[rottedham]

Hi,

I'm working on a download script and I was wondering if anyone could tell me how to store the amount of times a file has been downloaded to a database. I have an understanding of how to display it, I just don't know what to do to add up the hits. If anyone can help me, I'd greatly appreciate it.

Thanks.

[Klesti]

Originally Posted by rottedham

Hi,

I'm working on a download script and I was wondering if anyone could tell me how to store the amount of times a file has been downloaded to a database. I have an understanding of how to display it, I just don't know what to do to add up the hits. If anyone can help me, I'd greatly appreciate it.

Thanks.

Yor mysql table have to contain a field named hits
for example :
id
url
author
hits

in a file called download.php you will executate 2 mysql queries. i supposed that your mysql table is called downloads

PHP Code:

$sql = "SELECT * from downloads WHERE id=$id"; //Here $id is the id taken from the url f.e download.php?id=23

$result = mysql_query($sql);

$row = mysql_fetch_assoc($result);

$hits = $row['hits'];  //here is the number of hits 

//now increase the number of hits 

$hits = $hits + 1;

//lets upadate it

$sql = "UPADATE downloads SET hits="$hits" WHERE id=$id"; 

mysql_query($sql); 

[rottedham]

thank you SOO much. I've been trying to figure this out all night, just messing around with the code and you've told me exactly what i needed to know.

[rottedham]

I am having an issue running this script. I am getting a warning saying "mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource."

[dgm]

Or, you could do it in one query.

PHP Code:

$sql = "UPDATE downloads SET hits = hits+1 where id=$id"; 

That would be much more efficient.

[Klesti]

Maybe you didn't pass the $id variable.

for understanding if the result is taken :

PHP Code:

 $sql = "SELECT * from downloads WHERE id=$id"; //Here $id is the id taken from the url f.e download.php?id=23 

if (!$result = mysql_query($sql)) {



echo "Error: " . mysql_error();

}



$row = mysql_fetch_assoc($result); 

$hits = $row['hits'];  //here is the number of hits 

//now increase the number of hits 

$hits = $hits + 1; 

//lets upadate it 

$sql = "UPADATE downloads SET hits="$hits" WHERE id=$id"; 

if (!mysql_query($sql)) {

echo "error" . mysql_error();

} 

You should be secured that the variable $id is passed to the script because the sql query won't work without it and will show you that error