Question

我一直试图让这个脚本完成一段时间 - 我是一种新的php和mysql，但我一直试图让它检查数据库的用户名，然后如果用户名存在 - 停止检查db并且如果它不存在则将其添加到db。

这是我的代码：

                    //input from application
                    $test = "wheelsmanx";
                    // Create connection
                    $conn = new mysqli($servername, $username, $password, $dbname);
                    // Check connection
                    if ($conn->connect_error) {
                         die("Connection failed: " . $conn->connect_error);
                    }

                    $sql = "SELECT mainusername FROM CCCpro_test";
                    $result = $conn->query($sql);

                    if ($result->num_rows > 0) {

                        while($row = $result->fetch_assoc()) { 
                            if ($row["mainusername"] === $test) {
                                echo "User Name Already In Use.";


                            }if($row["mainusername"] !== $test){
                                echo "this statement";

                            [code that inserts into db i can do this part myself]   

                            }

                            }
                        $conn->close(); 

                    } else {
                         echo "0 results";
                    }

                    $conn->close();

Answer 1

您的代码存在的问题是您在已确认该用户存在的INSERT语句中执行了新名称的if。此外，我认为你通过选择所有用户搞砸了你的SELECT声明。

查看INSERT ON DUPLICATE以获得更好的方法，或修改您的代码，如下所示。

                $sql = "SELECT mainusername FROM CCCpro_test WHERE mainusername = $test";
                $result = $conn->query($sql);

                if ($result->num_rows > 0) {
                    echo "User Name Already In Use.";
                }
                else{  //no rows selected therefore the user doesn't exist
                        [code that inserts into db i can do this part myself]   
                }

                $conn->close();

请阅读我有一个地方可以去，所以我很懒，所以我没有绑定$test变量，因此请勿复制并粘贴此代码而不更新它以绑定{{ 1}}变量。请阅读有关PDO和变量绑定的this post以防止SQL注入。

Answer 2

这是我的完整工作代码，如果有人需要它 - 它使用post方法 - 从html表单....如果有人需要将其破解成其他东西

好吧，伙计们，我很感谢你们所有的帮助：D但是我已经找到了答案或解决方法我想 - 我一整天都想着它如何让它发挥作用我想出了这个

                    $servername = "127.0.0.1";
                    $username = "TESTUSER";
                    $password = "TESTPASS";
                    $dbname = "TESTDB";

                    $testusername = $_POST['mainusername'];
                    $testpassword = $_POST['mainpassword'];

                    //input from application
                    $test = $_POST['mainusername'];
                    $test2 = "0";
                    //Count switch 
                    $countswitch = "0";
                    // Create connection
                    $conn = new mysqli($servername, $username, $password, $dbname);
                    // Check connection
                    if ($conn->connect_error) {
                         die("Connection failed: " . $conn->connect_error);
                    }
                    $sql1 = "INSERT INTO CCCpro_test ( mainusername, mainpassword ) VALUES ('$testusername','$testpassword' )";
                    $sql = "SELECT mainusername FROM CCCpro_test";
                    $result = $conn->query($sql);

                    if ($result->num_rows > 0) {

                        while($row = $result->fetch_assoc()) { 
                            if ($row["mainusername"] === $test) {
                                echo "Im Sorry Username Already In Use";
                                 $countswitch ++;
                            }
                            }

                            if($countswitch == $test2){
                                echo "User Name Registered";

                                $db_handle = mysql_connect($servername, $username, $password);
                                $db_found = mysql_select_db($dbname, $db_handle);

                                if ($db_found) {

                                $result1 = mysql_query($sql1);

                                mysql_close($db_handle);

                                }
                            }
                            if ($countswitch == 3){
                                echo "this";
                            }   
                            } else {
                         echo "0 results";
                    }

                    $conn->close();

如果变量在db中，那么stop- if变量不是 - 然后输入它

2 个答案: