Re: [PHP-DB] Soft-Linking and includes
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 files in /www/dir1/subdir, or your include would need an explicit path. Jonathan Hilgeman wrote: I'm on a red hat system, and I've soft-linked two directories: /www/dir1/subdir /www/dir2/subdir -- /www/dir1/subdir Now, inside subdir is a file that tries to include(../info.php); which prints out some information about the file paths and my database stuff. So there's: /www/dir1/info.php /www/dir1/subdir/includer.php /www/dir2/info.php /www/dir2/subdir -- /www/dir1/subdir Now, when I run /www/dir2/subdir/includer.php, it SHOULD include the file ../info.php which translates into /www/dir2/info.php. However, the symbolic linking seems to have messed it up, and instead of running dir2's info.php, it seems to think it is in dir1, and instead includes dir1's info.php file. Has anyone run into this and/or know a fix for it? Thanks! - Jonathan -- Out beyond the ideas of wrong-doing and right-doing there is a field. I'll meet you there. -Rumi *(O)* [EMAIL PROTECTED] *** under construction *** http://centralcoasthealing.net Healing Resources on the Central Coast http://kornsnake.com Internet Development -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Looking for a good MySQL db abstraction layer
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 like it because it's easy to understand and play around with. Also has sql logging which can be switched on for debugging. Michael Zornek wrote: I'm looking for a good MySQL db abstraction layer. I'm just coming back to PHP/MySQL and rather writing my own I figured I'd try to see if any are already out there. The layer should obviously have the basics, of connect, query, and close, and possibly some error checking and handling. thanks, ~ Mike -- Out beyond the ideas of wrong-doing and right-doing there is a field. I'll meet you there. -Rumi *(O)* [EMAIL PROTECTED] *** under construction *** http://centralcoasthealing.net Healing Resources on the Central Coast http://kornsnake.com Internet Development -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[Fwd: Re: [PHP-DB] if variable is equal to 2 through 4]
$my_array = array (2, 3, 4);
if (in_array($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 == 2 through 4 ???
Should you have any questions, comments or concerns, feel free to call
me at 318-338-2034.
Thank you for your time,
Jay Fitzgerald, Design Director - CSBW-A, CPW-A, CWD-A, CEMS-A
==
Bayou Internet..(888)
30-BAYOUhttp://www.bayou.com
Mississippi Internet...(800)
MISSISSIPPI...http://www.mississippi.net
Vicksburg Online..(800)
MISSISSIPPIhttp://www.vicksburg.com
==
Tel: (318) 338-2034ICQ: 38823829
Fax: (318) 323-5053
--
Out beyond the ideas of wrong-doing and right-doing
there is a field. I'll meet you there.
-Rumi
*(O)*
[EMAIL PROTECTED]
*** under construction ***
http://centralcoasthealing.net Healing Resources on the Central Coast
http://kornsnake.com Internet Development
--
Out beyond the ideas of wrong-doing and right-doing
there is a field. I'll meet you there.
-Rumi
*(O)*
[EMAIL PROTECTED]
*** under construction ***
http://centralcoasthealing.net Healing Resources on the Central Coast
http://kornsnake.com Internet Development
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Re: [PHP-DB] Database Connection Properties
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 patrick gibson wrote: I'm new to PHP, and I have a question regarding the storage of connection properties to the database. I have a site on a shared (Unix) server. I'm adding some PHP to a section which will be database-driven. I'm concerned about storing the username and password to my database (some of the data contained within this database is sensitive) in my PHP files which are readable by the webserver, and thus by anyone on the system. Is there a recommended strategy for keeping my database username and password in a secure location so that no other user can access the files? Any help would be greatly appreciated. Thanks very much, Patrick ---| patrick gibson |---+ email: email @ patrickg.com url: http://patrickgibson.com/ + -- Out beyond the ideas of wrong-doing and right-doing there is a field. I'll meet you there. -Rumi *(O)* [EMAIL PROTECTED] *** under construction *** http://centralcoasthealing.net Healing Resources on the Central Coast http://kornsnake.com Internet Development -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP-DB] database/form help needed
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 results 9 at a time, and that works great. Now, I also have a checkbox called ID for each of the 9 entries and here is my problem. How can I dynamically assign a unique ID for my checkboxes? I am getting my checkboxes through a loop, and because of that the ID is called id - for all of my entries, so I can't select multiple as only one entry comes out on the results page. How can I dynamically create checkboxes with a unique id (Such as id1, id2 etc ...) so that I can make multiple selections from my DB? Thank you all so much - Happy New Year. Regards Chris Payne www.planetoxygene.com -- Out beyond the ideas of wrong-doing and right-doing there is a field. I'll meet you there. -Rumi *(O)* [EMAIL PROTECTED] *** under construction *** http://centralcoasthealing.net Healing Resources on the Central Coast http://kornsnake.com Internet Development -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP-DB] Problem with SQL query on several tables
