Lose the single quotes, quoting a single variable is nonsense.
Quotes are used to define strings, it is a feature (called substitution)
of PHP that allows you to use a variable within a string.
The reason it doesn't work is that this subtitution is only performed on
strings enclosed in "'s (dou
Hi, looking at your sql statement, it seems like you are always setting
my_job_id and job_id to the same value for each row in the table. "select
job_id AS my_job_id" is just putting the value of job_id in a variable
called my_job_id. Could you tell us the structure of your mysql table please
On Tue, 2002-02-05 at 16:23, Michael O'Neal wrote:
> Hi.
>
> I'm working on an "edit" page where the pull down menu is populated from a
> database. I can't figure out how to print "selected" when that particular
> record is the one associated with the current ID. Can anyone help?
>
> Here's my
Hi.
I'm working on an "edit" page where the pull down menu is populated from a
database. I can't figure out how to print "selected" when that particular
record is the one associated with the current ID. Can anyone help?
Here's my current code:
$title";
}
$num = @mysql_num_rows($result);
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