Select top 5 results: PHP/SQL

Question

I'm trying to select the top 5 results, but whenever adding order by votes desc limit 5 to this code:

$req2 = mysql_query('select id, url, name, description, banner, votes 
                     from topsite 
                     order by votes desc 
                     limit '.$first_message .','.$last_message); 
while($dnn2 = mysql_fetch_array($req2))

it won't work. How can I fix it?

Please dont use the `mysql_` database extensions, it is deprecated (gone for ever in PHP7) Especially if you are just learning PHP, spend your energies learning the `PDO` or `mysqli_` database extensions, [and here is some help to decide which to use](http://stackoverflow.com/questions/12859942/why-shouldnt-i-use-mysql-functions-in-php) — RiggsFolly, Nov 11 '15 at 16:06
What are the $first_message and $last_message? To get only the top 5, just order by descending order and limit to 5. — Ron Dadon, Nov 11 '15 at 16:07
@m02ph3u5, Whenever I refresh the page after adding `order by votes desc limit 5` it will just show a blank page. No errors. — Mattxd, Nov 11 '15 at 16:19
Maybe you got the *LIMIT* wrong. It's `LIMIT , `. Look at this example from the docs: `SELECT * FROM tbl LIMIT 5,10; # Retrieve rows 6-15` https://dev.mysql.com/doc/refman/5.6/en/select.html - `$last_message` is most probably not what it should be — m02ph3u5, Nov 11 '15 at 16:22
@m02ph3u5 Yes it wrong. He only actually need `LIMIT 5` as all he wants is the TOP5 — RiggsFolly, Nov 11 '15 at 16:29

score 1 · Answer 1 · edited May 23 '17 at 11:51

If you only want the TOP 5 then you only need one limiter like this

Also if you use double quotes around your query you can expand $variables automatically, which makes it a lot easier to read and debug.

Also if you want to see the errors produced by the mysql_ extension you have to look for them

$top_5_please = 5;

$req2 = mysql_query("select id, url, name, description, banner, votes 
                     from topsite 
                     order by votes desc 
                     limit $top_5_please"); 

// error check
if ( ! $req2 ) {
    echo mysql_error();
    exit;
}

while($dnn2 = mysql_fetch_array($req2))

FINAL NOTE

Please dont use the mysql_ database extensions, it is deprecated (gone for ever in PHP7) Especially if you are just learning PHP, spend your energies learning the PDO or mysqli_ database extensions, and here is some help to decide which to use

score 0 · Answer 2 · answered Nov 11 '15 at 16:07

You can try the following. Also note that mysql_* functions are deprecated. Use PDO instead.

$req2 = mysql_query('select id, url, name, description, banner, votes 
                     from topsite 
                     order by votes desc 
                     limit '.$your_limit .' OFFSET '.$your_offset);

score 0 · Answer 3 · edited Nov 11 '15 at 16:09

0

$req2 = mysql_query('select id, url, name, description, banner, votes 
                     from topsite 
                     order by votes desc 
                     limit 5');

you can't pass string into limit you have to use integer only in limit

edited Nov 11 '15 at 16:09

RiggsFolly

93,638
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answered Nov 11 '15 at 16:08

soniabhishek36

59
5

if you are passing parameter then the parameter also should be integer type and not concat that as string. – soniabhishek36 Nov 11 '15 at 16:12
As the query is a string it does not actually matter about the type of the `$limit` variable as long as it is either an integer or a string that holds an integer i.e. `"5"` – RiggsFolly Nov 11 '15 at 16:16
My comment adressed the answer, not your comment. – m02ph3u5 Nov 11 '15 at 16:28
@m02ph3u5 Oh my apologise – RiggsFolly Nov 11 '15 at 16:30

Select top 5 results: PHP/SQL

3 Answers3