RE: [PHP-DB] mssql_fetch_array problem

[Michael Rudel]

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 "<option value=" . $myrow["countryid"] . " selected>"
       . $myrow["countryname"]
       . "</option>"
       . "\n";
}
else
{
   print "<option value=" . $myrow["countryid"] . ">"
       . $myrow["countryname"]
       . "</option>"
       . "\n";
}


Hope this helps. If not, plz post what is going wrong.

Greetinx,
  Mike

Michael Rudel
- Web-Development, Systemadministration -
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> -----Original Message-----
> From: Jennifer Arcino Demeterio [mailto:[EMAIL PROTECTED]]
> Sent: Tuesday, June 19, 2001 6:18 AM
> To: [EMAIL PROTECTED]
> Subject: [PHP-DB] mssql_fetch_array problem
>
>
> hello all,
>
> hope someone could help me ....
>
> it seems that the mssql_fetch_array did not work as it should
> when the data
> type to be fetch is an integer ... is there any work around for this?
>
> Below is my script ...
>
> $country_sql = "select countryname, countryid from TBL_COUNTRY";
>     if ($country_query = mssql_query($country_sql)) {
>        while ($myrow = mssql_fetch_array($country_query))
>           if ($country == $myrow["countryid"]) {
>                print "<option value=" . $myrow["countryid"] .
> " selected>" .
> $countryid . "</option>";
>          } else {
>                print "<option value=" . $myrow["countryid"] . ">" .
> $countryid . "</option>";
>          }
>       }
>     }
>
>
> Thanks!
>
>
>
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