CODEWITHSUNDEEP
javascriptphpjquerymysqlajax
Dhruvi Mistry
Dhruvi Mistry

Reputation: 122

Insert data into mysql database using ajax in php

I am trying to insert value using ajax in php, but data is not inserted in database. I have taken this code from the questions answered in other question from this site. Can anyone suggest where am I making mistake..?

<script>
  $("#submit").click(function() {
                var name= $("#name").val();
                var password= $("#password").val();

                $.ajax({
                    type: "POST",
                    url: "insert.php",
                    data: "name=" + name+ "&password=" + password,
                    success: function(data) {
                       alert("sucess");
                    }
                });


            });

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.10.0/jquery.min.js"></script>

<?php
    //------insert.php------
     $servername = "localhost";
    $username = "username";
    $password = "password";
    $dbname = "dbname";

    // Create connection
    $conn = new mysqli($servername, $username, $password,$dbname);

    // Check connection
    if ($conn->connect_error) {
        die("Connection failed: " . $conn->connect_error);
    } 


          $name=$_POST['name'];
            $pass=$_POST['password'];
             $sql= mysqli_query($conn,"INSERT INTO insert_tbl(name,pass) VALUES('".$name."','".$pass."')");

 ?>

Upvotes: 0

Views: 59387

Answers (5)

Vinothini K
Vinothini K

Reputation: 1

     <script>
      var formData = new FormData($("#form")[0]);
      $.ajax({
    url: "emp_insert.php",
    type: "post",
    processData:false,
    contentType:false,
    data: formData,
    success: function (d) {
        // alert(d);

    swal(d, "", "success", { button: "ok" });

        $("#form")[0].reset();

    },

   });
    </script>

File upload php

     <?php
     session_start();
    // print_r($_POST);die;
    require_once 'db.php';
    //error_reporting(1);

    $emp_name = $_POST['emp_name'];
    $dob = $_POST['dob'];

    $aadhar = $_POST['aadhar'];
      $cid = $_POST['cid'];
      $woid = $_POST['woid'];
      $c_emp_id = $_POST['c_emp_id'];
      $wop_from = $_POST['wop_from'];
     $wop_to = $_POST['wop_to'];
     $doj=$_POST['doj'];
     $designation = $_POST['designation'];


        $filename = $_FILES["photo"]["name"];
    $target_directory = "upload/"; 
     $target_file = $target_directory.basename($_FILES["photo"]["name"]); 
  //name is to get the file name of uploaded file 
    $filetype = strtolower (pathinfo($target_file, PATHINFO_EXTENSION));
    $newfilename = time().".".$filetype;
    move_uploaded_file($_FILES["photo"]["tmp_name"], $target_file); // 
    tmp_name is the file temprory stored in the server //Now to check if 
    uploaded or not 
// if(move_uploaded_file($_FILES["file"]["tmp_name"], $newfilename)) 

  $sql="select * from 
 employee('$emp_name','$dob','$filename',$aadhar,$cid,$woid,'$wop_from',
'$wop_to','$doj','$c_emp_id','$designation')";
 $result = pg_query($sql);
 //move_uploaded_file($tmp, $target);
 if($result){

   echo 'Data stored sucess';
   }

 else{
   echo'Data not storedd';
    }

  exit();

?>

Upvotes: -1

SELVAMANI R
SELVAMANI R

Reputation: 1

Php DB connection(Postgres)

<?php
$db = pg_connect("host=localhost port=5432 dbname=demo user=postgres 
password=selva27");
?>

Php Insert

 <?php
// session_start();
// print_r($_POST);die;
require_once'db.php'; 

//error_reporting(0);

$name = $_POST['cname'];
$address = $_POST['address']; 
$name = $_POST['name'];
$email = $_POST['email'];
$phone = $_POST['phone']; 

$sql="insert into 
demo(col1,col2,col3,col4)values('val1',val2,val3,val4)";

$result = pg_query($sql);

if($result){

echo 'Data stored sucessfully';
  }

else{
  echo'Data not stored';
}

pg_close($db);
exit();

?>

Php Fetch

<?php
require_once'db.php';
 
$date='set datestyle=SQL,DMY';
$select="select * from manpower order by manid";
pg_query($date);
$result=pg_query($select);
$data=array();
    
    while($row=pg_fetch_assoc($result)){
        $data[]=$row;
 
