Re: [PHP] RE: Help with my first recursion menu
Hi, On Wed, Oct 28, 2009 at 5:22 PM, MEM tal...@gmail.com wrote: I've been told that stack is the way to go, so I'm trying to understand the following code: http://pastebin.com/m5616c88f I've commented every line so that any of you could see if I'm interpreting something wrong: I have two questions about this code, that hopefully someone on the list could explain: 1) Why do we need to remove the last array item? (on line 32): array_pop($urlStack); On line 20, $url creates an URL path that includes the $cat path. Once you've completed iteration on all children of $cat, the url path should no longer include $cat path, thus it is removed from $urlStack. 2) Why we print the closed tag of the li element, after the recursive call? (on line 29) echo /li\n; Line 29 is closing the li element that was opened on line 23 for $cat. Thanks a lot in advance, Márcio -Lex
Re: [PHP] supplied argument errors
On Tue, Jun 23, 2009 at 4:10 PM, PJ af.gour...@videotron.ca wrote: I think there is something I do not understand in the manual about mysql_fetch_assoc(), mysql_affected_rows() The code works, but I get these annoying messages. snippet: snip What are the warnings?
Re: [PHP] Check out system for Shopping Cart
On Thu, May 21, 2009 at 7:32 AM, Vernon St.Croix vstcr...@hotmail.comwrote: *Warning*: mysqli_query() expects parameter 1 to be mysqli, object given in *C:\wamp\www\draft\checkout.php* on line *26* ***for the script ?php include(mysql.class.php); include (header.php); //include (functions.php); $user = 1; $total = 178.93; include (mysql_connect.php); //Turn off autocommit //mysql_autocommit($con, FALSE); //Add orders to the order table $sql = INSERT INTO orders (user_id, total) VALUES ($user, $total); $r = mysqli_query($con, $sql);* The warning message is saying that it's expecting the first parameter you passed in mysqli_query() to be a mysqli link identifier, i.e., that the variable $con used in the line above be your connection to the database. Somewhere before this function call, you need to set $con: $con = *mysqli_connect* ($host, $username, $passwd, $dbname); -- Lex
Re: [PHP] Check out system for Shopping Cart
Vernon,
CC messages to php-general to keep responses on-list. See my comments
inline.
On Thu, May 21, 2009 at 12:06 PM, Vernon St.Croix vstcr...@hotmail.comwrote:
I have amended as you advised and the mysqli errors are now gone. Thanks
for that.
I however have another one I can't get rid of:
*Warning*: Invalid argument supplied for foreach() in *
C:\wamp\www\draft\checkout.php* on line *53*
Your order could not be processed.
*checkout.php*
?php
code snipped
include (mysql_connect.php);
//Turn off autocommit
$con = mysqli_connect (localhost, root, , rum);
mysqli_autocommit($con, FALSE);
//Add orders to the order table
$sql = INSERT INTO orders (user_id, total) VALUES ($user, $total);
$r = mysqli_query($con, $sql);
code snipped
* foreach ($_SESSION['cart'] as $id = $item){*
$qty = $item['quantity'];
$price = $item['price'];
mysqli_stmt_execute($stmt);
$affected += mysqli_stmt_affected_rows($stmt);
}
The reason you're getting this warning is that foreach is expecting
$_SESSION['cart'] to be an array. Do a print_r() or var_dump() of
$_SESSION['cart'] to see what is actually being passed to the function.
mysqli_stmt_close($stmt);
if ($affected ==count($_SESSION['cart'])) {
mysqli_commit($con);
echo'Thank you for your order';
}else{
mysqli_rollback($con);
echo'Your order could not be processed';
}
}else{
mysqli_rollback($con);
echo'System error';
}
mysqli_close($con);
include (footer.html);
?
