parameter count for mysql_fetch_row()

JAPH

what is the right parameter count for mysql_fetch_row() and mysql_num_rows()?

ir4z0r

what about php.net?

each time 1 p. ($result, the one returned by mysql_query())

btw it's ...row() without 's'

JAPH

thanks, but any clue why a bit of code like this:

$result = mysql_query("SELECT pass from ".$user_prefix."_users where uid='$uid'", $dbi);
list($pass) = mysql_fetch_row($result, $dbi);

isn't functioning????

ir4z0r

Originally posted by ir4z0r
1 p. ($result, the one returned by mysql_query())

well, I count 2 parameters in your mysql_fetch_row(), do you count 1?

JAPH

ok, it was an idiotic mistake on my part....but wouldn't something like $result be an invalid mysql resource with out the $dbi(conects to database) variable?

keith73

Originally posted by JAPH
ok, it was an idiotic mistake on my part....but wouldn't something like $result be an invalid mysql resource with out the $dbi(conects to database) variable?

not really. In the absence of a link identifier, it will assume the last open link. Most mysql commands in php make the link resource optional. most, but NOT all. Read the manual at php.net to find out which ones require it and which ones don't.

Of course, if you happen to have 2 link resource identifiers, then you would need to specify.

As for your code snippet:
print out the # of rows being returned before mysql_fetch_row and see how many results you get.

print "Rows: " . mysql_num_rows($result) . "<br>\n";

Ideally, it should be just 1, but is uid a unique # in your table?
Where is $uid being set?

if the # of rows is 0, then it found no matches, if # of rows is > 1 then uid is not unique but it should be.

keith

ir4z0r

mysql_fetch_result() doesn't need the link identifier because, the
query has already been processed and the result is now located
in memory (afaik).
hence $result is only kind of a pointer which points to the resultsets
in memory --> no need of a db link anymore

JAPH

right, well, that's a rather moot point since $result is being returned as an "invalid mysql result resource"??????

keith73

Originally posted by JAPH
right, well, that's a rather moot point since $result is being returned as an "invalid mysql result resource"??????

which means that the query isn't returning results.

keith

JAPH

which is more correct syntax: to use the code snippit seen above, or to use something like:

$query = "SELECT pass from ".$user_prefix."_users WHERE uid='$uid'", $dbi);
$result = mysql_query($query, $dbi);
$pass = mysql_fetch_row($result);

keith73

Originally posted by JAPH
which is more correct syntax: to use the code snippit seen above, or to use something like:

$query = "SELECT pass from ".$user_prefix."_users WHERE uid='$uid'", $dbi);
$result = mysql_query($query, $dbi);
$pass = mysql_fetch_row($result);

there is no such thin as "more correct syntax" it's either correct or incorrect.

in the code here, your first line is wrong. should be:

$query = "SELECT pass from ".$user_prefix."_users WHERE uid='$uid'";
$result = mysql_query($query, $dbi);
print "ROWS: " . mysql_num_rows($result) . "<br>";
$pass = mysql_fetch_row($result);

if the rows being returned is not 1, then your mysql_fetch_row will only return the first result. If it is 0 or gives an error, then your query isn't working.

your first code snippet is also incorrect, but toss that out anyway and use what you posted here.

keith

JAPH

thank you, the code works perfectly now.