 }
 
 
echo json_encode($data);
exit();

  ?>

Php Delete

<?php
require('db.php');
if(isset($_GET['id']))
{
 $sql = "DELETE FROM manpower WHERE manid=".$_GET['id'];
 pg_query($sql);
 echo 'Deleted successfully.';
}

?>

Php Edit

<?php
require_once'db.php';
if(isset($_GET['id'])){
$id=$_GET['id'];
$select="select * from manpower where manid=$id";
$result=pg_query($select);
$data=array();
    
    while($row=pg_fetch_assoc($result)){
        $data[]=$row;
 
 }
 
 }  
 echo json_encode($data);
 exit();
 ?>

Php Update

<?php
require_once'db.php';
// print_r($_POST); die;
$id=$_POST['id'];
$name=$_POST['name'];
$dob=$_POST['dob'];
$skill=$_POST['skill'];
$address=$_POST['address'];
$mobile=$_POST['mobile'];
$email=$_POST['email'];
$remarks=$_POST['remarks'];

$update="update manpower set name='$name',date_of_birth='$dob' skill_code=$skill
where manid=$id";
$result=pg_query($update);
if($result){
echo"Updated successfully";
}
else{
echo"Something wrong";
}
?>

Upvotes: 0

Sami Ullah
Sami Ullah

Reputation: 45

You can use this ajax to insert data in database.

 $(document).ready(function() {
        //register
        $("#register_btn").on("click", function() {
            $("#register_btn").html(
                'Please Wait ...'
            );
            $(".error").html("");
            $.ajax({
                type: "POST",
                url: "register-submit.php",
                dataType: "json",
                data: $("#register_form").serialize(),
                success: function(response) {
                    alert(response.mesage)
                    $("#register_btn").html("Sign Up");
                },
                error: function(error) {
                    console.log(error);
                    $("#register_btn").html("Sign Up");
                },
            });
        });
    })

At "register-submit.php" you will place your PHP code file name with path. you can check this tutorial for complete example.

Upvotes: 0

Sanju Yadav
Sanju Yadav

Reputation: 11

This is html form for insert data

<form id="frmrecord" method="post">
    <input type="text" name="txtusermame" />
    <input type="password" name="txtpassword" />
    <input type="submit" value="Insert" />
</form>

Use this code for call insert.php file to insert data

jQuery(document).ready(function ($) {

    $("#frmrecord").submit(function (event) {
                event.preventDefault();
                //validation for login form
        $("#progress").html('Inserting <i class="fa fa-spinner fa-spin" aria-hidden="true"></i></span>');

            var formData = new FormData($(this)[0]);
            $.ajax({
                url: 'insert.php',
                type: 'POST',
                data: formData,
                async: true,
                cache: false,
                contentType: false,
                processData: false,
                success: function (returndata) 
                {
                    //show return answer
                    alert(returndata);
                },
                error: function(){
                alert("error in ajax form submission");
                                    }
        });
        return false;
    });
});

After calling file you can receive data in php file Insert.php

<?php 
  $usernmae=$_POST['txtusername']; 
  $password=$_POST['password']; 
  $sql= mysqli_query($conn,"INSERT INTO insert_tbl(name,pass) 
    VALUES('".$usernmae."','".$password."')");
 ?>

Download Demo

Upvotes: 1

Tsotne Kekelia
Tsotne Kekelia

Reputation: 97

<script>
  $("#FORM_ID").submit(function() {
                var name= $("#name").val();
                var password= $("#password").val();

                $.ajax({
                    type: "POST",
                    url: "insert.php",
                    data: "name=" + name+ "&password=" + password,
                    success: function(data) {
                       alert("sucess");
                    }
                });


            });
</script>

and also either load 

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.10.0/jquery.min.js"></script>

before your script tag or use 

 <script>
$(document).ready(function(){
      $("#FORM_ID").submit(function() {
                    var name= $("#name").val();
                    var password= $("#password").val();

                    $.ajax({
                        type: "POST",
                        url: "insert.php",
                        data: "name=" + name+ "&password=" + password,
                        success: function(data) {
                           alert("sucess");
                        }
                    });


                });
        });
    </script>

Upvotes: 3

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