[PHP] PHP/MySQL Code
This code won't work. I'm trying to get an error if the username
is entered incorrectly but it won't go through the if brackets. I even
tried printing out $num and it's 0. I also tried
if ($num = "0") {
if ($num = '0') {
?php
include("connect.php");
$result=mysql_query("select * from table where
username='$username'",$connection) or die ("Can't do it");
$num = mysql_numrows($result);
if ($num = 0) {
print "htmlerror message etc.
exit;
}
Jeff Oien
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Re: [PHP] PHP/MySQL Code
"Jeff Oien" [EMAIL PROTECTED] wrote:
This code won't work. I'm trying to get an error if the username
is entered incorrectly but it won't go through the if brackets. I even
tried printing out $num and it's 0. I also tried
if ($num = "0") {
if ($num = '0') {
The single "=" sets the variable $num to that value. A double "==" is an
equality comparison operator - that's what you want in this case. Also,
since mysql_numrows() returns a number, not a string, you don't need quots
around the 0.
--
Steve Werby
President, Befriend Internet Services LLC
http://www.befriend.com/
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RE: [PHP] PHP/MySQL Code
The comparison operator is ==. = means "assign". Try:
if ($num == 0) {
foo(bar, baz);
}
AFAIK what your code means is "assign 0 to $num and then, if that worked,
print the error message."
HTH
Jon
-Original Message-
From: Jeff Oien [mailto:[EMAIL PROTECTED]]
Sent: 06 April 2001 17:51
To: PHP
Subject: [PHP] PHP/MySQL Code
This code won't work. I'm trying to get an error if the username
is entered incorrectly but it won't go through the if brackets. I even
tried printing out $num and it's 0. I also tried
if ($num = "0") {
if ($num = '0') {
?php
include("connect.php");
$result=mysql_query("select * from table where
username='$username'",$connection) or die ("Can't do it");
$num = mysql_numrows($result);
if ($num = 0) {
print "htmlerror message etc.
exit;
}
Jeff Oien
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Re: [PHP] PHP/MySQL Code
At 11:51 AM 4/6/01 -0500, Jeff Oien wrote:
if ($num = 0) {
print "htmlerror message etc.
exit;
}
This "conditional" will always run. you are assigning $num the value of
zero within your if() statement, and that assignation actually happens, so
the result is "true" and the braced statements run.
Try:
if ($num == 0) {
print "blah blah blah, etc.\n"
exit;
}
- Brian
-
Brian S. Dunworth
Sr. Software Development Engineer
Oracle Database Administrator
The Printing House, Ltd.
(850) 875-1500 x225
[EMAIL PROTECTED]
-
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RE: [PHP] PHP/MySQL Code
"Jeff Oien" [EMAIL PROTECTED] wrote:
This code won't work. I'm trying to get an error if the username
is entered incorrectly but it won't go through the if brackets. I even
tried printing out $num and it's 0. I also tried
if ($num = "0") {
if ($num = '0') {
The single "=" sets the variable $num to that value. A double "==" is an
equality comparison operator - that's what you want in this case. Also,
since mysql_numrows() returns a number, not a string, you don't need quots
around the 0.
--
Steve Werby
President, Befriend Internet Services LLC
http://www.befriend.com/
Ah, one of those repeated rookie mistakes. Thanks for helping me out.
Jeff Oien
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Re: [PHP] PHP/MySQL Code
I think the function should be mysql_num_rows. There's an mSQL function msql_numrows, but for MySQL it needs the second underscore in between 'num' and 'rows'. $num = mysql_numrows($result); -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP] PHP/MySQL Code
"long huynh" [EMAIL PROTECTED] wrote: I think the function should be mysql_num_rows. There's an mSQL function msql_numrows, but for MySQL it needs the second underscore in between 'num' and 'rows'. $num = mysql_numrows($result); Either will work. mysql_numrows() is the old name of the function from previous PHP versions. -- Steve Werby President, Befriend Internet Services LLC http://www.befriend.com/ -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
