[SOLVED] WHERE LIKE problem.

[Verbat]

This is the code i have -

 

$query8 = sprintf("SELECT * from test WHERE username LIKE '%a%'",$db_connect); 
$result8 = dbquery($query8);
while ($rpw = mysql_fetch_assoc($result8) ) {
$username = $rpw['username'];
echo "$username";
}

 

I tryed to work it with some VAR info but it didnt work so i made a simple search code like this and for some reason it doesnt work.

What am i doing wrong?

[trq]

What is your expected results?

[rajivgonsalves]

well what is dbquery did you write that function... for a matter of simplicity it should of been

 

$query8 = "SELECT * from test WHERE username LIKE '%a%'"; 
$result8 = mysql_query($query8);
while ($rpw = mysql_fetch_assoc($result8) ) {
$username = $rpw['username'];
echo "$username";
}

[Verbat]

rajivgonsalves - I tryed with mysql_query aswell and still it gives and error

 

 

thorpe - all usernames with a inside. Just a simple search i made to see if it's working.

[rajivgonsalves]

this

 

$result8 = mysql_query($query8);

 

change it too

 

$result8 = mysql_query($query8) or die("Cannot execute:".mysql_error());

 

and tell me if you get any errors

[Verbat]

This is the error i get -

 

Warning: sprintf() [function.sprintf]: Too few arguments in test.php on line 27

Cannot execute:Query was empty

[trq]

As rajivgonsalves pointed out, you do not need sprintf at all. Read the replies to your questions.

[Verbat]

As rajivgonsalves pointed out, you do not need sprintf at all. Read the replies to your questions.

 

Ohhh, i missed that part.

Sorry.

 

Thanks alot for the help.