Question

我在这里结束了我的智慧。我已尝试在所有不同的浏览器中仍然得到相同的触发器错误消息Notice: Duplicate entry for key 'PRIMARY'但是，它仍然插入数据库。我认为该页面同时执行了两次。代码永远不会到达存储会话数据的块。任何见解将不胜感激！

            <?php               
                if($_POST[submit])
                {                   
                    $username = "";
                    $password = "";
                    $hostname = "";
                    $dbname = "";

                    //connection to the database:
                    $db = mysqli_connect($hostname, $username, $password, $dbname, '3306')
                    or die("Unable to connect to MySQL");

                    $id = !empty($_POST[id]) ? "'$_POST[id]'" : "NULL";
                    $fname = !empty($_POST[fname]) ? "'$_POST[fname]'" : "NULL";
                    $lname = !empty($_POST[lname]) ? "'$_POST[lname]'" : "NULL";    

                    $sql = "INSERT INTO Table(id, fname, lname)
                                VALUES ($id, $fname, $lname)";

                    $result = $db->query($sql) OR trigger_error($db->error);

                    if ($result && $db->affected_rows > 0)
                    {
                        // store session data
                        $_SESSION['fileid']=$id;
                        echo 'Creating file '.$id.'<br/>
                            <a href = "partTwo.php" class="button">Continue</a>';
                    }
                    else
                        echo "No changes made.";
                }               
                else 
                    echo '<form method="post" action="createFile.php">
                            <div><span class = "formlabel">File number:</span><input type="Text" name="id"></div>
                            <div><span class = "formlabel">First name:</span><input type="Text" name="fname"></div>
                            <div><span class = "formlabel">Last name:</span><input type="Text" name="lname"></div>
                            <input class="submit" type="Submit" name="submit" value="Submit">
                        </form>';
            ?>

Answer 1

而不是if($_POST[submit])使用if(isset($_POST[submit]))

Answer 2

我已经编辑了你的代码，现在效果很好:)

     <?php               

        //connection to the database:

                            $username = "";
                            $password = "";
                            $hostname = "";
                            $dbname = "";
                            $db = mysqli_connect($hostname, $username, $password, $dbname, '3306')
                            or die("Unable to connect to MySQL");

                        if(isset($_POST[submit]))
                        {                   

                           if(($id = !empty($_POST['id']) || $fname = !empty($_POST['fname']) || $lname = !empty($_POST['lname']))
    {
       echo "error msg.";
    }else{
    $id = "";
    $fname = "";
    $lname = "";   

                            $sql = "INSERT INTO Table(id, fname, lname)
                                        VALUES ($id, $fname, $lname)";

                            $result = $db->query($sql) OR trigger_error($db->error);
}

                            if ($result && $db->affected_rows > 0)
                            {
                                // store session data
                                $_SESSION['fileid']=$id;
                                echo 'Creating file '.$id.'<br/>
                                    <a href = "partTwo.php" class="button">Continue</a>';
                            }
                            else
                                echo "No changes made.";
                        }               
                        else 
                            echo '<form method="post" action="createFile.php">
                                    <div><span class = "formlabel">File number:</span><input type="Text" name="id"></div>
                                    <div><span class = "formlabel">First name:</span><input type="Text" name="fname"></div>
                                    <div><span class = "formlabel">Last name:</span><input type="Text" name="lname"></div>
                                    <input class="submit" type="Submit" name="submit" value="Submit">
                                </form>';
                    ?>

SQL INSERT查询正在执行两次

2 个答案: