I record from an array?

rayge

I have a test script (very stripped down IE no Onsubmit for changing data ECT) here that SHows the countries in a dropdown and the currently selected one shows up in teh field.

What i need to do is take that same SELECTED variable name and PRINT it as text on the page somewhere as well.. I must have been working too hard on this stuff cause im having a brain fart here. I AM a newbie just so you know my level of understanding here.
Thanks for any help.

Code pasted below.

<?
include("config.php");

$objConn=mysql_connect($Host,$DBUser,$DBPass);
mysql_select_db($DBName2,$objConn);

$showcountry=${"showcountry"};
if ($showcountry=="")
{
$showcountry=224;
}
else
{
$showcountry=intval($showcountry);
}
?>

<input type="hidden" name="showcountry" value="<? echo ${"showcountry"}; ?>">
<Select name="country">
<?
$res_query=mysql_query(("Select * from country"),$objConn);
$res=mysql_fetch_array($res_query);

while(!($res==0))
{
$countryid=$res["id"];
$countryname=$res["countryname"];
?>

<option value="<? echo $countryid; ?>" <? if ($countryid==$showcountry)
{
?>selected<? } ?>><? echo $countryname; ?></option>
<?
$res=mysql_fetch_array($res_query);
}
?>
</select>
</form>
<?
mysql_close($objConn);
?>

pornost4r

first, i dont understand some things here:
- you wrote twice: $res=mysql_fetch_array($res_query);
- form is closed, but not opened
- what is this hidden input field?

u want to print out the selected value?

<?php

include("config.php");

if ($country)
{
echo "You choosed this $country";
}

else
{
$objConn=mysql_connect($Host,$DBUser,$DBPass);
mysql_select_db($DBName2,$objConn);

$res_query=mysql_query(("Select * from country"),$objConn);
$res=mysql_fetch_array($res_query);

echo "<form>\n";
echo "<select name=country>\n";

while(!($res==0))
{
$countryid=$res["id"];
$countryname=$res["countryname"];

if ($countryid==$showcountry) { $selected = "selected"; }

echo "<option value=\"$countryid\" $selected>
$countryname</option>\n";
}
echo "</select>\n";
echo "<input type=submit name=submit value=\"OK !\">\n";
echo "</form>";

mysql_close($objConn);

}

mike_y

you have to assign the country name to a temp. variable inside your if statement. Something like:

<option value="<? echo $countryid; ?>" <? if ($countryid==$showcountry)
{ $tempName = $countryname;
?>selected<? } ?>><? echo $countryname; ?></option>

and then display it... <?=$tempCountry?>
Is that what you mean ?

-Mike

rayge

That didnt help me or im just spacing more than i thought.
Let me expand the code a little more. and If you want to see it in action email me. maybe it might help.

<?
include("config.php");

$objConn=mysql_connect($Host,$DBUser,$DBPass);
mysql_select_db($DBName2,$objConn);

$showcountry=${"showcountry"};
if ($showcountry=="")
{

$showcountry=224;
}
else
{

$showcountry=intval($showcountry);
}

?>
<script>
function swapit(id){
document.regform.showcountry.value = id
document.regform.action = "seltest.php"
document.regform.submit()
}
</script>

<form name="regform" method="post" action="somefile.php">
<input type="hidden" name="showcountry" value="<? echo ${"showcountry"}; ?>">
<Select name="country" onchange="swapit(this.options[this.options.selectedIndex].value)">
<option value="-1"> chose a country </option>
<?
$res_query=mysql_query(("Select * from country"),$objConn);
$res=mysql_fetch_array($res_query);
while(!($res==0))
{

$countryid=$res["id"];
$countryname=$res["countryname"];
?>

<option value="<? echo $countryid; ?>" <? if ($countryid==$showcountry)
{
?>selected<? } ?>><? echo $countryname; ?></option>
<?
$res=mysql_fetch_array($res_query);

}
?>
</select>

<? if ($showcountry!="")
{
?>
<Select name="state_province">
<?
$res_query=mysql_query(("Select * from state_province where countryid = '".$showcountry."'"),$objConn);

$res=mysql_fetch_array($res_query);
while(!($res==0))
{

$areaid=$res["id"];
$state_province=$res["state_province"];
$countryid=$res["countryid"];

?>
<option value="<? echo $state_province; ?>"><? echo $state_province; ?></option>
<?
$res=mysql_fetch_array($res_query);

}
?>
</select>
<? } ?>

</form>

<?
mysql_close($objConn);
?>

rayge

YES :-) Thats exactly what i mean!

Thanks so much :-)

I hate being a newbie, it just makes me feel like an idiot LOL

rayge

One last question. Looking at the following code again(its the same code expanded a bit). How can i make it so if The state_province field is EMPTY it makes that WHole field DIssappear entirely?

<?
include("config.php");

$objConn=mysql_connect($Host,$DBUser,$DBPass);
mysql_select_db($DBName2,$objConn);

$showcountry=${"showcountry"};
if ($showcountry=="")
{

$showcountry=224;
}
else
{

$showcountry=intval($showcountry);
}

?>
<script>
function swapit(id){
document.regform.showcountry.value = id
document.regform.action = "seltest.php"
document.regform.submit()
}
</script>

$countryid=$res["id"];
$countryname=$res["countryname"];
?>

<option value="<? echo $countryid; ?>" <? if ($countryid==$showcountry)
{ $currentcountry = $countryname;
?>selected<? } ?>><? echo $countryname; ?></option>
<?
$res=mysql_fetch_array($res_query);

}
?>
</select>

$areaid=$res["id"];
$state_province=$res["state_province"];
$countryid=$res["countryid"];

?>
<option value="<? echo $state_province; ?>"><? echo $state_province; ?></option>
<?
$res=mysql_fetch_array($res_query);

}
?>
</select>
<? } ?>
<?=$currentcountry?>

</form>

<?
mysql_close($objConn);
?>

mike_y

Try this tell me what happens...
<?
$res_query=mysql_query(("Select * from state_province where countryid = '".$showcountry."'"),$objConn);
if (mysql_num_rows($res_query) > 0){
<Select name="state_province">
$res=mysql_fetch_array($res_query);
while(!($res==0))
{

$areaid=$res["id"];
$state_province=$res["state_province"];
$countryid=$res["countryid"];
?>
<option value="<? echo $state_province; ?>"><? echo $state_province; ?></option>
<?
$res=mysql_fetch_array($res_query);

}
</select>
}
?>

rayge

Lots o errors But it might just be syntax errors?

rayge

YAY!! It worked ;-) I fixed teh simple syntax errors and it works great.. YOU DA MAN! :-))