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 columns you need to process, indexed by datestamp, and append the data from the 3 tables-- then query the temp table. Approach #2 (php) Create an associative array (hash table) in php then append the data from the 3 tables, order as desired, and process. Which you choose would depend on which you find simpler-- the mysql approach would probably take a more time, but be easier to reuse on successive pages. ib Rosen wrote: Thanks, but I want wo JOIN data from three tables and then to order all data by datestamp. Can I Do It ? Thanks, Rosen Andrey Hristov wrote in message 071401c17759$f5873c80$0b01a8c0@ANDreY... Mysql says that in the join there are some columns with namer datestamp so you have to choose by which you want to order. Think about making LEFT JOIN. select * from f1 LEFT JOIN f2 ON f1.some_field=f2.some_field LEFT JOIN f3 ON f2.some_field=f3.some_field roder by f1.datestamp; Regards, Andrey Hristov IcyGEN Corporation http://www.icygen.com BALANCED SOLUTIONS - Original Message - From: Rosen [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Tuesday, November 27, 2001 5:34 PM Subject: [PHP-DB] Problem with SQL query on several tables Hi, I'm trying to execute SQL query: select * from f1,f2,f3 order by datestamp MySQL returns me an error: 1052 - Column: 'datestamp' in order clause is ambigious The tables f1,f2,f3 have some structure. The field datestamp is type datetime. Why I can't order query on 'datestamp' ? Thanks, Rosen -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED] -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP-DB] Result is not empty
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 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, returns false, and $row = NULL.
-ib
Why would that matter? He still sets the column variables on the first
iteration? Yes, he needs no while loop, but the variables that are
set inside the loop should remain set after the loop terminates.
Kevin Schaaps wrote:
Greetings once again,
Today I hope to have a challenge for you. :)
I've created a query which returns 1 record. This is confirmed when
testing
it in MySQL itself.
Now PHP sees that there is 1 record in the result, but is completely
unwilling to show the information.
The css responsible for that page does not change the color of the
text so
it should (like all the rest of the page is) be visible.
Please help :)
Yours,
Kevin
$query = SELECT concat(rank_tbl.abbreviation, ' ',
character_tbl.name, ' ',
character_tbl.surname) as CO,
character_tbl.ircnick, character_tbl.email FROM
character_tbl, rank_tbl, sim_tbl WHERE sim_tbl.co = character_tbl.id
AND character_tbl.rank = rank_tbl.id AND sim_tbl.co =
character_tbl.id AND character_tbl.rank = rank_tbl.id
AND sim_tbl.id = $sim_id;
$result = mysql_query($query);
$num_rows = mysql_num_rows($result);
if ($num_rows == 1)
{
while ($row = mysql_fetch_array($result));
{
$co = $row[CO];
$nick = $row[ircnick];
$email = $row[email];
};
echo
table
tr
td width=\50\/td
td width=\125\CO:td
tda href=\$email\$co/a/td
/tr
tr
td width=\50\/td
td width=\125\IRC NICKtd
td$nick/td
/tr
tr
td width=\50\/td
td width=\125\td
td/td
/tr
/table;
}
else
{
echo br p class=\medium\No Commanding Officer
assigned to
$name/p;
};
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Re: [PHP-DB] Result is not empty
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, returns false, and $row = NULL.
-ib
Kevin Schaaps wrote:
Greetings once again,
Today I hope to have a challenge for you. :)
I've created a query which returns 1 record. This is confirmed when testing
it in MySQL itself.
Now PHP sees that there is 1 record in the result, but is completely
unwilling to show the information.
The css responsible for that page does not change the color of the text so
it should (like all the rest of the page is) be visible.
Please help :)
Yours,
Kevin
$query = SELECT concat(rank_tbl.abbreviation, ' ', character_tbl.name, ' ',
character_tbl.surname) as CO,
character_tbl.ircnick, character_tbl.email FROM
character_tbl, rank_tbl, sim_tbl WHERE sim_tbl.co = character_tbl.id
AND character_tbl.rank = rank_tbl.id AND sim_tbl.co =
character_tbl.id AND character_tbl.rank = rank_tbl.id
AND sim_tbl.id = $sim_id;
$result = mysql_query($query);
$num_rows = mysql_num_rows($result);
if ($num_rows == 1)
{
while ($row = mysql_fetch_array($result));
{
$co = $row[CO];
$nick = $row[ircnick];
$email = $row[email];
};
echo
table
tr
td width=\50\/td
td width=\125\CO:td
tda href=\$email\$co/a/td
/tr
tr
td width=\50\/td
td width=\125\IRC NICKtd
td$nick/td
/tr
tr
td width=\50\/td
td width=\125\td
td/td
/tr
/table;
}
else
{
echo br p class=\medium\No Commanding Officer assigned to
$name/p;
};
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Re: [PHP-DB] (int)mysql_result?
(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, SUM); I do not find anything about the SUM either. Thanks Dave -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
