My limited understanding of Unix hard and soft links is that
/www/dir2/subdir is not a directory, it is a pointer to the inode that
is pointed to by /www/dir1/subdir. I think to accomplish what you want,
/www/dir2/subdir would have to be a true directory, filled with soft
links to each of the
PEAR, Metabase, and ADODB are all good db abstraction packages with
their respective merits. But if your looking for something simple,
efficient, for mysql only try:
http://www.phpclasses.org/browse.html/package/107.html
The package contains basic connection, sql command, and recordset
classes. I
$my_array = array (2, 3, 4);
if (in_array($variable, $my_array)) {
echo hello;
}
Jay Fitzgerald wrote:
i am currently using this code:
if ($variable == 2) || ($variable == 3) || ($variable == 4)
{
echo hello;
}
how would I write it if I wanted to say this:
if $variable
Store the connection strings in an include file preferably outside the
wed root. You can include files with a fully resolved path, or using the
php include_path variable if you have access to the configuration.
Alternately, if you don't have access to directories outside the
webroot, put the
You could setup an autoincremented unique integer field in your db
table(s), then when you fetch the row from the table use the id in
generating your checkbox name.
-ib
Chris Payne wrote:
Hi there everyone,
I have a loop which goes through my MySQL database and some PHP code which grabs
It sounds to me like what you're trying to do is APPEND data from three
similarly structured tables into one entity ordered by a column common
to all 3 tables called datestamp. If I'm mistaken, then please ignore
all of the following:
Approach #1 (mysql)
Create a temporary table with the
yes, of course you're right-- gotta stop working so late... the use of
the while statement where it wasn't needed threw me off, I guess.
Paul DuBois wrote:
Take out the while statement--
i.e. simply use:
$row = mysql_fetch_array($result);
the way you have things constructed now, the
Take out the while statement--
i.e. simply use:
$row = mysql_fetch_array($result);
the way you have things constructed now, the while statement evaluates true on the
first iteration and $row equals the result row from the query. Because the while
returned true, it is evaluated a second time,
(int)
forces (casts) result as integer--
see
http://www.php.net/manual/en/language.types.type-juggling.php#language.types.typecasting
SUM would be the field name
-ib
[EMAIL PROTECTED] wrote:
What does adding the (int) do to this statement:
$poll_sum = (int)mysql_result($poll_result, 0,
