mysql problems
Okay, I've got another one for you. I can not figure out why the mysql database won't insert these values! (I've taken out almost all the fields to make it easier).
if(isset($submit)){
//update the database with the information
$id = getUserID($cookie_name);
$sql = "INSERT into dailytracker (contactsmade,listingappointmentsmade) VALUES('$contactsmade','$listingappointmentsmade') where id=$id";
}So if the form is submitted, insert into table dailytracker collumns contactsmade and listingappointmentsmade where the username matches who is currently logged in.
<input type='text' style='text-align: right' size='3' maxlength='5' name='contactsmade' value=\"".$user_info['contactsmade']."\" >
<input type='text' style='text-align: right' size='3' maxlength='5' name='listingappointmentsmade' value=\"".$user_info['listingappointmentsmade']."\" >Here are the fields.
$user_info = getUserInfo(getUserID($cookie_name));'
Here is where I get the users info into the field values.
Then I submit. I don't know what I'm doing wrong here!!!
CptAwesome posted this at 09:40 — 18th December 2004.
He has: 370 posts
Joined: Dec 2004
$sql = "INSERT INTO `table_name` (`column1`,`column2`) VALUES ('$data1','$data2');
for a new entry, and you can't 'insert into' a row that exists in that case it's
$sql = "UPDATE `table_name` SET `column_name` = 'value' WHERE `id` = '$id';
just remember, fields, and tables need `` (below the tilde~) around them, and strings need ' ' around them.
I suggest getting MySQLadmin, and just look at some of the queries it runs when updating information.
Busy posted this at 09:47 — 18th December 2004.
He has: 6,151 posts
Joined: May 2001
You using the right one, INSERT vs UPDATE ?
do you get a value if you echo $sql ?
and do you have a $something = mysql_query($sql); ?
CptAwesome posted this at 10:10 — 18th December 2004.
He has: 370 posts
Joined: Dec 2004
echoing $sql will just get you the query obviously.
and you don't need to do a $something = for the mysql_query($sql); but you can then do:
if($something)
{
echo "YAY IT WORKED!";
}
else
{
echo "bunk, still didn't work";
}
Busy posted this at 21:43 — 18th December 2004.
He has: 6,151 posts
Joined: May 2001
Isn't a bad idea to go right back to basics, finding out if the variables actually have values can be the answer. Easier than echoing all variables, especially if you have more than a couple.
Again the mysql_query() is helpful for fault finding. there are other ways of doing it, sure, but the end of the day you need to find the problem whether it be a missing quote, comma, brace, value or whatever.
timjpriebe posted this at 14:31 — 18th December 2004.
He has: 2,667 posts
Joined: Dec 2004
I have to agree with what Busy said. The WHERE clause does not work with INSERT. It sounds like you're wanting to use UPDATE, not INSERT.
airoid3000 posted this at 00:47 — 19th December 2004.
They have: 71 posts
Joined: Mar 2004
Thanks for the help, I was able to get it in the database.
Here is my new problem. I am trying to load the form with all the values for that certain date. (I have a date field in the database table). How can I tell the page that I want it to load the values only for a selected date?
timjpriebe posted this at 12:46 — 20th December 2004.
He has: 2,667 posts
Joined: Dec 2004
SELECT field1, field2, field3 FROM tablename WHERE datefield = 'XX-XX-XXXX'
Then, of course, you'll want to use a while loop to go through the results for that date, unless your DB is set up so that there's only one entry per date.
Tim
http://www.tandswebdesign.com
CptAwesome posted this at 18:42 — 20th December 2004.
He has: 370 posts
Joined: Dec 2004
if the date field is mixed date/time that won't work, but you could query and pull all results (depending on the size, this could be a bad idea) and then use an ereg function, or some other comparison method to analyze the info.
airoid3000 posted this at 03:39 — 21st December 2004.
They have: 71 posts
Joined: Mar 2004
Thanks, sorry that was a stupid question.
timjpriebe posted this at 13:16 — 21st December 2004.
He has: 2,667 posts
Joined: Dec 2004
Good point, CptAwesome. You might have to do this:
SELECT field1, field2, field3 FROM tablename WHERE datefield LIKE 'XX-XX-XXXX %'
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