Hi,
My MAC users (2 so far, unknown OS versions) complain that my code
always fails here saying
From: les
Subject: Bugs
Message: Hi John, I couldn't get it to work! It didn't seem to
recognize my numbers as numbers. Is this because I am on a Mac (gasp)?
Les
Other users do not have this problem.
Any idea why this function would fail?
First we get the data:
$doag=$_GET['DOAG']; check100($doag, "doag");
Here is the function call. It should check for a valid number, then,
if it isn't valid, find the agency the user submitted and tell them
they need to send a number.
function check100($x, $ag) {
if (!is_numeric($x )){
mysql_connect(" localhost","dbn ame", "passwd");
mysql_select_db ("dbname");
$query = "select agency from agency where agency_id = '$ag';";
$result = mysql_query($qu ery) or die("Error, agency query
failed");
while ($row = mysql_fetch_arr ay($result))
{ $ag = $row['agency']; }
$e_code[6]=$e_code[6] + 1;
print "The value in $ag is not a number, please try again.";
}
=============== =============== =============== =============== ==
My server runs
Apache/1.3.37 Ben-SSL/1.57 (Unix) mod_gzip/1.3.26.1a mod_fastcgi/2.4.2
mod_throttle/3.1.2 Chili!Soft-ASP/3.6.2 FrontPage/5.0.2.2635 mod_perl/
1.29 PHP/4.4.6
Operating System
Name: Linux
Vendor: Debian
Version: CustomLinux
PHP Version: 4.4.6
MySQL Version: 5.0.30
My MAC users (2 so far, unknown OS versions) complain that my code
always fails here saying
From: les
Subject: Bugs
Message: Hi John, I couldn't get it to work! It didn't seem to
recognize my numbers as numbers. Is this because I am on a Mac (gasp)?
Les
Other users do not have this problem.
Any idea why this function would fail?
First we get the data:
$doag=$_GET['DOAG']; check100($doag, "doag");
Here is the function call. It should check for a valid number, then,
if it isn't valid, find the agency the user submitted and tell them
they need to send a number.
function check100($x, $ag) {
if (!is_numeric($x )){
mysql_connect(" localhost","dbn ame", "passwd");
mysql_select_db ("dbname");
$query = "select agency from agency where agency_id = '$ag';";
$result = mysql_query($qu ery) or die("Error, agency query
failed");
while ($row = mysql_fetch_arr ay($result))
{ $ag = $row['agency']; }
$e_code[6]=$e_code[6] + 1;
print "The value in $ag is not a number, please try again.";
}
=============== =============== =============== =============== ==
My server runs
Apache/1.3.37 Ben-SSL/1.57 (Unix) mod_gzip/1.3.26.1a mod_fastcgi/2.4.2
mod_throttle/3.1.2 Chili!Soft-ASP/3.6.2 FrontPage/5.0.2.2635 mod_perl/
1.29 PHP/4.4.6
Operating System
Name: Linux
Vendor: Debian
Version: CustomLinux
PHP Version: 4.4.6
MySQL Version: 5.0.30