CODEWITHSUNDEEP
phpmysqlcodeigniter
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Reputation: 165

codeigniter MVC inserting sensitive data in mysql table

I am using codeigniter to build a page where users can like the post of another user.

I have three pages in this example, view_page.php , model_page.php and controller_page.php.

My view_page.php looks like this 

<?php foreach($members as $value): ?>
    <tr>
        <td><?php echo $value['post_id']; ?></td>
        <td><?php echo $value['post_title']; ?></td>
        <td><a href="like/<?php echo $value['post_id']; ?>">Like</a></td>
    </tr>
<?php endforeach; ?>

Controller_page.php

function like($post_id) {
    $this->load->model('model_page');
    $this->model_page->like_user($post_id);
    redirect('controller_page/viewdata');
}

My question is, If a user with user_id = 1, posts something on the view_page.php (as shown in the image below), his post_id, post_title, and like button will show.

If another user with User_id = 22, likes that post, how do I add user_id=22 in the mysql table, to know which user liked the post?

I was thinking of passing the user_id=22 in the url of the like button, but then people can just manipulate the likes but changing user_id in the address bar.

How do I do it correctly?

Thanks in advance

enter image description here

Upvotes: 1

Views: 79

Answers (2)

Ketan Solanki
Ketan Solanki

Reputation: 697

The Whole code for your questions would ask many aspects , As of now the solution provided by Tanseer UL Hassan is the best DB design ,now since your question is how to implement it , you can code in the following manner:

1) Create the variable which is to be passed to the controller :

<input type="hidden" name="hidUserIdFromSession" id="hidUserIdFromSession" value="<?php echo $someUserIdFromSession;?>">

2) Use ajax to Insert data in the above-mentioned table, like this :

$('#yourButton').click(function() {
//Modify this according to your button name
var postId = "Get your post id here";
var userId = $("#hidUserIdFromSession").val();
     $.ajax({
     type: 'POST',
     url: '<?php echo base_url("your/url/comes/here")?>',
     data: "postId="+postId+"&userId="+userId,
     success:function(response){
         //Your succes functions code comes here(Alert or any other processing).
     }
  });
});

3) Get the values inside controller function like this:

public function functionName(){
  //This is the same function which is called inside ajax call
  $arrReturn = array();
  $postId = $this->input->post('postId');
  $userId = $this->input->post('userId');
  $inserted = $this->model->insertLikeData($postId,$userId);
  //this is a call to model function in which you will actually make the Query to insert the data.
  if($inserted){
   $arrReturn['status'] = "Success";
   $arrReturn['data'] = array();//In case if you want to send some data.
  }else{
   $arrReturn['status'] = "Error";
   $arrReturn['data'] = array();//In case if you want to send some data.
  }
  $arrReturn = json_encode($arrReturn);
  echo $arrReturn;
}

Upvotes: 0

Tanseer UL Hassan
Tanseer UL Hassan

Reputation: 156

You can make a table where you store all likes For Example.

id post_id user_id
1  1       22 
2  1       44

Upvotes: 2

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