Data insertion issues with HTML form and PHP to mySQL

Question

I have seen few posts in SO related to the same scenario as mine but did not find a proper resolution. So am posting question with my problem stuff.

I have an HTML form

 <form method="post" id="myForm">
      <label for="e_name">Name</label>
      <input name="e_name" id="emp_name" value="" type="text" data-theme="a">
      <label for="date">Date</label>
      <input name="date" id="emp_dob" value=""  data-theme="a">
      <label for="gender">Gender</label>
      <select name="gender" id="emp_gender" data-role="slider" data-theme="a" data-inline="true">
        <option value="male">Male</option>
        <option value="female">Female</option>        
      </select>
      <label for="address">Address</label>
      <textarea name="address" id="emp_address" value="" type="text" data-theme="a"></textarea><br><br>      
      <input type="button" id="insert" value="Submit">      
    </form>

  <div id="someElement"></div>

And I have the following to perform my form elements submission to a PHP page-

$(document).ready(function(){ 
    $("#insert").click(function(e) { 
    e.preventDefault();
    alert("am in the Insert function now");
        $.ajax({
        cache: false,
        type: 'POST',
        url: 'insert.php',
        data: $("#myForm").serialize(),
        success: function(d) {
            $("#someElement").html(d);
        }
        });
    }); 
});

Here is my PHP -

 <?php
    $con=mysqli_connect("localhost","root","root","employee");
    // Check connection
    if (mysqli_connect_errno())
      {
      echo "Failed to connect to MySQL: " . mysqli_connect_error();
      }
      $name =" ";
      $dob =" ";
      $gender =" ";
      $address =" ";

        if(isset($_POST['emp_name'])){ $name = $_POST['emp_name']; }
        if(isset($_POST['emp_dob'])){ $dob = $_POST['emp_dob']; }
        if(isset($_POST['emp_gender'])){ $gender = $_POST['emp_gender']; }
        if(isset($_POST['emp_address'])){ $address = $_POST['emp_address']; }

        echo $name;
        echo $dob;
        echo $gender;


echo $address;


$sql="INSERT INTO emp_details (emp_name, emp_gender, emp_address) VALUES ('$name', '$gender', '$address')";

if (!mysqli_query($con,$sql))
  {
  die('Error: ' . mysqli_error($con));
  }

echo "1 record added";

mysqli_close($con);
?>

Now what happens is, when I enter some values in my form and click on Submit, action performs well and a new row inserts in the database. But all the rows are empty. They're just empty, not even "NULL". I tried to echo my field values in the PHP but there is no output there. My "1 Record Added" has come-up well, but no form values appear here.

Kindly help me sorting this out.

Thanks in advance.

var_dump($_POST); in your php file & check you actually getting data when your form is submitting ? Also it seems you are passing serialize data from ajax call. — Rikesh
– Rikesh, Commented May 2, 2013 at 8:52
This code vulnerable to hack through sql injection. Use prepared statements and sanitize all data from user. — b1_
– b1_, Commented May 2, 2013 at 8:55

Menno · Accepted Answer · 2013-05-02 08:58:52Z

1

$_POST[] references to the name attribute of html-tag, not id.

For example, $_POST['emp_name'] should be $_POST['e_name']

Furthermore, don't encapsulate your variables with single quotes:

"INSERT INTO emp_details (emp_name, emp_gender, emp_address) VALUES ('$name', '$gender', '$address')";

Do this instead:

"INSERT INTO emp_details (emp_name, emp_gender, emp_address) VALUES ('" . $name . "', '" . $gender . "', '" . $address . "')";

Or use bind_param() from mysqli ofcourse!

edited May 2, 2013 at 8:58

answered May 2, 2013 at 8:53

Menno

12.8k14 gold badges59 silver badges92 bronze badges

Sign up to request clarification or add additional context in comments.

5 Comments

Lohith Korupolu Over a year ago

Done.. I made a silly mistake ... considered ID instead of name... am so dumb, I forgot this basic. It's been more than 3 years I worked on this stuff.. thanks for reminding me that I need more basics strength. :) :)

Menno Over a year ago

You're welcome. I infer from this comment you've managed to solve your question?

Lohith Korupolu Over a year ago

may I ask you for any good and easy/quick tutorials available online for PHP?

Menno Over a year ago

Well this depends on what you're trying to accomplish. PHP is used for a numerous purposes. Though it's a well documented language. In any case PHP.net is your best friend.

Menno Over a year ago

@user1397891 I don't know how fell advanced you are, but in case you're interested W3Schools has some excellent examples, well implemented.

웃웃웃웃웃 · Accepted Answer · 2013-05-02 08:54:02Z

Make the id and name of your input elements same

<form method="post" id="myForm">
      <label for="e_name">Name</label>
      <input name="emp_name" id="emp_name" value="" type="text" data-theme="a">
      <label for="date">Date</label>
      <input name="emp_dob" id="emp_dob" value=""  data-theme="a">
      <label for="gender">Gender</label>
      <select name="emp_gender" id="emp_gender" data-role="slider" data-theme="a" data-inline="true">
        <option value="male">Male</option>
        <option value="female">Female</option>        
      </select>
      <label for="address">Address</label>
      <textarea name="emp_address" id="emp_address" value="" type="text" data-theme="a"></textarea><br><br>      
      <input type="button" id="insert" value="Submit">      
    </form>

Otherwise change your $_POST array keys.Because you will get the keys of $_POST array will be the name of input elements.But i recommend you to mak ethe name and id same

Collectives™ on Stack Overflow

Data insertion issues with HTML form and PHP to mySQL

2 Answers 2

5 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

5 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related