Insert data through ajax into mysql database

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灰色年华
灰色年华 2020-11-30 06:33

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6条回答
  • 2020-11-30 06:39

    Try this:

      $(document).on('click','#save',function(e) {
      var data = $("#form-search").serialize();
      $.ajax({
             data: data,
             type: "post",
             url: "insertmail.php",
             success: function(data){
                  alert("Data Save: " + data);
             }
    });
     });
    

    and in insertmail.php:

    <?php 
    if(isset($_REQUEST))
    {
            mysql_connect("localhost","root","");
    mysql_select_db("eciticket_db");
    error_reporting(E_ALL && ~E_NOTICE);
    
    $email=$_POST['email'];
    $sql="INSERT INTO newsletter_email(email) VALUES ('$email')";
    $result=mysql_query($sql);
    if($result){
    echo "You have been successfully subscribed.";
    }
    }
    ?>
    

    Don't use mysql_ it's deprecated.

    another method:

    Actually if your problem is null value inserted into the database then try this and here no need of ajax.

    <?php
    if($_POST['email']!="")
    {
        mysql_connect("localhost","root","");
        mysql_select_db("eciticket_db");
        error_reporting(E_ALL && ~E_NOTICE);
        $email=$_POST['email'];
        $sql="INSERT INTO newsletter_email(email) VALUES ('$email')";
        $result=mysql_query($sql);
        if($result){
        //echo "You have been successfully subscribed.";
                  setcookie("msg","You have been successfully subscribed.",time()+5,"/");
                  header("location:yourphppage.php");
        }
         if(!$sql)
        die(mysql_error());
        mysql_close();
    }
    ?>
        <?php if(isset($_COOKIE['msg'])){?>
           <span><?php echo $_COOKIE['msg'];setcookie("msg","",time()-5,"/");?></span> 
        <?php }?>
    <form id="form-search" method="post" action="<?php echo $_SERVER['PHP_SELF'];?>">
        <span><span class="style2">Enter you email here</span>:</span>
        <input name="email" type="email" id="email" required/>
        <input type="submit" value="subscribe" class="submit"/>
    </form>
    
    0 讨论(0)
  • 2020-11-30 06:42

    ajax:

    $(document).on('click','#mv_secure_page',function(e) {
        var data = $("#m_form1").serialize();
        $.ajax({
            data: data,
            type: "post",
            url: "adapter.php",
            success: function(data){
            alert("Data: " + data);
            }
        });
    });
    

    php code:

    <?php
    /**
     * Created by PhpStorm.
     * User: Engg Amjad
     * Date: 11/9/16
     * Time: 1:28 PM
     */
    
    
    if(isset($_REQUEST)){
        include_once('inc/system.php');
    
        $full_name=$_POST['full_name'];
        $business_name=$_POST['business_name'];
        $email=$_POST['email'];
        $phone=$_POST['phone'];
        $message=$_POST['message'];
    
        $sql="INSERT INTO mars (f_n,b_n,em,p_n,msg) values('$full_name','$business_name','$email','$phone','$message') ";
    
        $sql_result=mysqli_query($con,$sql);
        if($sql_result){
           echo "inserted successfully";
        }else{
            echo "Query failed".mysqli_error($con);
        }
    }
    ?>
    
    0 讨论(0)
  • 2020-11-30 06:50

    The ajax is going to be a javascript snippet that passes information to a small php file that does what you want. So in your page, instead of all that php, you want a little javascript, preferable jquery:

    function fun()
    {
        $.get('\addEmail.php', {email : $(this).val()}, function(data) {
            //here you would write the "you ve been successfully subscribed" div
        });
    }
    

    also you input would have to be:

    <input type="button" value="subscribe" class="submit" onclick="fun();" />
    

    last the file addEmail.php should look something like:

    mysql_connect("localhost","root","");
    mysql_select_db("eciticket_db");
    
    error_reporting(E_ALL && ~E_NOTICE);
    
    $email=mysql_real_escape_string($_GET['email']);
    $sql="INSERT INTO newsletter_email(email) VALUES ('$email')";
    $result=mysql_query($sql);
    if($result){
    echo "You have been successfully subscribed.";
    }
     if(!$sql)
    die(mysql_error());
    
    mysql_close();
    

    Also sergey is right, you should use mysqli. That's not everything, but enough to get you started.

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  • 2020-11-30 06:53

    The available answers led to the fact that I entered empty values into the database. I corrected this error by replacing the serialize () function with the following code.

    $(document).ready(function(){
    
    // When click the button.
    $("#button").click(function() {
    
        // Assigning Variables to Form Fields
        var email = $("#email").val();
    
        // Send the form data using the ajax method
        $.ajax({
            data: "email=" + email,
            type: "post",
            url: "your_file.php",
            success: function(data){
                alert("Data Save: " + data);
            }
        });
    
    });
    
    });
    
    0 讨论(0)
  • 2020-11-30 06:57

    Why use normal jquery ajax feature. Why not use jquery ajax form plugin, which post the form data by ajax to the form action link.

    Check it here:

    http://malsup.com/jquery/form/#getting-started

    It is very easy to use and support several data formats including json, html xml etc. Checkout the example and you will find it very easy to use.

    Thank you

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  • 2020-11-30 06:58

    I will tell you steps how you can insert data in ajax using PHP

    AJAX Code

    <script type="text/javascript">
    function insertData() {
    var student_name=$("#student_name").val();
    var student_roll_no=$("#student_roll_no").val();
    var student_class=$("#student_class").val();
    
    
    // AJAX code to send data to php file.
        $.ajax({
            type: "POST",
            url: "insert-data.php",
            data: {student_name:student_name,student_roll_no:student_roll_no,student_class:s
             tudent_class},
            dataType: "JSON",
            success: function(data) {
             $("#message").html(data);
            $("p").addClass("alert alert-success");
            },
            error: function(err) {
            alert(err);
            }
        });
    
     }
    
    </script>
    

    PHP Code:

    <?php
    
    include('db.php');
    $student_name=$_POST['student_name'];
    $student_roll_no=$_POST['student_roll_no'];
    $student_class=$_POST['student_class'];
    
    $stmt = $DBcon->prepare("INSERT INTO 
    student(student_name,student_roll_no,student_class) 
    VALUES(:student_name, :student_roll_no,:student_class)");
    
     $stmt->bindparam(':student_name', $student_name);
     $stmt->bindparam(':student_roll_no', $student_roll_no);
     $stmt->bindparam(':student_class', $student_class);
     if($stmt->execute())
     {
      $res="Data Inserted Successfully:";
      echo json_encode($res);
      }
      else {
      $error="Not Inserted,Some Probelm occur.";
      echo json_encode($error);
      }
    
    
    
      ?>
    

    You can customize it according to your needs. you can also check complete steps of AJAX Insert Data PHP

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