simple sql help

gnomer

Hi,

I have a table in a database which has 2 fields, userid and impressions

what I want to do is specify the userid and read the impressions for that userid

what query would i use? thanks

forbairt

In SQL you'd do this

select impressions from tablenamehere where userid=50;

???

gnomer

thank you!

i dont no anything about sql

forbairt

no problem

gnomer

also this would just show the impression number if I executed this in PHP?

forbairt

forbairt wrote: »

select impressions from tablenamehere where userid=50;

<?php
$userid = 2;
$query = "select impressions from tablenamehere where userid=$userid";
$res = mysql_query($query);
$row = mysql_fetch_row($res))
echo $row['impressions'];
?> 

from my head ... may / may not work

gnomer

more trouble

heres my PHP:


$query = "select impressions from unpaidhits where userid='1'";

$result = mysql_query($query);
echo $result;

?>
$result should be 1 but it is Resource id #3

what am I doing wrong?

forbairt

gnomer wrote: »

more trouble

heres my PHP:


$query = "select impressions from unpaidhits where userid='1'";

$result = mysql_query($query);
echo $result;

?>
$result should be 1 but it is Resource id #3

what am I doing wrong?

you've sent the query to mysql ... its sent back a resource ... which you need to perform a mysql_fetch_row on .. and that will in turn send back an associative array ... and you'll want to extract that like in my example

gnomer

just saw your post

your line 17 is wrong somewhere

you have 2 close brackets and one open bracket and no ; and when i changed your last closed bracket to ; it gives me :


Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in /home/***/public_html/test.php on line 17

gnomer

oops myfault i forgot it fill out the tablenamehere with my table name.

when I run your example im just getting a blank page

gnomer

anyone?

forbairt

er in my example change the userid to 1 ?

gnomer

actually ive sorted it all out now

for anyone wanting to know what I did

<?php
$userid='someid';
$q="select * from yourtable where userid='$userid'";
$r=mysql_query($q);
while ($row = mysql_fetch_array($r)){
$userid=$row/COLOR][COLOR=#FF0000]'userid'[/COLOR][COLOR=#66CC66;
$impressions=$row/COLOR][COLOR=#FF0000]'impressions'[/COLOR][COLOR=#66CC66;
echo "Userid: $userid\n";
echo "Impressions: $impressions";
}
?>

forbairt

[php]<?php
$userid='someid';
$q="select * from yourtable where userid='$userid'";
$r=mysql_query($q);
while ($row = mysql_fetch_array($r)){
$userid=$row;
$impressions=$row;
echo "Userid: $userid\n";
echo "Impressions: $impressions";
}
?>
[/php]

er wasn't that pretty much what I said to do ???

gnomer

lol i dont know im no sql expert someone on another forum gave me that and it seems to work perfectly.

louie

you could do without the "while". Also what's going to be displayed id no records are found?

there is a revised code
[php]
<?php
$impressions = "";
$userid='someid';
if($userid != ""){//execute query
$q="select * from yourtable where userid='$userid'";
$r=mysql_query($q);
$row = mysql_fetch_array($r);
$userid=$row;
$impressions=$row;
}
if($impressions != ""){
echo "Userid: $userid\n".
"Impressions: $impressions";
}else{
echo "Welcome";
}

?>
[/php]