Ian Bert Tusil wrote:
is there something wrong with my SQL Query? I didnt get any results neither errors..
[snip]
mysql_query($SQLQuery,$link);
That's because you're not checking for errors...
mysql_query($SQLQuery,$link) or die(mysql_error());
--
---John Holmes...
Amazon Wishlist: www.amazon.co
is there something wrong with my SQL Query? I didnt get any results
neither errors..
$link = mysql_connect('localhost', '') or die('Cannot connect to
the database.');
mysql_select_db("profiles",$link);
$SQLQuery = "INSERT INTO profyles (username, fname, mname, lname, address,
country, gender,
is there something wrong with my SQL Query? I didnt get any results neither errors..
') or die('Cannot connect to the
database.');
mysql_select_db("profiles",$link);
$SQLQuery = "INSERT INTO profyles (username, fname, mname, lname, address, country,
gender, sexuality, status, bday, race, yahoo,
I seem to have run into another problem with a query that I am doing
exactly like the example in the MySQL documentation, I think. The following
query is not working on my PHP page, and it is not working from a MySQL
command line either. This looks exactly the same as an example in the M
on 7/26/01 11:11 AM, Steve Fitzgerald ([EMAIL PROTECTED]) wrote:
> ran the query in the MySQL client and got the ERROR 1054: Unknown column
> '$ContactID' in 'where clause'. I know the column ContactID exists.
Of course it did. It can't find the text $ContactID in the table. Try
putting in a
ContactID is setup as a tinyint so I dropped the single ticks. No luck. I
ran the query in the MySQL client and got the ERROR 1054: Unknown column
'$ContactID' in 'where clause'. I know the column ContactID exists.
Steve
"Josh Hoover" <[EMAIL PROTECTED]> wrote in message
E973048AB322D411AE9900
In your query, do you need the single ticks around the $ContactID? Is the
ContactID column a char/varchar field which would require the single ticks?
I'm wondering if that's causing your problem. Also, what if you do that
query via the MySQL client? What do you get then?
Josh Hoover
KnowledgeS
I'm trying to query my database to fill in data First Name Last Name. Using
the script below I get (depending on the ContactID I enter)
Contact: 1 1 or 2 2
If ContactID=0 I get
Contact:
Any ideas on what I am doing wrong?
Thanks.
Steve Fitzgerald
\n", mysql_result($result,"LastName"));
