date:20010903

Re: RE: [PHP-DB] Random

2001-09-03 Thread Russ Michell


Try:

$result = mysql_db_query ($DBName,select * from Contacts order by rand()
LIMIT 0,5 );

As your first example had the '0' and then the '5' after the LIMIT clause, the second 
didn't have 
th any '0'.

Cheers.
Russ

On Fri, 31 Aug 2001 20:59:16 +0100 Seb Frost [EMAIL PROTECTED] wrote:

 I know it's documented as rand() but you might want to try random() just in
 case... I'm sure I read that somewhere
 
 - seb
 
 
 -Original Message-
 From: John Hurleston [mailto:[EMAIL PROTECTED]]
 Sent: 31 August 2001 19:18
 To: [EMAIL PROTECTED]
 Subject: [PHP-DB] Random
 
 
 I would like to get the first 5 random entries out of my mysql database,
 what would be the correct SELECT option?
 
 I adjusted this line
 
 $result = mysql_db_query ($DBName,select * from Contacts LIMIT 0,5 );
 which works correctly. :)
 
 $result = mysql_db_query ($DBName,select * from Contacts order by rand()
 LIMIT 5 );
 Doesn't work at all. :(
 
 
 Can anyone through any light on this subject.
 
 Thanks.
 
 
 
 
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#---#

  Believe nothing - consider everything   
  
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  Anglia Polytechnic University Webteam
  Room 1C 'The Eastings' East Road, Cambridge
  
  e: [EMAIL PROTECTED]
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[PHP-DB] newbie having problem

2001-09-03 Thread George Pitcher


Hi all,

I'm doing some basic php/MySQL stuff (LinuxPPC on Mac7100). I'm gettin g the
following error when I search for somthing.

Supplied argument is not a valid MySQL result resource in
/usr/local/apache/htdocs/bizflyer/Bizflyer_R1.php on line 32

Here's my relevant code:

?php
mysql_connect ('pingu','root@localhost','');
mysql_select_db ('Bizplanes');
if ($Serial == )
 {$Serial = '%';}
if ($Type == )
 {$Type = '%';}
if ($Con == )
 {$Con = '%';}
$result=mysql_query (SELECT * FROM biz WHERE
  ID LIKE '%$Serial%' AND
  Type LIKE '%$Type%'
  Con LIKE '%$Con%'
  ORDER BY ID);
?

TABLE STUFF HERE

?php
if ($row=mysql_fetch_array($result)) { # this is line 32
do {
 print (TRTD);
 print $row['ID'];
 print (TD);
 print $row['Type'];
 print (TD);
 print $row['Con'];
 print (TD);
 print $row['Operator'];
 print (/TD/TR);
} while ($row=mysql_fetch_array($result));
} else {print (Sorry, no aircraft matching your criteria were found.);}
?
=

Any suggestions as to where I'm going wrong?


Regards

George Pitcher

Technical Manager
HERON Project
Napier University
Edinburgh EH10 5DT

[EMAIL PROTECTED]
[EMAIL PROTECTED]
[EMAIL PROTECTED]

http://www.heron.ac.uk

   programmer -  A device for transmuting caffeine into code.




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SV: [PHP-DB] newbie having problem

2001-09-03 Thread Torgil Zechel


Check if mysql_query returns false. If it does (and im pretty sure it do),
use mysql_error to check whats wrong... (Another way is to echo the query
and paste it into the mysql command line client..)

$result = mysql_query(...);

if( $result == FALSE )
{
echo mysql_error();
}

 I'm doing some basic php/MySQL stuff (LinuxPPC on Mac7100). I'm
 gettin g the
 following error when I search for somthing.
 
 Supplied argument is not a valid MySQL result resource in
 /usr/local/apache/htdocs/bizflyer/Bizflyer_R1.php on line 32
 
 Here's my relevant code:
 
 ?php
 mysql_connect ('pingu','root@localhost','');
 mysql_select_db ('Bizplanes');
 if ($Serial == )
  {$Serial = '%';}
 if ($Type == )
  {$Type = '%';}
 if ($Con == )
  {$Con = '%';}
 $result=mysql_query (SELECT * FROM biz WHERE
   ID LIKE '%$Serial%' AND
   Type LIKE '%$Type%'
   Con LIKE '%$Con%'
   ORDER BY ID);
 ?

