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corillo181
Forum Commoner
Posts: 76 Joined: Wed Apr 26, 2006 3:02 pm
Post
by corillo181 » Thu Dec 28, 2006 7:45 pm
this code work find to update a record that already exist, but the insert query doesn't work at all..
Code: Select all
<?php
$a_t_id=$artist['artist_id'];
$getcounter=mysql_query("SELECT counter FROM tra_music_top10 WHERE artist_id='$a_t_id'")or die(mysql_error());
$count=mysql_fetch_array($getcounter);
if(!$count['counter']){
echo 0;
}else{
$i=$count['counter'];
echo $i;
}
if($_POST['Submit']){
$voted=($i+1);
$update=mysql_query("UPDATE tra_music_top10 SET counter='$voted' WHERE artist_id='$a_t_id'")or die(mysql_error());
if(!$update){
$newrecord=mysql_query("INSERT into tra_music_top10(artist_id,counter)VALUES('$a_t_id','1')")or die(mysql_error());
}
}
?>
hawleyjr
BeerMod
Posts: 2170 Joined: Tue Jan 13, 2004 4:58 pm
Location: Jax FL & Spokane WA USA
Post
by hawleyjr » Thu Dec 28, 2006 8:58 pm
Your syntax looks okay, what error are you getting?
Edit:
On second thought...look into
mysql_num_rows()
corillo181
Forum Commoner
Posts: 76 Joined: Wed Apr 26, 2006 3:02 pm
Post
by corillo181 » Thu Dec 28, 2006 11:00 pm
i got it to wokr by putting a 0 when a artist name is included, so every artist got a vote of 0 so when someone vote for the artist it adds 0+1=1 wich works fine. thankx for the help.
timvw
DevNet Master
Posts: 4897 Joined: Mon Jan 19, 2004 11:11 pm
Location: Leuven, Belgium
Post
by timvw » Fri Dec 29, 2006 4:58 am
Imho, it would be a lot cleaner if you simply used
INSERT ON DUPLICATE KEY .
corillo181
Forum Commoner
Posts: 76 Joined: Wed Apr 26, 2006 3:02 pm
Post
by corillo181 » Fri Dec 29, 2006 1:02 pm
i tried this but now everythings update at the same time..
Code: Select all
<?php
$a_t_id=$artist['artist_id'];
$getcounter=mysql_query("SELECT counter FROM tra_music_top10 WHERE artist_id='$a_t_id'")or die(mysql_error());
$count=mysql_fetch_array($getcounter);
$i=$count['counter'];
echo $i;
if($_GET['vote']){
$voted=($i+1);
$update=mysql_query("UPDATE tra_music_top10 SET counter='$voted' WHERE artist_id='$a_t_id'")or die(mysql_error());
}
?>