PHP check if MySQL table contains value - php

Question

Okay so the title may be a bit misleading. What I am trying to do is add a favorite system to my site. I have one column for my favorite things and I set it up so after each item ID there is a :. How can I check the string returned from my database (1345:13456:232:524378:324) if it contains 232? If it does I would echo preRend else I would echo insert and insert that ID followed by a :. This is what I have so far:
<?php
session_start();
require_once(".conf.php");
$logged = $_SESSION['logged'];
$user = $_SESSION['user'];
$fwdfav = $_POST['id'];
$query = mysql_query("SELECT * FROM accountController WHERE user='$user'");
if ($logged == 1)
{
while ($row = mysql_fetch_array($query)) {
if ($row['fav-itms'] //This is where I got stuck. How to check if it contains a value.)
{
mysql_query("INSERT INTO accountController ('fav-itms') VALUES ('$fwdfav')");
echo 'inserted';
}
else
{
echo 'preRend';
}
}
}
else
{
echo 'nlog';
}
?>
Thank you so much! I am sure there are a lot of errors here as I am very tired.

Answer 1

The approach you are taking is extremely inefficient and does not take advantage of the fact that you are using a database.
(Btw... I hope this is just example code; you have a giant SQL injection vulnerability in your INSERT query.)
What I would do instead is create a second table that would look something like:
favorites (
id int(11) NOT NULL auto_increment,
user_id int(11),
fav_id int(11)
)
And have each row represent a user-favorite pair. Then you can let MySQL do the heavy lifting of figuring out whether a user has favorited something, e.g.,
SELECT COUNT(*) FROM favorites WHERE user_id = %d AND fav_id = %d;
// Substitute the actual look-up values in using prepared statements
You could also similarly quickly get the actual favorites for a user, etc.
Remember, a database is designed for the explicit purpose of storing and looking up information quickly. PHP is a general-purpose programming language. Where possible, let MySQL do the walking for you.
(This advice is general for a moderately scaled setup. If you need to handle millions of simultaneous users, far more optimization is obviously required, and conventional relational databases might not even be suitable. But I don't get the impression that's where you're at right now.)

Answer 2

You could explode it in array as check, like:
$yourArr = explode(":", $row['fav-itms']);
$checkFor = 232;
if(in_array($checkFor, $yourArr)) {
//it exists
}
else {
//does not exist
}
Did you mean something like this

Answer 3

I know this was posted a while ago but it came up when I did a search.
I have a database storing information for my portfolio, it holds locations for images.
I am working on a page to display the full view of the project. Within the page I need it to check the columns for the images and if any are empty I need it to not display anything.
This is how I've done it.
// connect, select database, query table relevant to page. I have done a query for a specific row.
if($row[columnName1]){
echo '<div> displaying value </div>';
}
if($row[columnName2]){
echo '<div> displaying value </div>';
}
what is happening, if columnName1 in selected row has a value display the value in div else nothing. then on to column 2.
if it is done like this
if(!$row[columnName1]){
//content displayed
}
and the column does not contain a value then what is in between the {} will be ran.
Works the way I needed it to, maybe this will help someone.