 TABLE STUFF HERE

 ?php
 if ($row=mysql_fetch_array($result)) { # this is line 32
 do {
  print (TRTD);
  print $row['ID'];
  print (TD);
  print $row['Type'];
  print (TD);
  print $row['Con'];
  print (TD);
  print $row['Operator'];
  print (/TD/TR);
 } while ($row=mysql_fetch_array($result));
 } else {print (Sorry, no aircraft matching your criteria were found.);}
 ?
 =

 Any suggestions as to where I'm going wrong?


 Regards

 George Pitcher

 Technical Manager
 HERON Project
 Napier University
 Edinburgh EH10 5DT

 [EMAIL PROTECTED]
 [EMAIL PROTECTED]
 [EMAIL PROTECTED]

 http://www.heron.ac.uk
 
programmer -  A device for transmuting caffeine into code.
 



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Re: [PHP-DB] newbie having problem

2001-09-03 Thread George Pitcher


I did this directly into MySQL on the LinuxPPC box:

mysql select * from biz where
- ID LIKE '%AXD%' AND
- Type LIKE '%' AND
- Con LIKE '%'
- ORDER BY ID;

and this is the response:


++-+-+--+-+---+
| ID | Type| Con | Operator | History | Notes |
++-+-+--+-+---+
 |DC | Piper Aztec | 27-4169 |  | G-AXDC  |   
 |DL | Piper Twin Comanche | 30-1856 |  | G-AXDL  |   
 |DM | BAE HS 125/400B | 25194   |  | G-AXDM  |   
++-+-+--+-+---+

In these records, what is actually in the ID field is also being displayed in 
'History'.

Not sure if this clarifies things?

George


- Original Message - 
From: Torgil Zechel [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Monday, September 03, 2001 11:38 AM
Subject: SV: [PHP-DB] newbie having problem


 Check if mysql_query returns false. If it does (and im pretty sure it do),
 use mysql_error to check whats wrong... (Another way is to echo the query
 and paste it into the mysql command line client..)
 
 $result = mysql_query(...);
 
 if( $result == FALSE )
 {
 echo mysql_error();
 }
 
  I'm doing some basic php/MySQL stuff (LinuxPPC on Mac7100). I'm
  gettin g the
  following error when I search for somthing.
  
  Supplied argument is not a valid MySQL result resource in
  /usr/local/apache/htdocs/bizflyer/Bizflyer_R1.php on line 32
  
  Here's my relevant code:
  
  ?php
  mysql_connect ('pingu','root@localhost','');
  mysql_select_db ('Bizplanes');
  if ($Serial == )
   {$Serial = '%';}
  if ($Type == )
   {$Type = '%';}
  if ($Con == )
   {$Con = '%';}
  $result=mysql_query (SELECT * FROM biz WHERE
ID LIKE '%$Serial%' AND
Type LIKE '%$Type%'
Con LIKE '%$Con%'
ORDER BY ID);
  ?
 
  TABLE STUFF HERE
 
  ?php
  if ($row=mysql_fetch_array($result)) { # this is line 32
  do {
   print (TRTD);
   print $row['ID'];
   print (TD);
   print $row['Type'];
   print (TD);
   print $row['Con'];
   print (TD);
   print $row['Operator'];
   print (/TD/TR);
  } while ($row=mysql_fetch_array($result));
  } else {print (Sorry, no aircraft matching your criteria were found.);}
  ?
  =
 
  Any suggestions as to where I'm going wrong?
 
 
  Regards
 
  George Pitcher
 
  Technical Manager
  HERON Project
  Napier University
  Edinburgh EH10 5DT
 
  [EMAIL PROTECTED]
  [EMAIL PROTECTED]
  [EMAIL PROTECTED]
 
  http://www.heron.ac.uk
  
 programmer -  A device for transmuting caffeine into code.
  
