td class="annotation">Muss identisch sein zum ersten eingegebenen Passwort</td>
                    </tr>
                    <tr>
                        <td>Berechtigungslevel:</td>
                        <td>
                            <input type="radio" name="addUser_permission" value="1" checked />1<br />
                            <input type="radio" name="addUser_permission" value="2" />2
                        </td>
                    </tr>
                </table>
                <input type="hidden" name="startCheck" value="1" />
                <input type="submit" value="Nutzer hinzufügen" class="button" />
            </form>
        </section>
    </body>
</html>

如果对我的问题的描述不够清楚,我很乐意提供所需的任何其他信息.

解决方法:

我建议您不要获取所有用户名,并通过PHP检查是否有相同的用户名.您可以像这样简单地进行查询

$result=$connection->query("SELECT username FROM user WHERE username='".$connection->real_escape_string($addUser_name)."';"); 

然后检查查询是否返回给您任何行

if($result -> num_rows > 0)
{
  //the username is already in use
}
else
{
  //the username is unique
}

标签：mysqli,html-form,html,mysql,php
来源： https://codeday.me/bug/20191028/1950170.html