VEE
-- Lex
Re: [PHP] Non-Object errors
Terion, On Tue, Apr 28, 2009 at 1:14 PM, Miller, Terion tmil...@springfi.gannett.com wrote: Here is my code now: $query = select name, age, warrant, bond, wnumber, crime FROM warrants where .$warranttype. = .$warranttype. OR .$searchname. = .$searchname. ;$result = mysql_query($query);$row = mysql_fetch_assoc($result);$num_results = mysql_num_rows ($result); Should this post of gone on the Db list maybe? One thing I noticed about your code above has to do with your WHERE clause. Say, $warranttype = 'warrant' $searchname = 'Terion' Your $query will become: SELECT name, age, warrant, bond, wnumber, crime FROM warrants WHERE warrant = warrant OR Terion = Terion The first time you use $warranttype and $searchname should be the column names rather than the value you would like that column to be equal to. -- Lex
Re: [PHP] difficult select problem
PJ, On Tue, Apr 7, 2009 at 11:37 AM, PJ af.gour...@videotron.ca wrote: $SQL = SELECT b.*, c.publisher, a.first_name, a.last_name FROM book AS b LEFT JOIN book_publisher as bp ON b.id = bp.bookID LEFT JOIN publishers AS c ON bp.publishers_id = c.id LEFT JOIN book_author AS ba ON b.id = ba.bookID LEFT JOIN author AS a ON ba.authID = a.id WHERE LEFT(last_name, 1 ) = '$Auth' ; (PLEASE LET ME KNOW IF THERE IS SOMETHING WRONG WITH THE CODE) Let me try to clarify what I'm saying about your query. The above query will ONLY return authors who match the WHERE condition, thus have last name starting with A. This query will never find the second author for those books, unless that author's last name also starts with an A. That's why you first need to get a list of the book IDs that match your WHERE condition, and then grab the authors related to those book IDs (whether through two queries or using a sub-query). You've been shown several different ways of doing this through the responses provided. Do the provided queries not work for your test data? - Lex
Re: [PHP] difficult select problem
PJ,
On Mon, Apr 6, 2009 at 12:32 PM, PJ af.gour...@videotron.ca wrote:
I am trying to limit the search for books to only those that start with
A (does not have to be case sensitive); but within that selection
there may be a second author whose name may start with any letter of the
alphabet.
First off, are you trying to search for book TITLES that start with A or
book AUTHORS that start with A?
If it's authors, it might make more sense to do a select that retrieves the
book ids that have an author whose last name starts with an A and once you
have those IDs, do a second query that will retrieve all authors for those
books. So something like below (UNTESTED):
//Find out which books are written by an author with a last name starting
with $Auth
$SQL = SELECT b.id FROM book AS b
LEFT JOIN book_author AS ba ON b.id = ba.bookID
LEFT JOIN author AS a ON ba.authID = a.id
WHERE LEFT(last_name, 1) = '$Auth' ;
Then do the following select statement with a foreach loop that retrieves
the book IDs above
$query = SELECT b.id, GROUP_CONCAT(CONCAT(' ', a.first_name, a.last_name)
ORDER BY a.last_name)
FROM book AS b
LEFT JOIN book_author AS ba ON b.id = ba.bookID
LEFT JOIN author AS a ON ba.authID = a.id
WHERE b.id = $value
GROUP BY b.id ; // $value is the b.id for each iteration of the foreach
loop
More information about the GROUP_CONCAT() function can be found in the MySQL
reference (
http://dev.mysql.com/doc/refman/5.0/en/group-by-functions.html#function_group-concat
)
- Lex
Re: [PHP] difficult select problem
On Mon, Apr 6, 2009 at 2:05 PM, PJ af.gour...@videotron.ca wrote:
My code already has selected the books whose authors last names start
with A as well as the authors themselves.
Within the results some books have 2 authors.
If you already have the book IDs where this happens, then for each bookID
(probably a foreach loop in your code) simply do a query at that point that
will obtain ALL authors for that bookID -- don't include the condition that
last name start with an A in this query.
$query = SELECT b.id http://b.id, GROUP_CONCAT(CONCAT(' ',
a.first_name, a.last_name) ORDER BY a.last_name)
I believe this should be (CONCAT_WS(' ', a.first_name, a.last_name) or
(CONCAT(a.first_name,' ', a.last_name)
Sorry, yes that should be
GROUP_CONCAT( CONCAT(' ', a.first_name, a.last_name) ORDER BY a.last_name)
so that it concatenates each author as FirstName LastName and then groups
all authors for a given book separated by commas, so bookID=1 would have
something like Author1FirstName LastNameAuthor1, FirstNameAuthor2
LastNameAuthor2
Problem here is that I need to use the ba.ordinal somewhere in the cod
to distinguish between the authors - if ba.ordinal is 1, the author is
listed as the first author and if there is no ba.ordinal = 2 for the
ba.bookID, then there is only 1 author. I have not included this in my
query because it only returns the ordinal number for the author which
does not help me at all. I need to know who the second author is.
In effect, it would be great if I could get 2 joins on the same table
(author) for
1. (CONCAT(first_name, ' ', last_name) AS Author WHERE ab.ordinal = 1
LEFT(last_name, 1) = $Auth
2. (CONCAT(first_name, ' ', last_name) AS Author1 WHERE ab.ordinal = 2
The problem you'd run into with #1 above, is you'd only get authors whose
last name starts with A if they're the primary author. Is that the desired
functionality you're looking for?
