Hi Batalf,
So I would run that again with different names for Previous records too?
So, I'd run 3 queries in total?
Thanks heaps!
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Trying to get next record from MySQL db
Hi All,
This is probably extremely simple, but I cant work it out.
I have build a site for a client, Under the 'collections' link this displays all records from that 'collection' in thumbnail form, clicking one of them passes the id in the URL to display that product. Now, I have done $next = $id + 1; but of course, if the next record has been deleted, it wont return anything. How can I write a loop to keep checking until it finds the next record and pass the id of that record though the URL?
Thanks in Advance,
Christian
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Answers
Glad I could help!
I just thought of another option which also should work.
What if you have a link like this:
id=$id&next=true
i.e. send the current id and a variable called "next" to the current page. Then you could perform the query for next or previous record there, a query like this:
if(isset($_GET['next'])){
$sql = "select id from yourTable where id>'$id' order by id limit 1";
$res = mysql_query($sql) or die ("Error in query: " . mysql_error());
if($inf = mysql_fetch_array($res)){ // A next record is found
$id= $inf["id"]; // Set $id to the next ID
}
}
The same rules should also been applied for a variable called "previous".
The advantage of this solution is that you don't have to make 3 queries on each page.
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by: BatalfPosted on 2005-04-14 at 01:46:02ID: 13779658
You only need one query:
$sql = "select id from yourTable where id>'$id' order by id limit 1";
$res = mysql_query($sql) or die ("Error in query: " . mysql_error());
if($inf = mysql_fetch_array($res)){ // A next record is found
$next = $inf["id"];
}
You search for the first id which is bigger than the variable $id. This is the next id in the table.
"yourTable" is just an example name.
Batalf