I think it I am the right path now, it has to do with $_GET or $_POST not
passing the information on to mysql.
But I will look at the log shortly.
Chuck
----- Original Message -----
From: "Thalis A. Kalfigopoulos" <[EMAIL PROTECTED]>
To: "Chuck Payne" <[EMAIL PROTECTED]>
Cc: "PHP General" <[EMAIL PROTECTED]>
Sent: Sunday, March 17, 2002 10:00 PM
Subject: Re: [PHP] Brain Dead...
> Is it possible to check the Mysql log and tell us its point of view?
>
>
> On Sun, 17 Mar 2002, Chuck Payne wrote:
>
> > When I go to post, I get the following alert...."Couldn't execute query"
so
> > I know I can connect, so it must be my SQL statement. That why I
wondering
> > how to pass the menu information to php so that it can do the SQL
statement.
> >
> > I don't have $abc in the menu but I do have abc. So I am trying to find
out
> > how to do that. Maybe I am going about it the wrong way.
> >
> > Chuck
> > ----- Original Message -----
> > From: "Thalis A. Kalfigopoulos" <[EMAIL PROTECTED]>
> > To: "Chuck Payne" <[EMAIL PROTECTED]>
> > Cc: "PHP General" <[EMAIL PROTECTED]>
> > Sent: Sunday, March 17, 2002 9:40 PM
> > Subject: Re: [PHP] Brain Dead...
> >
> >
> > > What does "I get that it can't do the query" mean? What error are you
> > getting?
> > >
> > >
> > > On Sun, 17 Mar 2002, Chuck Payne wrote:
> > >
> > > > Ok, but I have this as an example....
> > > >
> > > > <select name="abc" size="1">
> > > > <option value="a" selected>A</option>
> > > > <option value="b">B</option>
> > > > <option value="c">C</option>
> > > > <option value="d">D</option>
> > > > </select>
> > > >
> > > > Are you saying that abc should be $abc? I not clear.
> > > >
> > > > I have for my sql as this...
> > > >
> > > > db_name = "xxxinks";
> > > >
> > > > $table_name = "ssslinks";
> > > >
> > > > $connection = @mysql_connect("mybox.com","anyusers",!@#$%) or die
> > ("Couldn't
> > > > connect");
> > > >
> > > > $db = @mysql_select_db($db_name, $connection) or die ("Couldn't
select
> > > > database.");
> > > >
> > > > $sql = "INSERT INTO $table_name
> > > > (abc, key_word, links, name)
> > > > VALUES
> > > > ('$abc', '$keywords', '$links', '$name')";
> > > >
> > > > $result = @mysql_query($sql, $connection) or die("Couldn't execute
> > query.");
> > > >
> > > > but I get that it can't do the query. So I am think it not passing
> > abc...
> > > >
> > > > Chuck
> > > > ----- Original Message -----
> > > > From: "Thalis A. Kalfigopoulos" <[EMAIL PROTECTED]>
> > > > To: "Chuck "PUP" Payne" <[EMAIL PROTECTED]>
> > > > Cc: "PHP General" <[EMAIL PROTECTED]>
> > > > Sent: Sunday, March 17, 2002 9:25 PM
> > > > Subject: Re: [PHP] Brain Dead...
> > > >
> > > >
> > > > > Dropdownmenu is a <select name="ddmenu"><option>....</select>
> > > > > So the php variable that will hold the user's choice is stored in
> > variable
> > > > $ddmenu.
> > > > > Thus, no "input" needed.
> > > > >
> > > > >
> > > > > cheers,
> > > > > --thalis
> > > > >
> > > > >
> > > > > On Sun, 17 Mar 2002, Chuck "PUP" Payne wrote:
> > > > >
> > > > > > I am trying to get a form to work with pull down menu. But I am
so
> > brain
> > > > > > dead I forgot how to get the input from the pull down menu.
Where do
> > I
> > > > put
> > > > > > input to pass on to php?
> > > > > >
> > > > > > Chuck Payne
> > > > > >
> > > > > >
> > > > > > --
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> > > > > >
> > > > >
> > > > >
> > > >
> > > >
> > >
> > >
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> > >
> > >
> >
> >
> >
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