Script doesn't working

Question

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Hi. At the outset, let me say that just starting to program and sorry for my bad English. Anyway.
This code is used to like system. I don't know why this code doesn't working.

Code:

PHP

<?php
    $mysqli = new mysqli('localhost', 'root','');
    $mysqli->select_db('like');

    if(isset($_POST['liked']))
    {
        $postid = $_POST['postid'];
        $result = mysqli_query("SELECT * FROM posts WHERE postid=$postid");
        $row = mysqli_fetch_array($result);
        $n = $row['likes'];

        mysqli_query("UPDATE posts SET likes=$n+1 WHERE id=$postid ");
        mysqli_query("INSERT INTO likes(userid, postid) VALUES(1,$postid)");
        exit();
    }
?>

<!DOCTYPE html>
<html>
<head>
    <title>LIKE and UNLIKE</title>
    <style type="text/css">
        .content
        {
            width:50%;
            margin:100px auto;
            border:1px solid #cbcbcb;
        }
        .post
        {
            width:80%;
            margin:10px auto;
            border:1px solid #cbcbcb;
            padding:10px;
        }
    </style>
</head>
<body>
    <div class="content">
      <?php
         $query = $mysqli->query("SELECT * FROM posts");
    
        while($row = mysqli_fetch_array($query)) { ?>
        <div class="post">
            <?php echo $row['text']; ?> <br>

            <?php      
                $result = mysqli_query($mysqli,"SELECT * FROM likes WHERE userid='1' AND postid=".$row['id']."");
                
                if(mysqli_num_rows($result) == 1) { ?>
                
                    <span><a href="" class="unlike" id="<?php echo $row['id']; ?>">unlike</a></span>
          <?php } else{ ?>
                         <span><a href="" class="like" id="<?php echo $row['id']; ?>">like</a></span>
           <?php } ?>
        </div>
       <?php } ?>

    </div>
<script type="text/javascript" src="jquery.min.js"></script>
<script type="text/javascript">
        $(document).ready(function(){       
            $('.like').click(function(){
                var postid = $(this).attr('id');
                $.ajax({
                    url:'index.php',
                    type: 'post',
                    async: false,
                    data:{
                        'liked': 1,
                        'postid': postid
                    }
                    success:function(){

                    }
                });
            });
        });
</script>
</body>
</html>

Table sql:

likes.sql

SQL

CREATE TABLE `likes` (
  `id` int(11) NOT NULL,
  `userid` int(11) NOT NULL,
  `postid` int(11) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1

posts.sql

SQL

CREATE TABLE `posts` (
  `id` int(11) NOT NULL,
  `text` text NOT NULL,
  `likes` int(11) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1

users.sql

SQL

CREATE TABLE `users` (
  `id` int(11) NOT NULL,
  `name` varchar(55) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1;

What I have tried:

I tried change code:

PHP

<?php

    if(!isset($_POST['liked']))
    {
        $postid = $_POST['postid'];
        $result = mysqli_query($mysqli,"SELECT * FROM likes WHERE postid='$postid'");
        $row = mysqli_fetch_array($result) or die("ERROR: ".mysqli_error($mysqli));
        $n = $row['likes'];

        $result = mysqli_query($mysqli,"UPDATE posts SET likes=$n+1 WHERE id=$postid ");
        $result = mysqli_query($mysqli,"INSERT INTO likes (userid, postid) VALUES(1,$postid)");
        exit();
    }
    ?>

Posted 17-Mar-17 8:44am

Mike CJ

Updated 18-Mar-17 6:27am

Add a Solution

Comments

ZurdoDev 17-Mar-17 14:51pm

You have to explain what is not working. What should it do? What is it not doing? What error is there? Click Improve question and add information.

Patrice T 17-Mar-17 16:13pm

Define "doesn't working"

2 solutions

Solution 2

index.php

PHP

<?php
    $mysqli = new mysqli('localhost', 'root','');
    $mysqli->select_db('like');   
?>
<!DOCTYPE html>
<html>
<head>
    <title>LIKE and UNLIKE</title>
    <style type="text/css">
        .content
        {
            width:50%;
            margin:100px auto;
            border:1px solid #cbcbcb;
        }
        .post
        {
            width:80%;
            margin:10px auto;
            border:1px solid #cbcbcb;
            padding:10px;
        }
    </style>
</head>
<body>
    <div class="content">
      <?php
         $query = $mysqli->query("SELECT * FROM posts");
    
        while($row = mysqli_fetch_array($query)) { ?>
        <div class="post">
            <?php echo $row['text'] . " " . $row['likes']; ?> <br>

