I am building a custom wallboard system for a restaurant chain in my area. After I got the main system up and running, I am now trying to redesign the system in order to minimize called to the database, and re-structure the data handling to be a bit more efficient.
With that being said, I was able to setup the new database and get it to automatically fill the data I would like. I was also able to pull the data that I need, 12 different rows to be exact. But now I am having a problem with sorting the data to the correct variables to be displayed later in the script.
This is what code I have for this operation so far:
$queryRecords = $conn->query("SELECT * FROM records WHERE store_id = '".$storeId."' order by input_date desc limit 12;") or die();
while ($row = mysqli_fetch_assoc($queryRecords)) {
$date.$row["record_id"] = $row["date"];
$team.$row["record_id"] = $row["team"];
$record.$row["record_id"] = $row["record"];
print($record.$row["record_id"]);
But when I run this code I get a 500 error in the console, then looking into the error logs, this is what I get:
[Wed Jul 17 18:24:56.068254 2019] [php7:emerg] [pid 1377] [client 73.95.135.142:59969] PHP Parse error: syntax error, unexpected â;â in /var/www/html/t.php on line 21
Any help would be greatly appreciated, I am fairly new to php especially when it comes to arrays.
Thank you in advance!
would have the ârecord_idâ of âmdaâ, so after getting all of the rows from the database, it sets them to variables based on this ID to be used later in the script. But in all actuality even though they are technically different data types, they fit into the same criteria which is why they are in the same table.
) but when I try to just print the ârecord_idâ it prints the value for ârecordâ any thoughts as to why this might be happening? I think it might be with the way that I am assigning variables, since I have no idea how