Resource ID #14

[norek]

My original code was below and it would give me error "RESOURCE ID #14".

if ($search_city) {
$sc1 = mysql_query("SELECT sid FROM CITY_TABLE WHERE descript LIKE '%$search_city%'");
$sc2 = $sc1'
}

**************************************************

Then after reading posts on the site, I modified it to what's below, but now the SQL is returning the value $sc2 as "Array"

if ($search_city) {
$sc1 = mysql_query("SELECT sid FROM CITY_TABLE WHERE descript LIKE '%$search_city%'");
$sc2 = mysql_fetch_array($sc1);
$sc3 = $sc2;
}

**************************************************

Or if I user the mysql_fetch_object or mysql_fetch_field, it returns "Object"

if ($search_city) {
$sc1 = mysql_query("SELECT sid FROM CITY_TABLE WHERE descript LIKE '%$search_city%'");
$sc2 = mysql_fetch_object($sc1);
$sc3 = $sc2;
}

**************************************************

I'm have no idea what else might be wrong... any ideas?

[dreamcatcher]

norek, welcome to Webmaster World. :)

You seem a little unclear on how to use the sql syntax.

Try:

if ($search_city) {
$sc1 = mysql_query("SELECT sid FROM CITY_TABLE WHERE descript LIKE '%$search_city%'");
$sc2 = mysql_fetch_array($sc1);
echo $sc2['sid'];
}

or for mysql_fetch_object:

if ($search_city) {
$sc1 = mysql_query("SELECT sid FROM CITY_TABLE WHERE descript LIKE '%$search_city%'");
$sc2 = mysql_fetch_object($sc1);
echo $sc2->sid;
}

dc

[norek]

Dreamcather, the reason I didn't use the echo command is because there is a lot more to the script then what I posted...

The problem was that I needed $sc2->sid;

You fixed hours of being stumped, I think I'm in love with you :P

THANKS!

[ergophobe]

Look for the thread in the forum library called "The basics of extracting data with MySQL" or some such title. That should help you out.

[webmasterworld.com...]

[dreamcatcher]

You fixed hours of being stumped, I think I'm in love with you :P

You`re welcome. And if you are single, female and Chinese I`ll give you my phone number. LOL!

dc