Passing variable from MySql

Radiot88 · December 21, 2018, 2:09am

This script broke with php7 having to do with global variables. The gist is: if there is an ID number, query and give the details. If there is no ID number, query and print the appropriate list. My script prints the list, and produces the correct URL, but never runs the top half of the script.

How does one get the variable from the PHP_SELF at the bottom to query at the top of the page .

<?

if ($idr) 
	{
		$photo_result = mysqli_query($db,"SELECT id, photos FROM TABLE WHERE id=$idr");
	}
	
 else
	{
		$result = mysqli_query($db,"
			if($r = mysqli_fetch_array($result))
				{do
			  		{
					printf("<a href=\"%s?idr=%s\">%s</a><br>\n", $PHP_SELF, $r["ID"],  $r["ID"]);
			   		}
				while ($r = mysqli_fetch_array($result));	
				}
			else 
			{
				echo "Sorry, no dice! ";
		  	}
	}
?>

Thanks in advance for any suggestions.
Sincerely
R

astonecipher · December 21, 2018, 2:31am

Looks like you have an issue with quotes.

As a general rule however, programming works in the following series,
INPUT
PROCESSING
OUTPUT

Always.

Radiot88 · December 21, 2018, 3:02am

Thanks Astonecipher,
That makes sense, so I how to I get the input into the top? Drop the if / else and pass the variable down the script?

The script generates a url like: domain.com/index.php?idr=1
Which used to produce all the details on 1.

Thanks again

astonecipher · December 21, 2018, 4:26pm

Where is that variable? I don’t see it being set anywhere in the posted code.

Radiot88 · December 21, 2018, 10:04pm

HI AStoneCipher,
Thanks again. After sleeping on it, I figured it out:

$idr=$_GET[‘idr’];

That pulled the ID from the URL and processed it through the top part of the script.

Thanks for trying.
Best,
R