Reputation: 318
Is there a way to check if a value exists in a mysql column? I have table songs, and there are some columns, one of them is called 'agent_ip' where i will put a list/array of all user ip's that will visit the site. I need to check if current user ip is present in column 'agent_ip'. Here is some of my code:
public function voteSong($song_id, $case, $agent_ip) {
$query = $this->link->prepare("SELECT * FROM songs WHERE id = ? LIMIT 1");
$query->bindValue(1, $song_id);
$query->execute();
$rowcount = $query->rowCount();
if ($rowcount != 0)
{
if (!in_array($agent_ip, $r['ip']))
{
if ($case === 'like')
{
while($r = $query->fetch())
{
$vote = $r['votes'] + 1;
}
}
elseif ($case === 'dislike')
{
while ($r = $query->fetch())
{
if ($r['votes'] > 0)
{
$vote = $r['votes'] - 1;
}
else
{
$vote = 0;
}
}
}
$query = $this->link->prepare("UPDATE songs SET datetime = ?, votes = ?, agent_ip = ? WHERE id = ?");
$query->execute(array(date("Y-m-d H:i:s"), $vote, $agent_ip, $song_id));
}
}
}
The line if(!in_array($agent_ip, $r['ip']))
contains the wrong function which won't work, but i need an alternative for mysql. $r['ip'] variable is data from the 'agent_ip' column which look like this 127.0.0.1, 127.0.0.1, 127.0.0.1 (using 127.0.0.1 just for example, every 127.0.0.1 is a different ip)
Upvotes: 2
Views: 3130
Reputation: 226
First you need to deserialize/decode the data from the column to the proper php array
and then you can use in_array
function. In your post edit you stated that you have a comma separated list of IP's, so to convert it to array you need to use an explode
function:
$ip_list = explode(', ', $r['ip']);
now you can use the in_array
function on the new array:
if(!in_array($agent_ip, $ip_list))
Upvotes: 1
Reputation: 4079
If you're only checking against a single IP, why don't you just modify your query from:
"SELECT * FROM songs WHERE id = ? LIMIT 1"
To:
"SELECT * FROM songs WHERE id = ? AND agent_ip = ? LIMIT 1"
It seems a bit wasteful to query your whole result set when you are only querying against a specific IP and returning a single row.
EDIT: Your current method would be extremely inefficient, you are passing a unique agent_ip
each time you want to query a song to check if the IP exists, that would be fine, but you are creating a new DB connection every time from which you pull back all info which belongs to that song.
Lets say we have 1 song, and 3IP's, currently the application would work like this:
1) Call the method, passing IP_1
2) Query the database getting all songs for ID1
3) Check if IP_1 is in the result set and do process
4) Call the method, passing IP_2
5) Query the database getting all songs for ID1
6) Check if IP_2 is in the result set and do process
7) Call the method, passing IP_3
8) Query the database getting all songs for ID1
9) Check if IP_2 is in the result set and do process
As you can see, there is a lot of repetition here which is going to hinder your apps performance as it scales, you would be so much better modifying your current function to accept a list of results for a song which is pre-queried only once and then recursively call a check function by passing that result array with your unique IP address.
UPDATE You stated I understand that i need to have 2 tables(1 = songs; 2 = votes). But i cannot imagine how i will get songs from database, arranged by votes quantity.
You should read SQL's JOIN
documentation, the concept is simple - JOIN allows you to pull back a more detailed set of information based on what you want to query, in your example you may want to find out how many votes a specific song has.
Your tables may look like:
Songs
SONG_ID Primary Key
SONG_TITLE
SONG_DURATION
SONG_TAGS
Votes
VOTE_ID Primary Key
SONG_ID Foreign Key - (references the song_id table)
VOTE_RES Bool (either 0 for no, 1 for yes)
AGENT_IP Who sent the vote
You could then find out how many people said they liked the song by performing a join:
SELECT * FROM songs
JOIN votes
ON songs.song_id = votes.song_id
WHERE songs.song_id = 1
AND votes.vote_res = 1;
This would return all the song with the id of 1 and all of its associated likes. Hope that helps a bit :)
Upvotes: 2