Question

我想使用ajax获取表单数据，然后将其提交到我的php页面，然后将其插入我的数据库中。成功插入后，我想显示＆＃34;您已成功注册＆＃34;并且失败＆＃34;无法注册＆＃34;如果您已经是会员，我会显示＆＃34;已经是会员，只需登录＆＃34;。

出于某种原因，我的ajax代码无法正常工作。它跳到最后一部分（否则声明甚至没有关心if语句）。我的PHP代码是正确的。因为我已经单独测试过它并且工作正常。

有人可以向我建议解决方案或其他方法来实现此目的请记住，我不想转到另一个页面。

这是我的JavaScript / jQuery代码：

$(document).ready(function){
    //Signup Js
    $("#Register").click(function(){ //#Register is my submit button id
        var action = $("#sp-form").attr('action'); //#sp-form is the register form id.
        var spform_data = {
            myusername: $("#myusername").val(),
            myemail: $("#myemail").val(),
            mypassword: $("#mypassword").val() //Getting data from the input forms using their respective id's
        }

        $.ajax({
            type: "POST",
            url: action,
            data: spform_data,
            success: function(outcome)
            {
                if(outcome == "success"){
                    $("#sp-form").slideUp('slow', function(){
                        $("#message_sp").html('<p class="success">You have been registered successfully!</p><p style="color:#1c1c1c;">Redirecting....</p>');
                    });
                }
                else if(outcome == "alreadymember") {
                    $("#modal_signup").css("height","380");
                    $("#message_sp").html('<p class="error">ERROR: Email match found,Just Sign in.</p>');
                }
                else if(outcome == "failed"){
                    $("#modal_signup").css("height","380");
                    $("#message_sp").html('<p class="error">ERROR:  Failed to Register</p>');
                }
                else{
                    $("#modal_signup").css("height","380");
                    $("#message_sp").html('<p class="error">ERROR:  Unknown   ERROR</p>');
                }                
            }            
        });
        return false;
    });
});

这是我的PHP / MySQLi代码

<?php
if ($_POST)
{
    $host = "localhost";
    $user= "root";
    $password = " ";

    try {
        $conn = mysqli_connect($host , $user , $password) or die("unable to connect to Local Host"); //Establishing connection with the server
        $sql = mysqli_select_db($conn,'audacity') or die("unable to connect to database"); //selecting a database from the server #after connecting
    } catch (Exception $e) {
        echo $e->getMessage();
    }
    $name = mysqli_real_escape_string($conn,$_POST['myusername']);
    $mail = mysqli_real_escape_string($conn,$_POST['myemail']);
    $pass = mysqli_real_escape_string($conn,$_POST['mypassword']);

    // Query of SQL
    $Query =mysqli_query($conn,"INSERT INTO  registration (Username,Email,Password) VALUES ('$name', '$mail' , '$pass')")) //if successful outcome is true
    if($Query ==true){
        echo "success";
    }
    elseif ($Query==false) {  #if failed
        $sqlbd = "SELECT Email FROM registration WHERE Email='$mail' ";
        $result = mysqli_query($conn,$sqlbd);

        if(mysqli_num_rows($result)==1){  //possible rsn for failure
            echo "alreadymember";             
        }
        else{
            echo "failed";  //unknown reason.
        }
    }
    mysqli_close($conn);
?>

问题在哪里

Answer 1

我认为你必须在你的ifs语句的开头检查num_rows是否= = 1 ...你的php只需输入email然后将num行检查为else语句...你首先检查num行然后设置if语句或者您已为电子邮件列设置了唯一的内容？

Answer 2

我可以确认您的代码有很多错误。我附上了修正版的代码;但是，我将假设（来自php代码）目标是防止多次注册;如果是这种情况，则应检查您的逻辑。您在检查之前插入，因此永远不会运行检查代码。所以在尝试插入之前进行检查。整体代码看起来不太好。我只修复了错误，你应该研究逻辑。

JQuery的 $（document）.ready（function（）{

            $("#sp-form").submit(function (e) {
                return false;
            });
            $("#Register").click(function () {
                var action = $("#sp-form").attr('action'); //#sp-form 
is the register form id.

                var spform_data = {
                    myusername: $("#myusername").val(),
                    myemail: $("#myemail").val(),
                    mypassword: $("#mypassword").val()         
//Getting data from the input forms using their respective id's
                }


                $.ajax({
                    type: "POST",
                    url: action,
                    data: spform_data,
                    success: function (outcome)
                    {
                        if (outcome.indexOf("success")>=0) {
                            $("#sp-form").slideUp('slow', function () {
                                $("#message_sp").html('<p 
class="success">You have been registered successfully!</p><p 
style="color:#1c1c1c;">Redirecting....</p>');
                            });
                        }
                        else if (outcome.indexOf("alreadymember")>=0) {


                            $("#modal_signup").css("height", "380");
                            $("#message_sp").html('<p 
class="error">ERROR: Email match found,Just Sign in.</p>');


                        }

                        else if (outcome.indexOf("failed")>=0) {


                            $("#modal_signup").css("height", "380");
                            $("#message_sp").html('<p 
class="error">ERROR:  Failed to Register</p>');


                        }
                        else {


                            $("#modal_signup").css("height", "380");
                            $("#message_sp").html('<p 
class="error">Unknown Error</p>');
                        }

                    }

                });
                return false;


            });
        });

// Php
 $host = "localhost";
        $user = "root";
        $password = "";
        $conn = "";
        try {
            $conn = mysqli_connect($host, $user, $password) or 
die("unable to connect to Local Host"); //Establishing connection with 
the server

            $sql = mysqli_select_db($conn, 'audacity') or die("unable 
to connect to database"); //selecting a database from the server #after 
connecting
        } catch (Exception $e) {
            echo $e->getMessage();
        }


        $name = mysqli_real_escape_string($conn, $_POST['myusername']);
        $mail = mysqli_real_escape_string($conn, $_POST['myemail']);
        $pass = mysqli_real_escape_string($conn, $_POST['mypassword']);

        // Query of SQL


        $Query = mysqli_query($conn, "INSERT INTO  registration 
(Username,Email,Password) VALUES ('$name', '$mail' , '$pass')"); //if 
successful outcome is true
        if ($Query == true) {

            echo "success";
        } elseif ($Query == false) {  #if failed
            $sqlbd = "SELECT Email FROM registration WHERE 
Email='$mail' ";
            $result = mysqli_query($conn, $sqlbd);

            if (mysqli_num_rows($result) == 1) {  //possible rsn for 
failure
                echo "alreadymember";
            } else {

                echo "failed";  //unknown reason.
            }
        }

        mysqli_close($conn);
    }

如何使用Ajax和PHP将表单数据输入数据库

2 个答案: