[PHP-DB] stuck on stupid...can't find the bloody error...

2004-04-25 Thread Dan Bowkley
okay, please help the newbie idiot.  Forgive me, for I cannot code my way out of a 
paper bag.  

What, O great Oracle of PHP Goodness is wrong with this picture?  When I punch in a 
work order, it spits back a blank page and does nothing whatsoever with the db.  Even 
if I deliberately enter a work order number that already exists.  Help this 
feeble-minded programmer wannabe?

Oh, as to the \' madness down yonder: should I or shouldn't I backslash the single 
quotes?

TIA
Dan



html

headtitleThe Board Lady - Work Order Database 0.1a/title/head

body

?php

define ('DB_USER', '');

define ('DB_PASSWORD', '*');

define ('DB_HOST', 'localhost');

define ('DB_NAME', 'boards');

$dbc = mysql_connect (DB_HOST, DB_USER, DB_PASSWORD) OR die ('Could not connect to 
database: ' . mysql_error());

@mysql_select_db (DB_NAME) OR die ('Could not connect to database: ' . mysql_error());

$page_req=$HTTP_GET_VARS['action'];

if ($page_req == ) {$page_req=0;}

if ($page_req == 0) {

echo SWORD data entrybr\n;

echo form action=\add.php\ method=\get\;

echo Work Order #: input type=\text\ name=\wo_num\br\n;

echo Customer Name: input type=\text\ name=\name\ Phone: input type=\text\ 
name=\phone\br\n;

echo Email Addy: input type=\text\ name=\email\ Date In: input type=\text\ 
name=\date\br\n;

echo Board Type and SN: input type=\text\ name=\board_type\ Last 3 of SN: 
input type=\text\ name=\last_three\br\n;

echo Weight In: input type=\text\ name=\weight_in\ Weight Out: input 
type=\text\ name=\weight_out\br\n;

echo input type=\hidden\ name=\action\ value=\1\\n;

echo INPUT type=\submit\ value=\Add Work Order\ INPUT type=\reset\br\n;

} 

if ($page_req == 1) {

$wo_num=$HTTP_GET_VARS['wo_num'];

$name=$HTTP_GET_VARS['name'];

$phone=$HTTP_GET_VARS['phone'];

$email=$HTTP_GET_VARS['email'];

$date=$HTTP_GET_VARS['date'];

$board_type=$HTTP_GET_VARS['board_type'];

$last_three=$HTTP_GET_VARS['last_three'];

$weight_in=$HTTP_GET_VARS['weight_in'];

$weight_out=$HTTP_GET_VARS['weight_out'];

$query_testingforadupe = SELECT job_no FROM boards WHERE job_no == $job_no ORDER BY 
job_no ASC;

$result_testingforadupe = @mysql_query ($query_testingforadupe);

if ($result_testingforadupe) {

echo That's a duplicate work order number, you ditz. Try again, this time without 
screwing it all up.brbr\n;

echo form action=\add.php\ method=\get\;

echo Work Order #: input type=\text\ name=\wo_num\br\n;

echo Customer Name: input type=\text\ name=\name\ Phone: input type=\text\ 
name=\phone\br\n;

echo Email Addy: input type=\text\ name=\email\ Date In: input type=\text\ 
name=\date\br\n;

echo Board Type and SN: input type=\text\ name=\board_type\ Last 3 of SN: 
input type=\text\ name=\last_three\br\n;

echo Weight In: input type=\text\ name=\weight_in\ Weight Out: input 
type=\text\ name=\weight_out\br\n;

echo input type=\hidden\ name=\action\ value=\1\\n;

echo INPUT type=\submit\ value=\Add Work Order\ INPUT type=\reset\br\n;

}

else {

$query_insert = INSERT INTO boards (wo_num, name, phone, email, date, board_type, 
last_three, weight_in, weight_out) VALUES (\'$wo_num\', \'$name\', \'$phone\', 
\'$email\', \'$date\', \'$board_type\', \'$last_three\', \'$weight_in\', 
\'$weight_out\');

$result_insert = @mysql_query ($query_insert);

if ($result_insert == ) {

echo input type=\hidden\ name=\action\ value=\0\\n;

echo INPUT type=\submit\ value=\Continue\\n;

}

else {echo OOPS! Your programmer is an idiot!\n;}

}}

mysql_close();

?

/body

/html




Re: [PHP-DB] stuck on stupid...can't find the bloody error...

2004-04-25 Thread Rachel Rodriguez
 
 echo form action=\add.php\ method=\get\;
 
 
echoform action=\add.php?action=$page_req\
method=\get\;

Try that first, and see if it fixes your issue.  I'm
shutting down for the day and unfortunately, I didn't
have time to parse through the remainder of script.

~Rachel

=
~Rachel




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[PHP-DB] Using $_SESSION array for large number of variables. Any recommendations?

