Record and not added into the database using Php ajax

Question

i am creating the simple crud system using php ajax.in the crud system i add the image also.but when i fill the data and browse image and click add button record is not added in to the database.what i tried so far i attached below.no error shown on the console.i think might be the problem with jquery send the form values and images data: {form_data: form_data,data: data},

Form Design

 <form role="form" id="frmcompany" class="card" enctype="multipart/form-data">
        <div align="left">
            <h3>Company</h3>
        </div>

        <div align="left">
            <label class="form-label">Company name</label>
            <input type="text" class="form-control" placeholder="Patient No" id="cno" name="cno" size="30px" required>
        </div>

        <div align="left">
            <label class="form-label">Country</label>
            <input type="text" class="form-control" placeholder="Patient Name" id="coutry" name="coutry" size="30px" required>
        </div>

        <div align="left">
            <label class="form-label">Currency</label>
            <input type="text" class="form-control" placeholder="Phone" id="currency" name="currency" size="30px" required>
        </div>

        <div align="left">
            <label class="form-label">Address</label>
            <input type="text" class="form-control" placeholder="Address" id="address" name="address" size="30px" required>
        </div>
            <div align="left">
                <div class="fileuploader fileuploader-theme-default">
                    <input type="hidden" name="fileuploader-list-files_" value="[]">
                    <input type="file" id="file" name="file" >
                    <div class="fileuploader-items">
                        <ul class="fileuploader-items-list"></ul>
                    </div>

                </div>
            </div>

        </br>

        <div align="right">
            <button type="button" id="save" class="btn btn-info" onclick="addPatient()">Add</button>
            <button type="button" id="clear" class="btn btn-warning" onclick="reset()">Reset</button>

        </div>


    </form>

jquery

            function addPatient()
{
    var upload_date = $('#file').prop('files')[0];
    var form_data = new FormData();
    form_data.append('file', upload_date);


    if($("#frmcompany").valid())
    {
        var url = '';
        var data = '';
        var method = '';

        if(isNew == true)
        {
            url = 'php/add_patient.php';
            data = $('#frmcompany').serialize() ;
            method = 'POST';
        }


        $.ajax({
                type : method,
                url : url,
                dataType : 'JSON',
            cache: false,
            contentType: false,
            processData: false,
               // data:  data,

            data: {form_data: form_data,data: data},

            success:function(data)
            {




                    if (isNew == true) {
                        alert("Company Addedd");
                    }



            }

        });

    }

}
$('#file').fileuploader
({
    limit: 1,
});

            <div align="right">
                <button type="button" id="save" class="btn btn-info" onclick="addPatient()">Add</button>
                <button type="button" id="clear" class="btn btn-warning" onclick="reset()">Reset</button>

            </div>

        </form>

php code

<?php

$servername = "***";
$username = "***";
$password = "***";
$dbname = "***";
$conn = new mysqli($servername,$username,$password,$dbname);

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

if($_SERVER['REQUEST_METHOD'] == 'POST')
{

   if (0 < $_FILES['file']['error'])
    {
        echo 'Error: ' . $_FILES['file']['error'] . '<br>';
    }
    else
   {
        move_uploaded_file($_FILES['file']['tmp_name'], 'uploads/' . $_FILES['file']['name']);
   }

    $stmt = $conn->prepare("insert into company(companyname,coutry,currency,address,image) VALUES (?,?,?,?,?)");
    $stmt->bind_param("ssss",$companyname,$coutry,$currency,$address,$image);
    $companyname = $_POST['cno'];
    $coutry = $_POST['coutry'];
    $currency =  $_POST['currency'];
    $address =  $_POST['address'];
    $image = $_FILES['file']['name'];

    if($stmt->execute())
    {
        echo 1;
    }
    else
    {
        echo 0;
    }
$stmt->close();
}
?>

`data: {form_data: form_data,data: data},` will give in php `$_POST["form_data"]` and `$_POST["data"]` — Cid, Sep 06 '19 at 15:06
You do `$stmt->bind_param("ssss",$companyname,$coutry,$currency,$address,$image);` and **after** `$companyname = $_POST['cno'];` and so on... Your variables are not set when you are binding — Cid, Sep 06 '19 at 15:08
Possible duplicate of [How do I get PHP errors to display?](https://stackoverflow.com/questions/1053424/how-do-i-get-php-errors-to-display) — Cid, Sep 06 '19 at 15:09
You have a `success` member, but no `error` one to check is inserting failed. — Cid, Sep 06 '19 at 15:12
@Your I feel that the "mysqli" tag should be added back, since there is an issue with the prepared statement. Look at it again and you will see it. Unless you feel that this (typo) is "php" related and not "mysqli"? — Funk Forty Niner, Sep 06 '19 at 16:22
@Cid `. Your variables are not set when you are binding` - it's OK — Your Common Sense, Sep 06 '19 at 16:24
@FunkFortyNiner May be I am overlooking something but there is nothing to indicate there is a problem with mysqli. As far as I can tell it's a problem with javascript — Your Common Sense, Sep 06 '19 at 16:25
@YourCommonSense Look at `VALUES (?,?,?,?,?)` and the number of `s`'s in the binding ;-) — Funk Forty Niner, Sep 06 '19 at 16:26
@FunkFortyNiner thank you I see now. So technically yes, but given tags are for the question re-use and there are no answers related to mysqli, this tag will be misleading for anyone having a problem with mysqli. Let's agree on just closing the question :) — Your Common Sense, Sep 06 '19 at 16:28

score 0 · Answer 1 · answered Sep 06 '19 at 15:18

 function addPatient()
{
    var upload_date = $('#file').prop('files')[0];
    var form_data = new FormData();
    form_data.append('file', upload_date);

      $.ajax({
            url: 'php/add_patient.php',
            cache: false,
            contentType: false,
            processData: false,
            data: form_data,
            type: 'POST',
            success: function(response) {
                console.log(response);
            },
            error: function(error) {
                console.log(error);
            }
        });
}

`form_data` doesn't have any of the other form fields. – Barmar Sep 06 '19 at 15:19 — Barmar, Sep 06 '19 at 15:19

score 0 · Accepted Answer · answered Sep 06 '19 at 15:18

0

You can't send a FormData object inside another object. All the POST parameters have to be in form_data.

function addPatient() {
  if ($("#frmcompany").valid()) {
    var url = '';
    var data = '';
    var method = '';
    var form_data = new FormData(document.getElementById("frmcompany"));
    var upload_date = $('#file').prop('files')[0];
    form_data.append('file', upload_date);

    if (isNew == true) {
      url = 'php/add_patient.php';
      data = $('#frmcompany').serialize();
      method = 'POST';
    }

    $.ajax({
      type: method,
      url: url,
      dataType: 'JSON',
      cache: false,
      contentType: false,
      processData: false,
      data: form_data,
      success: function(data) {
        if (isNew == true) {
          alert("Company Addedd");
        }
      }
    });
  }
}

answered Sep 06 '19 at 15:18

Barmar

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sir your one is working record is added into the database. why sir my browse image is not passing to the upload folder pls check sir thanks – nishan kishan Sep 06 '19 at 15:26
What does `var_dump($_FILES)` show? – Barmar Sep 06 '19 at 15:35
not shown anything sir at the console – nishan kishan Sep 06 '19 at 15:38
You have to go to the Network tab of DevTools to see the debugging output. – Barmar Sep 06 '19 at 15:42

Record and not added into the database using Php ajax

2 Answers2