 
 
 
  _
  Do You Yahoo!?
  Get your free @yahoo.com address at http://mail.yahoo.com
 
 
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[PHP-DB] Re: Search Results - Next 10 records

2001-09-03 Thread Marcus Tobias


Hi Dennis!

You can add the parameter LIMIT to your SQL query.
Syntax: LIMIT $start_record,$record_count
Example: $query_string = select * from $table where keywords like
'%$keyword%' limit 5,3;
This will return 3 records started with the 5. found record.

The startrecord has to increase on next page.
Store the current startrecord in the form and send it to the next page.

bye Marcus


Dennis [EMAIL PROTECTED] schrieb im Newsbeitrag
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
 Trying to figure out how to limit search results to say 10 records and
produce
 a button to show the next 10.  Do you use PHP to do the logic, or should I
make
 another table column for a temporary count?  This is what I'm up to so
far...

 Thanks in advance


 function show_all($table, $keyword=0)// initialize to zero
 {

global $start_record;

if ( !isset( $start_record ) ) $start_record = 0;

   $query_string = select * from $table where keywords like '%$keyword%';

$query_string = select * from $table where keywords like '%$keyword%' limit
$start_record,10;


   $entire_table = mysql_query($query_string);
   $count = 0;
   print(TABLE BORDER=1 width=700\n);
 while ($cols_rows = mysql_fetch_array($entire_table)) {
  $count++;
   if ($count  4){
 print(tr);
 print(td bgcolor=azure);
 print(font size=3);
 print ($cols_rows['num']);
 print(/tdtd bgcolor=azure);
 print(font size=2);
 print ($cols_rows['site_name']);
 print(/tdtd bgcolor=azure);
 print(font size=2);
 print ($cols_rows['keywords']);
 print(/tdtd bgcolor=azure);
 print(font size=2);
 print ($cols_rows['url']);
 print(P/td);
   }

 if ($count  3){
 print(tr);
 print(td bgcolor=silver);
 print(font size=3);
 print ($cols_rows['num']);
 print(/tdtd bgcolor=silver);
 print(font size=2);
 print ($cols_rows['site_name']);
 print(/tdtd bgcolor=silver);
 print(font size=2);
 print ($cols_rows['keywords']);
 print(/tdtd bgcolor=silver);
 print(font size=2);
 print ($cols_rows['url']);
 print(/td);
 $count++;
   }
   }

 print (FORM METHOD='POST' ACTION='search1.php');

print(INPUT TYPE='text' NAME='start_record' VALUE='$start_record+10');

 print (INPUT TYPE='submit' VALUE='Next'/FORM);
   print $count;
 print(/TABLE\n);
 }


 ?




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Re: [PHP-DB] Searching sql database using php

2001-09-03 Thread Marcus Tobias


Hi Devon!

Take a look in the PHP documentation in the function reference for mysql
function.
There you'll find a set of function for connecting to the database and doing
SQL - Querys over the database. If you are not famular with SQL (SELECT -
Statements)
you have to learn :)

bye Marcus

Devon [EMAIL PROTECTED] schrieb im Newsbeitrag
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
 I have a database of only about 25 fields written in MYSQL for our
internal
 client database, I need to be able to search each off these fields using
 php, can some one help or point me in the correct direction?


 Doug Semig [EMAIL PROTECTED] wrote in message
 [EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
  Yes.
 
  If you have a specific question, please post it and I'm sure folks will
be
  willing to try to help you out.
 
  Doug
 
  At 12:50 PM 9/3/01 +1000, Devon wrote:
  As per topic anyone know how to do it?
  
  Cheers
 
  --Doug's Signature File
  ~ ~ ~ I'm proud to work for GLISnet,  Michigan's best ISP ~ ~ ~
  --1.888.445.4763---8am to 8pm every day---http://www.glis.net--
  The views and opinions expressed in this email are my
  own and may not be those of my employer.
 





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[PHP-DB] Re: How can you hide database login passwd in your script?

2001-09-03 Thread Marcus Tobias


Hi Michael!

