From: Robert Sossomon [EMAIL PROTECTED]
And here is the additem.php file where it check for the info in the
field and it is erroring out.
It would be rather useful to know the error...
$get_iteminfo = select * from GCN_items where 'item_num' LIKE
'%$_POST[sel_item_id]%';
Take away the
; [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Query Error
From: Robert Sossomon [EMAIL PROTECTED]
And here is the additem.php file where it check for the info in the
field and it is erroring out.
It would be rather useful to know the error...
$get_iteminfo = select * from GCN_items where
From: Robert Sossomon [EMAIL PROTECTED]
The errors as it prints are:
The query I just ran was: select * from GCN_items where `item_num` =
'%fm-a294%'
The query I just ran was: Resource id #2
0
Those aren't errors; it's just what you asked the script to display. Your
query simply isn't
!
-Original Message-
From: CPT John W. Holmes [mailto:[EMAIL PROTECTED]
Sent: Tuesday, November 04, 2003 1:31 PM
To: Robert Sossomon; [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Query Error
From: Robert Sossomon [EMAIL PROTECTED]
The errors as it prints are:
The query I
if you are using mysql, it doesn't support the sub-select.
Wilmar Perez
If it's on MySQL, then the problem is that it doesn't support nested queries
:)
If it's on something else, i'm afraid i don't know :/
Back to MySQL, you can get around that by doing your select author_code
from authorxcat component query, then put the results into 'x', 'y', 'z'
format, then do
On Sunday 05 May 2002 14:44, erich wrote:
when i perform query an mysql db, to insert a new record, the PHP says:
Warning: Supplied argument is not a valid MySQL result resource in
g:\wwwroot\phpusermanager-1.0\add_order.php on line 24
the snippet is as follows
?
// connect to the
Echo your INSERT statement, so that you know what it looks like, if you
have the values you expect, field names are spelled correctly. that you're
not inserting char into int, etc.
Use mysql_errno() and mysql_err() (Check those against manual to determine
what error you are getting. Use
Try putting single quotes around your variables in your update statement
instead of the escaped double quotes, they are easier to read.
Also, were you wanting to update all entries in that table to the same,
c_name, s_addy, city, state, zip and phone? If not, you will need to
add 'where
Ok jas
Ok what is wrong with this sql statement?
$sql = UPDATE $table_name SET
c_name=\$c_name\,s_addy=\$s_addy\,city=\$city\,state=\state\zip=\zi
p\,phone=\$phone\;
insufficient checking:
comma missing between state and zip
incorrectly escaped before phone
...
I have looked on MySQL.com
Did you type this out or cut/paste? There was a comma missing between
state\zip . Should work when you add the comma.
Add a WHERE condition for the update, otherwise every row will be set to
these values.
Common practice is to use single quotes around the char variables, saves a
lot of
Looks fine, now retrieve the rows from the query.
-Original Message-
From: Randy Johnson [mailto:[EMAIL PROTECTED]]
Sent: Thursday, January 25, 2001 12:46 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] query error
When I run any kind of query this is the value of result
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