2004-04-25 Thread Ross Honniball
Hi all,

System : Windows XP / Apache / MySql / PHP

I'm considering using session variables to store a large amount of data. 
How much is 'large'? To be honest, I have no idea at this point - very 
early days.

For arguments sake, lets say the system consisted of 100 users who will all 
store between them say 1000 different variables and objects for a period of 
1 or 2 days.

In a nutshell, reason for storing this info is to save the state of php 
programs.

Is this approach advisable, or are there limits / other considerations when 
using session variables?

I would greatly appreciate peoples comments on potential problems pursuing 
this approach from anyone with extensive experience using session vars.

Thanks ... Ross

. Ross Honniball  JCU Bookshop Cairns Supervisor
. James Cook Uni, McGreggor Rd, Smithfield, Qld. 4878, Australia
. Ph:07.4042.1157  Fx:07.4042.1158   Em:[EMAIL PROTECTED]
. There are no problems. Only solutions.
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[PHP-DB] Re: stuck on stupid...can't find the bloody error...

2004-04-25 Thread Andy Ladouceur


Dan Bowkley wrote:

else {

$query_insert = INSERT INTO boards (wo_num, name, phone, email, date, board_type, last_three, weight_in, weight_out) VALUES (\'$wo_num\', \'$name\', \'$phone\', \'$email\', \'$date\', \'$board_type\', \'$last_three\', \'$weight_in\', \'$weight_out\');
No need to escape the single quotes. Not entirely sure if it's invalid 
or not, but it's unneccessary.
$result_insert = @mysql_query ($query_insert);

if ($result_insert == ) {

echo input type=\hidden\ name=\action\ value=\0\\n;

echo INPUT type=\submit\ value=\Continue\\n;

}

else {echo OOPS! Your programmer is an idiot!\n;}
Try this:

if (([EMAIL PROTECTED]($query_insert)) != FALSE) {
... echo HTML ...
}
else {echo Oops! Your programmer is broken. Oh and here's what MySQL 
said: br /\n.mysql_error();}

Hopefully that works for you. I think the problem may be the escapes 
quotes. Cheers,

Andy

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[PHP-DB] RE: stuck on stupid...can't find the bloody error..

2004-04-25 Thread J. Alejandro Ceballos Z.
Isnt easy to use

	if (! $page_req)

or

	if (strcmp($page_req,0))

instead of

	if ($page_req == )

or

	if ($page_req == 0)

?

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Re: [PHP-DB] Re: Using $_SESSION array for large number of variables. Any recommendations?

2004-04-25 Thread Kelly Hallman
Apr 25 at 7:48pm, Jimmy Brock wrote:
 I would recommend cookies, not session.
 
 Cookies are stored on the users machine, whereas session data is stored on
 the server. If you have a power failure or have to reboot the server session
 data will be lost -- since session data is stored in memory.

First, the default session handler stores data on disk, not in memory.
Not sure if restarting the webserver trashes them or not.

Cookies would not suffice for this much data. The size of each cookie and
the number of cookies allowed per site is limited. Not sure on the actual
limitations in modern browsers, but you probably shouldn't store more than
20 cookies, and they probably should not exceed 1024 bytes per cookie.  
Furthermore, all the cookie data is transmitted from the client on each
hit, so there are practical limitations with bandwidth as well.

-- 
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[PHP-DB] Re: Using $_SESSION array for large number of variables. Any recommendations?

2004-04-25 Thread Jimmy Brock
I would recommend cookies, not session.

Cookies are stored on the users machine, whereas session data is stored on
the server. If you have a power failure or have to reboot the server session
data will be lost -- since session data is stored in memory.

Ross Honniball [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
 Hi all,

 System : Windows XP / Apache / MySql / PHP

 I'm considering using session variables to store a large amount of data.
 How much is 'large'? To be honest, I have no idea at this point - very
 early days.

 For arguments sake, lets say the system consisted of 100 users who will
all
 store between them say 1000 different variables and objects for a period
of
 1 or 2 days.

 In a nutshell, reason for storing this info is to save the state of php
 programs.

 Is this approach advisable, or are there limits / other considerations
when
 using session variables?

 I would greatly appreciate peoples comments on potential problems pursuing
 this approach from anyone with extensive experience using session vars.

 Thanks ... Ross

 . Ross Honniball  JCU Bookshop Cairns Supervisor
 . James Cook Uni, McGreggor Rd, Smithfield, Qld. 4878, Australia
 . Ph:07.4042.1157  Fx:07.4042.1158   Em:[EMAIL PROTECTED]
 . There are no problems. Only solutions.

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RE: [PHP-DB] Re: Using $_SESSION array for large number of variables. Any recommendations?

