Re: [PHP-DB] First letter
Cornelia Boenigk wrote:
Hi Matt
You can teat a string like an Array
$first = $string[0];
$second = $string[1]
and so on.
Actually this method is deprecated. The current best way is to use curly
braces.
$first = $string{0};
$second = $string{1};
Lang
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[PHP-DB] Re: Categories and Items query
Having a Standard only, Deluxe only and Both will lead to problems in the
future. (i.e. what if you add in a Basic feature set?) What I would do is
1. Get rid of the Both row in feature sets.
2. Have another table called feature_set_features or something better. The
table has two columns, The Feature id and the Feature set ID. If a
feature is in one feature set, there in one row in the table. if a Feature
is in Both feature sets, then there are two rows in this table, one for
each feature set.
3. To query this, join features and feature_set_features on feature_id and
use where feature_set_id = 'standard' or whatever if is.
Hopefully I haven't confused you too much.
Lang
Ali Van Doren wrote:
Hello,
I am pretty new to PHP and MySQL, and am
struggling with this particular page I would like
to create.
I am building a page which lists categories of
house features (e.g. concrete, framing,
foundation, etc.) and then the particular features
that fall into those categories (e.g. the concrete
items are driveway, garage floor, sidewalk,
basement; the framing items include manufactured
floor joist, 1/2 OSB on roof, etc.) The tricky
part is that there are two lists of features I
need to produce: standard and deluxe. Some
features span both types, some are particular to
only one feature list.
I have created 3 tables to handle this:
- the feature_sets table has three items: both(1),
standard(2) only and deluxe(3) only
- the feature_categories table holds the 19
categories of features (concrete, framing, etc.)
- the features table holds all of the features (63
total.)
All tables have primary keys, and the features
table has 2 foreign keys corresponding to
featureset and the featurecategory.
What I would like to be able to do is to have the
category appear, and then list the corresponding
features under it, depending on whether it's the
deluxe of standard feature list. Here's my code:
?php
require_once
('/usr/home/fahomes/mysql_connect.php');
$query = SELECT category, feature_description
FROM features as f, feature_categories as fc WHERE
f.feature_category_id=fc.feature_category_id AND
f.featureset_id = '1' OR f.featureset_id = '3'
ORDER BY f.feature_category_id ASC;
$result = @mysql_query ($query);
if ($result) {
while ($row = mysql_fetch_array($result,
MYSQL_NUM)) {
echo tr
td align=\left\$row[0]/td
td
align=\left\ulli$row[1]/ul/td
/tr\n;
}
mysql_free_result ($result);
} else {
echo 'pThe features could not be displayed due
to a system error. Please contact the a
href=\mailto:[EMAIL PROTECTED]webmaster/a./pp'
. mysql_error() . '/p';
}
mysql_close();
?
What I am getting is initially it's creating a row
for each category/feature pair, so the categories
that have more than one feature listed appear in
more than one row. It works fine until record 34
(id 33) when it starts puking out that feature for
10 rows then the next feature for 1 row, then it
jumps back to the feature 33 for another 12 rows,
this time cycling through the categories. It's
quite bizarre:
http://roku.pair.com/fahomes/test/homes_dlx_features.php
So, I actually have two questions:
1) Does anyone know why my query is misbehaving so
badly? When I copy and paste the query into
MyPHPAdmin, it produces the same results, so I
know it's not the php. On thing I do notice is if
I change the query slightly (so that it looks for
records WHERE
f.feature_category_id=fc.feature_category_id AND
f.featureset_id = '1' OR '3' instead of
f.featureset_id = '1' OR f.featureset_id = '3'
it produces equally messed up, but different results.
2) Is there a way that I can display the data the
way I want to display it (list the category once,
then corresponding features underneath, before
going to the next category)? I would imagine it's
a WHILE loop that's inside another WHILE loop -- I
am just not sure how to go about it, and haven't
been able to track down anything in my books.
Thanks for any advice!
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Re: [PHP-DB] $_POST in MySQL query issue...
The reason curly braces are needed are because you are putting an array
element into a string. Consider..
$sBlah = 'blah';
echo $sBlahfoo$sBlahfoo;
PHP will attempt to echo the variable $sBlahfoo twice. if you use
echo {$sBlah}foo{$sBlah}foo;
you have told php explicitly when to look for variable substitution. so the
result will be
blahfooblahfoo
PHP will not look for array elements to substitute in strings, unless you
explicitly tell it to by using curly braces.
$aBlah['blah'] = 'foo';
echo $aBlah['blah'];
result Parse Error
echo {$aBlah['blah']};
result :
foo
Once again, try the manual for more information
http://php.net/manual/en/language.types.string.php
Hope this helps
Lang
Adam Reiswig wrote:
A couple of days ago I placed a post regarding using the $_POST[]
variable in an insert sql query. Both
$sql=insert into $table set Name = '.$_POST['elementName'].';
and
$sql=insert into $table set Name = '{$_POST['elementName']}';
worked perfectly. Thanks to everyone for your help. My question now is
regarding the curly brackets in the 2nd example. Can anyone describe
why using the curly brackets works and/or how php processes them. I
have read quite a bit about php and never come accross thier use in this
way. Thanks again.
-Adam Reiswig
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[PHP-DB] Re: $_POST in MySQL query issue...