            <?php      
                $result = mysqli_query($mysqli,"SELECT * FROM likes WHERE userid='1' AND postid=".$row['id']."");
                
                if(mysqli_num_rows($result) == 1) { ?>
                
                    <span><a href="" class="unlike" id="<?php echo $row['id']; ?>">unlike</a></span>
          <?php } else { ?>
          
                         <span><a href="" class="like" id="<?php echo $row['id']; ?>">like</a></span>
           <?php } ?>

        </div>
       <?php } ?>

    </div>
<script type="text/javascript" src="jquery.min.js"></script>
<script type="text/javascript">
        $(document).ready(function(){       
            $('.like').click(function(){
                var postid = $(this).attr('id');
                $.ajax({
                    url:'index_post.php',
                    type: 'post',
                    async: false,
                    data:{
                        'liked': 1,
                        'postid': postid
                    },
                    success:function(){
                    }
                });
            });
        });
        </script>
        
        <script type="text/javascript">
        $(document).ready(function(){                         
        $('.unlike').click(function(){
        var postid = $(this).attr('id');
        $.ajax({
        url:'index_unpost.php',
        type: 'post',
        async: false,
        data:{
        'unliked': 1,
        'postid2': postid
        },
        success:function(){

         }
      });
   });
});
</script>
</body>
</html>

index_post.php

PHP

<?php
    $mysqli = new mysqli('localhost', 'root','');
    $mysqli->select_db('like');

     $postid = $_POST['postid'];
     $result = mysqli_query($mysqli, "SELECT * FROM posts WHERE id=$postid");
     while($row2 = $result->fetch_array())
	 {
		$n = $row2['likes'];
		
		mysqli_query($mysqli, "UPDATE posts SET likes=$n+1 WHERE id=$postid");
	    mysqli_query($mysqli, "INSERT INTO likes(userid, postid) VALUES(1,$postid)");
	  }	
        exit();
?>

index_unpost.php

PHP

<?php
    $mysqli = new mysqli('localhost', 'root','');
    $mysqli->select_db('like');

     $postid = $_POST['postid2'];
     $result = mysqli_query($mysqli, "SELECT * FROM `posts` WHERE id=$postid");
	while($row3 = $result->fetch_array())
	  {
		$n = $row3['likes'];

        
		mysqli_query($mysqli, "UPDATE `posts` SET likes=$n-1 WHERE postid=$postid AND userid=1");
        mysqli_query($mysqli, "DELETE FROM `likes` WHERE id='$postid' userid=1 ");
	  }	
        exit();
?>

Posted 18-Mar-17 6:27am

Mike CJ

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Bryian Tan · Accepted Answer · 2017-03-17T18:25:00

ok, assuming the connection to MySQL database is working.
Here are my recommendations:

1. Syntax error on the AJAX call, need to have a comma (,) after the data
2. Move the update/insert into a separate script
3. There is no postid in table post, should be id
4. the mysqli_query, you did it correctly in the index.php by including the connection string in the function. but not sure why you wouldn't do the same in the post/update
5. there are lots of other improvement you need to make such as, prevent SQL injection or XSS, place the connectionstring into a separate file, etc...

Here are the updated code:

PHP

<?php
    $mysqli = new mysqli('localhost', 'root','');
    $mysqli->select_db('like');   
?>
<!DOCTYPE html>
<html>
<head>
    <title>LIKE and UNLIKE</title>
    <style type="text/css">
        .content
        {
            width:50%;
            margin:100px auto;
            border:1px solid #cbcbcb;
        }
        .post
        {
            width:80%;
            margin:10px auto;
            border:1px solid #cbcbcb;
            padding:10px;
        }
    </style>
</head>
<body>
    <div class="content">
      <?php
         $query = $mysqli->query("SELECT * FROM posts");
    
        while($row = mysqli_fetch_array($query)) { ?>
        <div class="post">
            <?php echo $row['text'] . " " . $row['likes']; ?> <br>

            <?php      
                $result = mysqli_query($mysqli,"SELECT * FROM likes WHERE userid='1' AND postid=".$row['id']."");
                
                if(mysqli_num_rows($result) == 1) { ?>
                
                    <span><a href="" class="unlike" id="<?php echo $row['id']; ?>">unlike</a></span>
          <?php } else{ ?>
                         <span><a href="" class="like" id="<?php echo $row['id']; ?>">like</a></span>
           <?php } ?>
        </div>
       <?php } ?>

    </div>
<script type="text/javascript" src="jquery.min.js"></script>
<script type="text/javascript">
        $(document).ready(function(){       
            $('.like').click(function(){
                var postid = $(this).attr('id');
                $.ajax({
                    url:'index_post.php',
                    type: 'post',
                    async: false,
                    data:{
                        'liked': 1,
                        'postid': postid
                    },
                    success:function(){
                    }
                });
            });
        });
</script>
</body>
</html>