CODEWITHSUNDEEP
phpjquerymysql
Anuj
Anuj

Reputation: 333

MySQL INSERT automatically adding duplicate rows

I'm trying to update a database using html form + jquery + php

When I click on the Submit button on the form it works but it is adding two same records in the database for each click.

I can't find out what is going wrong and therefore I'm posting the whole script I'm using here. If anyone can find the problem then please point it out.

Here are the scripts I'm using

HTML

<div id="result" class="results"></div>
<form id="person-form" class="person-form" method="post">
 <fieldset>
  <legend><strong>Add a new member</strong></legend>
  <table>
  <tr><td><label for="Name" >Name</label></td>
  <td colspan="2"><input type="text" name="name" value="Enter the Name, 55 char max." /></td></tr>
  <tr><td><label for="email">Email</label></td>
  <td colspan="2"><input type="text" name="email" value="Enter the email" /></td></tr>
  <tr><td><label for="subscribe">Subscribe</label></td>
  <td><input type="checkbox" name="subscribe" value="Yes" /></td>
  <td><input type="submit" value="Add member" id="add-member" class="add-member"/></td></tr>
  </table>
  <div style="clear:both;"></div>
 </fieldset>
</form>

jQuery

$("#add-url").live("click", function() {
    var name = $('input[name=name]').val();
    var email = $('input[name=email]').val();
    var subs = $('input[name=subscribe]:checkbox:checked').val();
    var data = 'name='+name+'&email='+email+'&subs='+subs;

    $.post(add_member_script.ajaxurl, data,  function(data) {
   $('#results').html(data);

});
  return false;
});

And PHP

$connect = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);

if (!$connect){
    die('Could not connect: ' . mysql_error());
}
mysql_select_db(DB_NAME, $connect);


if (empty($_POST['name']) || empty($_POST['email'])){
    echo 'enter some value';
}  else {

      if($_POST['subs']== 'yes') {
        $sql="INSERT INTO $table (name, email, subs) VALUES ('$_POST[name]','$_POST[email]','$_POST[subs]')";
      } else {
        $sql="INSERT INTO $table (name, email) VALUES ('$_POST[name]','$_POST[email]')";
      }

    if (!mysql_query($sql,$con)){
        mysql_error();
    } else {
        list_links_table ();
    }
    mysql_close($con);
    die();
}

Upvotes: 0

Views: 2670

Answers (3)

Anuj
Anuj

Reputation: 333

Finally I found it, it has no error at all in the scripts I posted above. The problem was with a WordPress hook that I registered twice and because of that it was sending 2 requests with each click. I removed it and the problem solved. Thank you @Erenor Paz, @Mark Eirich and @slugonamission for your kind efforts. With that I finished my first WordPress plugin :)

Upvotes: 0

Mark Eirich
Mark Eirich

Reputation: 10114

It looks like the form may be getting submitted both via AJAX and the "normal" way. To make sure, add onsubmit="return false" to the form tag:

<form id="person-form" class="person-form" method="post" onsubmit="return false">

Alternately, you can remove all your JS and allow the form to submit normally, and see if the duplication persists.

Upvotes: 1

Erenor Paz
Erenor Paz

Reputation: 3161

As said in the comments, i would suggest you to use <input type="button" value="Add member" id="add-member" class="add-member"/> instead of the "type=submit", to make sure the form will never be uploaded via HTML, but only using the Javascript code :)

Upvotes: 0

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