How to view input based on if condition in PHP MySQLi

Question

How to make the if statement, if the the post belongs to the user, then only the owner of the post has a like/view/edit/delete button available. If the post does not belong to the user, it only displays like and view buttons.

<?php
$get_post = "SELECT * FROM posts";
$run_post = mysqli_query($connect, $get_post);
if($run_post && mysqli_num_rows($run_post) > 0)
{
    while($row_post = mysqli_fetch_array($run_post))
    {
        $post_title  = $row_post['post_title'];
        $get_user    = "SELECT * FROM users WHERE user_id='$user_id'";
        $run_user    = mysqli_query($connect, $get_user);
        $check_user  = mysqli_fetch_array($run_user);
        $user_name   = $check_user['name'];
        $user_images = $check_user['images'];

        echo "<div id='post_wrap'>
            <p>$post_title</p>
            <a href=''><button>Like</button></a>
            <a href=''><button>View</button></a>
            <a href=''><button>Delete</button></a>
            <a href=''><button>Edit</button></a>
            </div>";
    }
    mysqli_free_result($run_post);
}
else
{
    echo "No post yet";
}
?>

users table (user_id, name, email) , posts (post_id, user_id, title) — Rzj Hayabusa
– Rzj Hayabusa, Commented Oct 18, 2016 at 4:06

Gayan · Accepted Answer · 2016-10-18 09:20:58Z

2

You have to store user_id as a session value. Then you can checked current user posts this way, $_SESSION['user_id'] == $row_post['user_id']

 echo "<div id='post_wrap'>
           <p>$post_title</p>
           <a href=''><button>Like</button></a>
           <a href=''><button>View</button></a>";
           if($_SESSION['user_id'] == $row_post['user_id']){
                  echo "<a href=''><button>Delete</button></a>
                        <a href=''><button>Edit</button></a>";
            }            

     echo "</div>";

edited Oct 18, 2016 at 9:20

answered Oct 18, 2016 at 4:06

Gayan

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5 Comments

Gayan Over a year ago

how you get current user id, have you used sessions?

Rzj Hayabusa Over a year ago

i use 'email' column field in the table as session

Gayan Over a year ago

you have to store user_id also as session.

Rzj Hayabusa Over a year ago

i can echo the value of 'email' session . but when i try to add if statement inside echo "" . it didnt work at all . :(

Gayan Over a year ago

Let us continue this discussion in chat.

Arshid KV · Accepted Answer · 2016-10-18 04:08:57Z

0

Please check following answer.

                 $html = "<a href=''><button>Like</button></a>
                            <a href=''><button>View</button></a>";
                if($user_id == $row_post['user_id']){
                   $html .= "<a href=''><button>Delete</button></a>
                         <a href=''><button>Edit</button></a>"
                }            
  echo $html;

answered Oct 18, 2016 at 4:08

Arshid KV

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1 Comment

Rzj Hayabusa Over a year ago

thanks . let me try the code first :) . i'll let you know

php-india · Accepted Answer · 2016-10-18 05:23:23Z

0

Use Following, check if user id and database id are same so that will add extra HTML in $login_user else will be empty. And echo $login_user in your View.

<?php

    $get_post = "SELECT * FROM posts";
    $run_post = mysqli_query($connect, $get_post);
        if($run_post && mysqli_num_rows($run_post) > 0 )
        {
            while($row_post = mysqli_fetch_array($run_post))
            {
                $post_title      = $row_post['post_title'];

                $get_user   =  "SELECT * FROM users WHERE user_id='$user_id'";
                $run_user   = mysqli_query($connect, $get_user);
                $check_user = mysqli_fetch_array($run_user);

                    $user_name  = $check_user['name'];
                    $user_images = $check_user['images'];   

                    $login_user = ($user_id == $row_post['user_id'])? "<a href=''><button>Delete</button></a>
                                    <a href=''><button>Edit</button></a>": "";
                        echo "<div id='post_wrap'>
                                    <p>$post_title</p>
                                    <a href=''><button>Like</button></a>
                                    <a href=''><button>View</button></a>".
                                $login_user . "
                                </div>";
            }
            mysqli_free_result($run_post);          
        }
    else
    {
        echo "No post yet";
    }
?>

answered Oct 18, 2016 at 5:23

php-india

366 bronze badges

2 Comments

Rzj Hayabusa Over a year ago

thanks for the reply :) . but i use 'email' field column in database as $session . but thanks. i'll try your suggestion first .. see if its work or not .

Rzj Hayabusa Over a year ago

hye . thanks for the response . :) . it's ok . i already found the solutions . stackoverflow.com/questions/40120634/…

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