Old Question

[Anon]

Now in my PHP/MYSQL script:

$result = mysql_query("select distinct customer.ID,customer.countryID,Links".
"from Database where $cond order by customer.ID,customer.countryID desc limit $offset,$limit" );

customer.ID can be from 1 - 10;
customer.countryID can be from 1 - 10;

How can I list the "select" result "IF" customer.countryID=5 at the top first?

I have tried

$result = mysql_query("select distinct customer.ID,customer.countryID,Links".
"from Database where $cond order by IF(customer.countryID='5','1','0'), customer.ID,customer.countryID desc limit $offset,$limit" );

but failed.... any friend help?

[Anon]

Hi Mike

I don't think this can be done with an SQL statement as the "order by field in('5','1','0',*)" doesn't work on MySQL.

I think the only way is during the readback loop to analyse the field number and act accordingly, a bit like taking out recurring data when doing reports.

eg:

while ($row = mysql_fetch_array($result)) {
if (eregi($row["countryID"],"5,1,0")) {
... do something
} else {
... do something else
}
}

Hope this helps

James

[Anon]

Hi James,

Thanks for your help.

By Refering:

http://www.phpbuilder.com/forum/read.php3?num=5&id=1725&thread=1712

ORDER BY IF(email='','0','1');

I have modified my part and tried:

$result = mysql_query("select distinct customer.ID,customer.countryID,Links".
"from Database where $cond order by IF(customer.countryID='5','1','0'), customer.ID,customer.countryID desc limit $offset,$limit" );

But it return the error $result is zero result.

Any idea?

[Anon]

Any body help?