You can put all passwords and loging into one settings file.
This file will included by all scripts using this datas.
this include file you can put in an seperate directory on the
webserver. this directory you can protect by the webserver.
with apache with .htaccess files with IIS in the admin area.
so your passwords are protected if php crashes. If php crashes
and your webserver is running, the webserver is transfering the source
of php files.

bye Marcus


Michael [EMAIL PROTECTED] schrieb im Newsbeitrag
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
 How can you hide database login passwd in your script?

 I hate having to have scripts with passwords in it that can be viewed by
 others.






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[PHP-DB] Re: Dropping tables in MySQL

2001-09-03 Thread Marcus Tobias


Hi Ali!

- export the database into a dump file.
- strip the table from the dump.
- drop the database
- import the new dump

bye Marcus


Ali Nayeri [EMAIL PROTECTED] schrieb im Newsbeitrag
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
 Hi guys

 I had a question. How do you drop a table in a database in MySQL that has
 spaces in it's name. I created some tables in M$ Access and then exported
 them using MyODBC to MySQL. One of the tables had whitespaces in the name.
 Now MySQL is all screwed up. I want to drop the table but nothing works. I
 tried , '', `', [] but nothing works. Any solutions?

 Thanks

 Ali


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Re: [PHP-DB] Re: Search Results - Next 10 records

2001-09-03 Thread hassan el forkani


hi

...and to be able to know the exact number of pages
do
$result = mysql_query(select count(*));
$nb_pages = $result/$nb_records_per_page;
et voila! add this to marcus's suggestion and you have a full paging script,
if you want some inspiration check out http://phpclasses.upperdesign.com i 
believe they have a class that does exactly that;

regards

At 14:02 03/09/01, Marcus Tobias wrote:
Hi Dennis!

You can add the parameter LIMIT to your SQL query.
Syntax: LIMIT $start_record,$record_count
Example: $query_string = select * from $table where keywords like
'%$keyword%' limit 5,3;
This will return 3 records started with the 5. found record.

The startrecord has to increase on next page.
Store the current startrecord in the form and send it to the next page.

bye Marcus


Dennis [EMAIL PROTECTED] schrieb im Newsbeitrag
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
  Trying to figure out how to limit search results to say 10 records and
produce
  a button to show the next 10.  Do you use PHP to do the logic, or should I
make
  another table column for a temporary count?  This is what I'm up to so
far...
 
  Thanks in advance
 
 
  function show_all($table, $keyword=0)// initialize to zero
  {

 global $start_record;

 if ( !isset( $start_record ) ) $start_record = 0;

$query_string = select * from $table where keywords like '%$keyword%';

$query_string = select * from $table where keywords like '%$keyword%' limit
$start_record,10;


$entire_table = mysql_query($query_string);
$count = 0;
print(TABLE BORDER=1 width=700\n);
  while ($cols_rows = mysql_fetch_array($entire_table)) {
   $count++;
if ($count  4){
  print(tr);
  print(td bgcolor=azure);
  print(font size=3);
  print ($cols_rows['num']);
  print(/tdtd bgcolor=azure);
  print(font size=2);
  print ($cols_rows['site_name']);
  print(/tdtd bgcolor=azure);
  print(font size=2);
  print ($cols_rows['keywords']);
  print(/tdtd bgcolor=azure);
  print(font size=2);
  print ($cols_rows['url']);
  print(P/td);
}
 
  if ($count  3){
  print(tr);
  print(td bgcolor=silver);
  print(font size=3);
  print ($cols_rows['num']);
  print(/tdtd bgcolor=silver);
  print(font size=2);
  print ($cols_rows['site_name']);
  print(/tdtd bgcolor=silver);
  print(font size=2);
  print ($cols_rows['keywords']);
  print(/tdtd bgcolor=silver);
  print(font size=2);
  print ($cols_rows['url']);
  print(/td);
  $count++;
}
}
 
  print (FORM METHOD='POST' ACTION='search1.php');

print(INPUT TYPE='text' NAME='start_record' VALUE='$start_record+10');

  print (INPUT TYPE='submit' VALUE='Next'/FORM);
print $count;
  print(/TABLE\n);
  }
 
 
  ?
 



