Jennifer,
Try this code..I am using the same stuff as you with the exception of mysql instead of
sql..but it may still work:
";
}
}
?>
It's kind of a "save as" type of statment but it works for me! I've had no problems
with that sql statement.
Angie
>>> "Jennifer Arcino Demeterio"
Ok lets start from the beginning, apologies if this these have been
mentioned previousely I haven't seent the rest of the thread:
* Make sure 'countryid' is spelled the same in your script as it is in
your DB table (I've done this and spent hours banging my head against
a brick wall)
* En
Thanks for all your answers, but, they still seem not to work. In my query,
the value of countryname is retrieved as expected but the value of countryid
is not. There seems to be no value retrieved for countryid. I am using
PHP4 on IIS4 on an NT machine with SQL Server 7 as the database.
"coun
Hello!!!
Try using the second argument of mysql _fetch_array
MYSQL_NUM
This will insure that an integer is used..
If not let me knowI'll do testing...
Dan Brunner
On Monday, June 18, 2001, at 11:18 PM, Jennifer Arcino Demeterio wrote:
> hello all,
>
> hope someone could h
try:
if ($country == intval($myrow["countryid"]))
mssql_fetch_array is found to have some weird behavior.
On Tue, 19 Jun 2001, Jennifer Arcino Demeterio wrote:
> hello all,
>
> hope someone could help me
>
> it seems that the mssql_fetch_array did not w
Not being able to see your code and not knowing exactly what problem you are
having I can only throw a few guesses here. If you could elaborate more on
what you are expecting and what you are getting that would be a big help.
$country_sql = "select countryname, countryid from TBL_COUNTRY";
Yo
Hi Jennifer,
I think that you want to display the Countryname in the DropDownBox, won't you ?
Try this one (in your while-loop):
if ($country == $myrow["countryid"])
{
print ""
. $myrow["countryname"]
. ""
. "\n";
}
else
{
print ""
. $myrow["countryname"]
