coding help-easy one me thinks...

[Anon]

I have a page that lists countries from a drop down menu then assigns the selected value (two letter country code) to $country, this form then passes its variables to a new form for verification before submission. I want to keep the countrycode as the value (for that is how I want it stored in the dbase) but SHOW the actual country name in the table on the review page.

here is what I have:

<OPTION VALUE="<?php echo $country; ?>" SELECTED>

<?php

$db = mysql_connect("hostnamet","username","password");

mysql_select_db("databasename",$db);

$result = mysql_query("SELECT countryname FROM country WHERE countrycode=$country",$db);

echo $result;

?>
</OPTION>

countryname is the field I WANT, country is the table name, and countrycode is the VALUE attribute in the previous form.

yet, when I select a country and it throws me to the review page there is just a blank spot for the selected value in country. everything else seems to work fine with no errors.

suggestions, ideas.

I am probably doing something simple wrong, but can't figure it out.

thanks,
/frank

[Anon]

I work mostly with odbc, but the one thing that strikes me as missing is: how are you getting the result into a variable? Casting it into an array, i.e.,
$row = mysql_fetch_array(mysql_db_query( DATABASE, $arg ));

or looking at the result set and manipulating it, i.e.,
$row=mysql_fetch_row($result);

while ($row) {
do something...;
}

.....?

[Anon]

thanks, i wasn't...:-)

I changed the code to:

<?php

$db = mysql_connect("host","user","pass");

mysql_select_db("database",$db);

$result1 = mysql_query("SELECT * FROM country WHERE country=$country",$db);

while ($row1 = mysql_fetch_array($result1)) {

printf("%s\n", $row1[countryname]);

}
?>

and it worked fine!!!

thanks for the suggestion.

/frank