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[PHP-DB] Re: Search Results - Next 10 records

2001-09-03 Thread Dennis Butler


Thanks very much Marcus.  It worked like a charm.

Dennis

On Mon, 3 Sep 2001 14:02:07 +0200, [EMAIL PROTECTED] (Marcus Tobias) wrote:
 Hi Dennis!
 
 You can add the parameter LIMIT to your SQL query.
 Syntax: LIMIT $start_record,$record_count
 Example: $query_string = select * from $table where keywords like
 '%$keyword%' limit 5,3;
 This will return 3 records started with the 5. found record.
 
 The startrecord has to increase on next page.
 Store the current startrecord in the form and send it to the next page.
 
 bye Marcus
 
 
 Dennis [EMAIL PROTECTED] schrieb im Newsbeitrag
 [EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
  Trying to figure out how to limit search results to say 10 records and
 produce
  a button to show the next 10.  Do you use PHP to do the logic, or should I
 make
  another table column for a temporary count?  This is what I'm up to so
 far...
 
  Thanks in advance
 
 
  function show_all($table, $keyword=0)// initialize to zero
  {
 
 global $start_record;
 
 if ( !isset( $start_record ) ) $start_record = 0;
 
$query_string = select * from $table where keywords like '%$keyword%';
 
 $query_string = select * from $table where keywords like '%$keyword%' limit
 $start_record,10;
 
 
$entire_table = mysql_query($query_string);
$count = 0;
print(TABLE BORDER=1 width=700\n);
  while ($cols_rows = mysql_fetch_array($entire_table)) {
   $count++;
if ($count  4){
  print(tr);
  print(td bgcolor=azure);
  print(font size=3);
  print ($cols_rows['num']);
  print(/tdtd bgcolor=azure);
  print(font size=2);
  print ($cols_rows['site_name']);
  print(/tdtd bgcolor=azure);
  print(font size=2);
  print ($cols_rows['keywords']);
  print(/tdtd bgcolor=azure);
  print(font size=2);
  print ($cols_rows['url']);
  print(P/td);
}
 
  if ($count  3){
  print(tr);
  print(td bgcolor=silver);
  print(font size=3);
  print ($cols_rows['num']);
  print(/tdtd bgcolor=silver);
  print(font size=2);
  print ($cols_rows['site_name']);
  print(/tdtd bgcolor=silver);
  print(font size=2);
  print ($cols_rows['keywords']);
  print(/tdtd bgcolor=silver);
  print(font size=2);
  print ($cols_rows['url']);
  print(/td);
  $count++;
}
}
 
  print (FORM METHOD='POST' ACTION='search1.php');
 
 print(INPUT TYPE='text' NAME='start_record' VALUE='$start_record+10');
 
  print (INPUT TYPE='submit' VALUE='Next'/FORM);
print $count;
  print(/TABLE\n);
  }
 
 
  ?
 
 
 



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Re: [PHP-DB] Use the User Input Critia as part of Query! (Mysql)

2001-09-03 Thread Jason Wong


- Original Message -
From: Jack [EMAIL PROTECTED]
To: [EMAIL PROTECTED]; [EMAIL PROTECTED]
Sent: Monday, September 03, 2001 10:35 PM
Subject: [PHP-DB] Use the User Input Critia as part of Query! (Mysql)


 Dear all
 I had a form which ask user to input the Staff Number to be as part of the
 query, but when i execute it, it seems not working at all

 here is what i wrote in php
 The user input box is : $StaffNum

 $query=select name,department,Leave_From,Leave_To,Leave_Total,Reason from
 leaverequest where Staff_Number=?print($StaffNum);? and authorized
is
 null;

 But said there is error on this line, which is what i expected!!

 So could anyone pls tell me how i can get the varaible into part of the
 Query of Mysql?