$sql=insert into $table set Name = '$_POST[elementName]';
The problem with this is that you need to use curly braces around the
variable being substituted in the string. Also use single quotes around the
array index.
$sql=insert into $table set Name = '{$_POST['elementName']}';
See the manual.. (Variable parsing section)
http://php.net/manual/en/language.types.string.php
Lang
Adam Reiswig wrote:
Greetings to all. I am trying for the life of me to place a $_POST[]
variable in my MySQL query. I am running the latest stable versions of
PHP, MySQL and Apache 2 on my Win2kPro machine. My register_globals are
set to off in my php.ini. My code I am attempting create is basically
as follows:
$table=elements;
$sql=insert into $table set Name = '$elementName';
This works with register_globals set to on. But, I want to be able to
turn that off. My code then, I am guessing, be something as follows:
$table=elements;
$sql=insert into $table set Name = '$_POST[elementName]';
Unfortunately this and every other combination I can think of,
combinations of quotes that is, does not work. I believe the source of
the problem is the quotes within quotes within quotes. I also tried:
$sql='insert into $table set Name = '.$_POST[elementName];
or
$sql=insert into $table set Name = .$_POST['elementName'];
and several other variations.
Can anyone give me some pointers to inserting $_POST[] statements inside
of query statements? I am sure there must be a way but I have spent a
lot of time on this and am really stumped here. Thanks for any help.
-Adam Reiswig
PS if anything here is not clear to you, please let me know and I'll
clarify as I can. Thanks again.
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[PHP-DB] Re: PHP crashing with using Oracle
I had to change ext/oci8/config.m4 Remove the line highlighted in red on this page. http://cvs.php.net/diff.php/php-src/ext/oci8/config.m4?login=2r1=1.37.2.6r2=1.37.2.7ty=h The line is AC_DEFINE(HAVE_OCI8_SHARED_MODE,1,[ ]) Then do configure/make/make install again. That will get the oci8 support to work. I haven't tried the oracle functions. Lang Reuben D. Budiardja wrote: Hello, I installed PHP-4.3.2 with oci8 and oracle support (using --with-oci8 --with-oracle). I recompile apache-1.3.27 to use the libpthread as mentioned here: http://ww.php.net/oci8 Now, I got crashes when using oracle with php. In my apache error logs, I get: [Sat Jul 5 17:24:28 2003] [notice] child pid 17496 exit signal [Segmentation fault (11) [Sat Jul 5 17:24:29 2003] [notice] child pid 17558 exit signal [Segmentation fault (11) and the application would just do nothing (I tried reload, submit, etc). if this happens, the only way I can fix it is by restarting apache. Could anyone give me pointer what's going on here and how to fix it? In case this matters, the Oracle 9i DB server is also in the same box. Somehow it has it's own Apache, but I don't use that. Thanks for any help. RDB -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: Oracle query returning values in sqlplus, but not php.
Upon further investigation, I've found that this was a permissions problem.
In oracle, a user can see everybody's 'PACKAGE', but only their own
'PACKAGE BODY'.
Lang
Lang Sharpe wrote:
Hi all
Here is a test version of my script. It queries the data dictionary view
ALL_SOURCE. In sqlplus it returns all souce code in the database. In php
it returns no rows.
The only odd thing about it is that ALL_SOURCE.TYPE is VARCHAR2(12), and
'PACKAGE BODY' is twelve characters long. When I change 'PACKAGE BODY' to
'PACKAGE', it does return rows.
?php
error_reporting(E_ALL);
$oradb = OCILogon('','','');
$packageStmt = OCIParse($oradb,
SELECT * FROM ALL_SOURCE WHERE TYPE = 'PACKAGE BODY');
OCIExecute($packageStmt);
if(OCIFetchInto($packageStmt,$aLine,OCI_ASSOC)) {
echo Cool Bananas\n;
var_dump($aLine);
} else {
echo No Bananas\n;
}
OCILogoff($oradb);
?
Im using PHP 4.3.2 Linked to Oracle 9.2.0.1.0 libraries connecting to
oracle 9.2.0.1.0 database on another machine. Can someone confirm it is
only me?
Lang
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[PHP-DB] Oracle query returning values in sqlplus, but not php.
Hi all
Here is a test version of my script. It queries the data dictionary view
ALL_SOURCE. In sqlplus it returns all souce code in the database. In php it
returns no rows.
The only odd thing about it is that ALL_SOURCE.TYPE is VARCHAR2(12), and
'PACKAGE BODY' is twelve characters long. When I change 'PACKAGE BODY' to
'PACKAGE', it does return rows.
?php
error_reporting(E_ALL);
$oradb = OCILogon('','','');
$packageStmt = OCIParse($oradb,
SELECT * FROM ALL_SOURCE WHERE TYPE = 'PACKAGE BODY');
OCIExecute($packageStmt);
if(OCIFetchInto($packageStmt,$aLine,OCI_ASSOC)) {
echo Cool Bananas\n;
var_dump($aLine);
} else {
echo No Bananas\n;
}
OCILogoff($oradb);
?
Im using PHP 4.3.2 Linked to Oracle 9.2.0.1.0 libraries connecting to oracle
9.2.0.1.0 database on another machine. Can someone confirm it is only me?
Lang
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