2004-04-25 Thread Peter Lovatt
Hi,

I would suggest a mysql table for the data. For large amounts it is probably
more efficient - not sure how much resource is needed to manage 100K pieces
of information using $_SESSION, but an indexed mysql table is probably much
quicker.

table

CREATE TABLE `session_data` (
`userID` VARCHAR( 20 ) NOT NULL ,
`var_name` VARCHAR( 100 ) NOT NULL ,
`var_data` VARCHAR( 250 ) NOT NULL ,
PRIMARY KEY ( `userID` , `var_name` )
);

you then retrieve the data at the beginning of each page request.

This is also persistent, so it will be there for as long as needed. If you
have a login system this also allows the user to switch machines or
browsers - $_SESSION is linked to the browser.

From experience this works well.

hth

Peter


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 -Original Message-
 From: Jimmy Brock [mailto:[EMAIL PROTECTED]
 Sent: 26 April 2004 00:48
 To: [EMAIL PROTECTED]
 Subject: [PHP-DB] Re: Using $_SESSION array for large number of
 variables. Any recommendations?


 I would recommend cookies, not session.

 Cookies are stored on the users machine, whereas session data is stored on
 the server. If you have a power failure or have to reboot the
 server session
 data will be lost -- since session data is stored in memory.

 Ross Honniball [EMAIL PROTECTED] wrote in message
 news:[EMAIL PROTECTED]
  Hi all,
 
  System : Windows XP / Apache / MySql / PHP
 
  I'm considering using session variables to store a large amount of data.
  How much is 'large'? To be honest, I have no idea at this point - very
  early days.
 
  For arguments sake, lets say the system consisted of 100 users who will
 all
  store between them say 1000 different variables and objects for a period
 of
  1 or 2 days.
 
  In a nutshell, reason for storing this info is to save the state of php
  programs.
 
  Is this approach advisable, or are there limits / other considerations
 when
  using session variables?
 
  I would greatly appreciate peoples comments on potential
 problems pursuing
  this approach from anyone with extensive experience using session vars.
 
  Thanks ... Ross
 
  . Ross Honniball  JCU Bookshop Cairns Supervisor
  . James Cook Uni, McGreggor Rd, Smithfield, Qld. 4878, Australia
  . Ph:07.4042.1157  Fx:07.4042.1158   Em:[EMAIL PROTECTED]
  . There are no problems. Only solutions.

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[PHP-DB] How to display table columns

2004-04-25 Thread andy amol
hi,
  I want to display the table coulumns along with those values below it. I am only 
able to display the table values, now I want to diplay the corresponding table 
attribut above the value.
 
eg : name age sex
   abc1m
   xyz 2f
 
I want to display name, age and sex.
 
I am using mysql as database.
 
thanks in advance.
 


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[PHP-DB] Re: How to display table columns

2004-04-25 Thread Andy Ladouceur
I think what you may be looking for is the DESCRIBE command. It allows 
you to get information about table columns.

http://dev.mysql.com/doc/mysql/en/DESCRIBE.html

Andy

Andy Amol wrote:
hi,
  I want to display the table coulumns along with those values below it. I am only able to display the table values, now I want to diplay the corresponding table attribut above the value.
 
eg : name age sex
   abc1m
   xyz 2f
 
I want to display name, age and sex.
 
I am using mysql as database.
 
thanks in advance.
 


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Re: [PHP-DB] How to display table columns

2004-04-25 Thread Marcjon Louwersheimer
Well, you would create a table like this
echo tabletrtdname/tdtdage/tdtdsex/td/tr;
Then, you would do a query (assuming you've already made a connection to
your database)...
$result = mysql_query(SELECT name,age,sex FROM database.usertable);
Then, run through  the results...
while ($row = mysql_fetch_assoc($result))
{ 
  echo
  trtd.$row[name]./tdtd.$row[age]./tdtd.$row[sex]./td/tr;
}
Then, end the table like this
echo /table;
There you have it.

- Original message -
From: andy amol [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Date: Sun, 25 Apr 2004 19:47:37 -0700 (PDT)
Subject: [PHP-DB] How to display table columns

hi,
  I want to display the table coulumns along with those values below it.
  I am only able to display the table values, now I want to diplay the
  corresponding table attribut above the value.
 
eg : name age sex
   abc1m
   xyz 2f
 
I want to display name, age and sex.
 
I am using mysql as database.
 
thanks in advance.
 


-
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Yahoo! Photos: High-quality 4x6 digital prints for 25¢
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Re: [PHP-DB] How to display table columns

2004-04-25 Thread John W. Holmes
andy amol wrote:
hi,
  I want to display the table coulumns along with those values below it. I am only able to display the table values, now I want to diplay the corresponding table attribut above the value.
 
eg : name age sex
   abc1m
   xyz 2f
 
I want to display name, age and sex.
http://us2.php.net/manual/en/function.mysql-field-name.php

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