- Lex
Re: [PHP] and and weird results
PJ, On Wed, Apr 1, 2009 at 1:40 PM, PJ af.gour...@videotron.ca wrote: SELECT * FROM book WHERE id IN (SELECT bookID FROM book_author WHERE authID IN (SELECT author.id FROM author WHERE LEFT(last_name, 1 ) = '$Auth' LEFT(last_name, 1 ) = '$Auth1' LEFT(last_name, 1 ) = '$Auth2')); Say $Auth = I, $Auth1 = J, $Auth2 = K. The above sub-query is stating select author ids where the first digit in last_name is I AND the first digit in last_name is J AND the first digit in last_name is K. For any given last_name (thus, author id), the first digit can only be one of those options, thus this sub-query will return ZERO results. To get a sub-query that will only return results that start with I, J, or K,you can instead use the following sub-query: SELECT author.id FROM author WHERE LEFT(last_name, 1) = '$Auth' OR LEFT(last_name, 1) = '$Auth1' OR LEFT(last_name, 1) = '$Auth2' The above will return all author IDs where the surname starts with I, J, or K. - Lex
Re: [PHP] validation inserts not working
PJ,
snip
$sql1 = INSERT INTO book ( title, sub_title, descr,
comment, bk_cover, copyright, ISBN, language, sellers )
VALUES ('$titleIN', '$sub_titleIN', '$descrIN', '$commentIN',
'$bk_coverIN', '$copyrightIN', '$ISBNIN', '$languageIN',
'$sellersIN');
$result1 = mysql_query($sql1, $db);
$autoid = mysql_insert_id($result1);
You're actually sending mysql_insert_id() the wrong parameter. It should be:
$autoid = mysql_insert_id($db); // you send the resource of your MySQL
connection (http://ca3.php.net/manual/en/function.mysql-insert-id.php)
This should be corrected in everywhere you call mysql_insert_id()
-Lex
Re: [PHP] validation inserts not working
On Wed, Mar 11, 2009 at 5:44 PM, PJ af.gour...@videotron.ca wrote:
haliphax wrote:
On Wed, Mar 11, 2009 at 4:29 PM, PJ af.gour...@videotron.ca wrote:
Lex Braun wrote:
PJ,
  snip
  $sql1 = INSERT INTO book ( title, sub_title, descr,
        comment, bk_cover, copyright, ISBN, language,
sellers )
      VALUES ('$titleIN', '$sub_titleIN', '$descrIN',
'$commentIN',
        '$bk_coverIN', '$copyrightIN', '$ISBNIN',
'$languageIN',
        '$sellersIN');
    $result1 = mysql_query($sql1, $db);
    $autoid = mysql_insert_id($result1);
You're actually sending mysql_insert_id() the wrong parameter. It
should be:
$autoid = mysql_insert_id($db); // you send the resource of your MySQL
connection (http://ca3.php.net/manual/en/function.mysql-insert-id.php)
This should be corrected in everywhere you call mysql_insert_id()
-Lex
I tried this (from the link above)
$result1 = mysql_query($sql1, $db);
  $autoid = mysql_insert_id();
  echo $autoid;
works...
but now, I have another problem... I am trying to debug this thing by
going 1 query at a time ( I comment out the rest): every time I do an
INSERT INTO book... the insert works fine but the ID is increased not
from the highest id value but from the last id inserted. So my highest
at the moment is 11; I insert test data and it goes in as id = 15; I
delete this field and redo an insert and it goes in as id = 16. How can
I change this so the auto_increment continues from the 11 ?
In MySQL,
ALTER TABLE [tablename] AUTO_INCREMENT = [value];
HTH,
Ok, but why is it doing that? I like to understand - I must have asked
too many questions when I was a kid... it stuck. :-)
It's doing that because an auto-incrementing column keeps incrementing with
each insertion. Thus, say you insert an entry into ID #10 and delete it
before another entry is made. MySQL will still have the auto_increment set
to 11 (one more than the last insertion) rather than subtracting one because
you deleted #10. It's the database method of ensuring you aren't
over-writing pre-existing auto_increment values.
NOTE: This doesn't take into account someone entering their own value into
the ID field greater than the current auto_increment value.
