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Posted by Robert Cummings on 07/20/05 05:03
On Tue, 2005-07-19 at 21:46, Jochem Maas wrote:
> Matt Darby wrote:
> > George B wrote:
> >
> >> Jay Blanchard wrote:
> >>
> >>> [snip]
> >>> $money -= 10;
> >>>
> >>> saves some chars ;)
> >>> [/snip]
> >>>
> >>> This actually requires two trips to the database, once to get $money and
> >>> once to update $money. Do it in the query instead...once.....
> >>>
> >>> $sqlUpdate = "UPDATE `myDatabase`.`myTable` SET `myMoney` =
> >>> (`myMoney`-10) WHERE `myCharacter` = `characterName` ";
> >>> $doUpdate = mysql_query($sqlUpdate, $myConnection);
> >>>
> >>> saves lots of characters ;)
> >>
> >>
> >> Hey, Look I made a new database just for this money testing stuff.
> >> This is the table:
> >>
> >> CREATE TABLE `money` (
> >> `money` varchar(255) NOT NULL default ''
> >> ) TYPE=MyISAM;
> >>
> >> Now, I use this code
> >>
> >> $sqlUpdate = "UPDATE `myDatabase`.`myTable` SET `myMoney` =
> >> (`myMoney`-10) WHERE `myCharacter` = `characterName` ";
> >> $doUpdate = mysql_query($sqlUpdate, $myConnection);
> >>
> >> And it should work, And it gives no error but it is not writing
> >> anything into the database.
> >>
> >
> > Lots of extra characters in that one... try this:
> >
> > $q=mysql_query("update myTable set myMoney=(myMoney-10) where
> > myCharacter='characterName'");
> > if(!$q){echo mysql_error();}
>
> so wasteful :-)
>
> $q=mysql_query("update u set m=(m-10) where c='$c'");if(!$q)die(mysql_error());
Ugh... ;)
if(!($q=mysql_query("update u set m=(m-10) where
c='$c'"))die(mysql_error());
Cheers,
Rob.
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