Re: [PHP] validation inserts not working
PJ,
On Tue, Mar 10, 2009 at 3:46 PM, PJ af.gour...@videotron.ca wrote:
snip
$sql1 = INSERT INTO book ( title, sub_title, descr,
comment, bk_cover, copyright, ISBN, language, sellers )
VALUES ('$titleIN', '$sub_titleIN', '$descrIN', '$commentIN',
'$bk_coverIN', '$copyrightIN', '$ISBNIN', '$languageIN',
'$sellersIN');
$result1 = mysql_query($sql1, $db);
$autoid = mysql_insert_id($result1);
}/* --- IF THIS LIKE IS DELETED, THE PAGE DOES NOT DISPLAY
So, if I select insert on the page, only the first query is executed
since the rest is commented out. Thus, I get onle the book table
inserted but not other tables, like author, book_author, or publishers,
book_publisher or categories, book_categories...
Is there something wrong with what follows immediately...
like, do I have the brackets right? I've tried about every
combination possible with no change.
//Check if Author is entered exists
if( (strlen($_POST[first_nameIN]) 0) (strlen($_POST[last_nameIN])
0) ) {
$sql2 = SELECT (first_name, last_name)
FROM author WHERE (first_name LIKE '$first_nameIN'
last_name LIKE '$last_nameIN)';
LIKE is going to do full-text search, which isn't what you want when
searching for a specific author. You can have the query return the ID for
that specific author (to use in your $sql2a query).
$sql2 = SELECT id FROM author WHERE first_name = '$first_nameIN' AND
last_name = '$last_nameIN' ;
$result2 = mysql_query($sql2);
if (mysql_num_rows($result2) 0) {
$row = mysql_fetch_assoc($result2); // gives you the row return
from $sql2
$sql2a = INSERT INTO book_author (authID, bookID, ordinal)
VALUES (author.id WHERE (first_name LIKE '$first_nameIN'
last_name LIKE '$last_nameIN'),
book.ID WHERE book.title LIKE '$titleIN'), '1';
With the change in $sql2 and the fact that the bookID is stored in $autoid,
this becomes:
$sql2a = INSERT INTO book_author (authID, bookID, ordinal) VALUES ( .
$row['id'] . , . $autoid . , '1');
$result2a = mysql_query($sql2a, $db);
}
elseif (mysql_num_rows($result2) = 0) {
$sql2b = INSERT INTO author (first_name, last_name)
VALUES ('$first_nameIN', '$last_nameIN');
$result2b = mysql_query($sql2b, $db);
$authorID = mysql_insert_id($result2b); // gives you the id
of the newly inserted author for use in your book_author table
$sql2c = INSERT INTO book_author (authID, bookID, ordinal)
VALUES (author.id WHERE (first_name LIKE '$first_nameIN'
last_name LIKE '$last_nameIN'), book.ID
WHERE book.title LIKE '$titleIN'), '1';
With the addition of $authorID and the fact that bookID is stored in
$autoid, your $sql2c should now be:
$sql2c = INSERT INTO book_author (authID, bookID, ordinal) VALUES ( .
$authorID . , . $autoid . , '1');
$result2c = mysql_query($sql2c, $db);
}
}
- Lex
[PHP] Re: [PHP-DB] Re: [PHP] Re: [PHP-DB] Re: Problems with displaying results
On Thu, Mar 5, 2009 at 10:30 AM, Terion Miller webdev.ter...@gmail.comwrote:
Still having problems with getting this script to work, the first part of
the query does now work since I used the suggested JOIN, so my results are
there and I can echo them but now I can't seem to get them to display
neatly
somehow:
CODE THUS
FAR
$query = SELECT admin.UserName, admin.AdminID, workorders.WorkOrderID,
workorders.CreatedDate, workorders.Location, workorders.WorkOrderName,
workorders.FormName, workorders.STATUS, workorders.Notes, workorders.pod
FROM admin LEFT JOIN workorders ON (admin.AdminID = workorders.AdminID)
WHERE admin.Retail1 = 'yes'
;
$result = mysql_query ($query) ;
//$row = mysql_fetch_array($result);
while ($row = mysql_fetch_row($result)){
$admin_id = $row['AdminID'];
mysql_fetch_row() returns a numerical array (
http://ca2.php.net/manual/en/function.mysql-fetch-row.php), but then you are
trying to assign $admin_id using an associative array. Thus, you need to
either return your row as an associative array (
http://ca2.php.net/manual/en/function.mysql-fetch-assoc.php) or assign
$admin_id as a numerical array:
Method 1: Use a numerical array
$result = mysql_query($query);
while($row = mysql_fetch_row($result)) {
$admin_id = $row[1]; // since it's the 2nd item in your SELECT
...
}
OR
Method 2: Use an associative array
$result = mysql_query($query);
while($row = mysql_fetch_assoc($result)) { // returns result row as an
associative array
$admin_id = $row['AdminID'];
}