Try something like:

$query=select name,department,Leave_From,Leave_To,Leave_Total,Reason from
leaverequest where Staff_Number='$StaffNum' and authorized is
null;


hth
--
Jason Wong
Gremlins Associates
www.gremlins.com.hk
Tel: +852-2573-5033
Fax: +852-2573-5851




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Re: [PHP-DB] newbie having problem

2001-09-03 Thread Jason Wong



- Original Message -
From: George Pitcher [EMAIL PROTECTED]
To: Torgil Zechel [EMAIL PROTECTED]; [EMAIL PROTECTED]
Sent: Monday, September 03, 2001 11:19 PM
Subject: Re: [PHP-DB] newbie having problem


  if( $result == FALSE )
  {
  echo mysql_error();
  }
 Yes - progress - its telling me that no database was selected.

 My code again:
   ?php
   mysql_connect ('pingu','root@localhost','');
   mysql_select_db ('Bizplanes');
   if ($Serial == )
{$Serial = '%';}
   if ($Type == )
{$Type = '%';}
   if ($Con == )
{$Con = '%';}
   $result=mysql_query (SELECT * FROM biz WHERE
 ID LIKE '%$Serial%' AND
 Type LIKE '%$Type%'
 Con LIKE '%$Con%'
 ORDER BY ID);
   ?

You haven't specified a password when connecting to the server -- does it
require one?

Have you tried the code as per manual:

$link = mysql_connect(mysql_host, mysql_login, mysql_password)
or die (Could not connect);
print (Connected successfully);
mysql_select_db (my_database)
or die (Could not select database);

to ensure that you *are* able to connect to the server and also *select* the
database?

hth
--
Jason Wong
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www.gremlins.com.hk
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[PHP-DB] Use the User Input Critia as part of Query! (Mysql)

2001-09-03 Thread Jack


Dear all
I had a form which ask user to input the Staff Number to be as part of the
query, but when i execute it, it seems not working at all

here is what i wrote in php
The user input box is : $StaffNum

$query=select name,department,Leave_From,Leave_To,Leave_Total,Reason from
leaverequest where Staff_Number=?print($StaffNum);? and authorized is
null;

But said there is error on this line, which is what i expected!!

So could anyone pls tell me how i can get the varaible into part of the
Query of Mysql?

Thx
jack
[EMAIL PROTECTED]



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Re: [PHP-DB] Use the User Input Critia as part of Query! (Mysql)

2001-09-03 Thread Hugh Bothwell



Jason Wong [EMAIL PROTECTED] wrote in message
021701c13489$245b1f40$[EMAIL PROTECTED]">news:021701c13489$245b1f40$[EMAIL PROTECTED]...
 - Original Message -
 From: Jack [EMAIL PROTECTED]
 To: [EMAIL PROTECTED]; [EMAIL PROTECTED]
 Sent: Monday, September 03, 2001 10:35 PM
 Subject: [PHP-DB] Use the User Input Critia as part of Query! (Mysql)

  $query=select name,department,Leave_From,Leave_To,
  Leave_Total,Reason from leaverequest where
  Staff_Number=?print($StaffNum);? and authorized
  is null;
 
 Try something like:
 $query=select name,department,Leave_From,Leave_To,
 Leave_Total,Reason from leaverequest where Staff_Number
 ='$StaffNum' and authorized is null;

Also, for security, you would be wise to cast $StaffNum
to int before using it:

$StaffNum = (int) $StaffNum;
$query =
SELECT 
.name,department,Leave_From,Leave_To,
.Leave_Total,Reason 
.FROM leaverequest 
.WHERE Staff_Number=$StaffNum 
.AND authorized IS NULL;

HTH



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[PHP-DB] $3000 Per Hour or $1 Million in 26 weeks? ? How Is This Possible??

2001-09-03 Thread recount665t7y


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Re: [PHP-DB] Warning: Page has Expired

2001-09-03 Thread olinux o


I have the same troubles. I believe that the only
solution is to use GET rather than POST as your FORM
METHOD. There may be another way, but this may work
fine, as long as you are not working with
passwords/sensitive info.

olinux


--- Mad Nas [EMAIL PROTECTED] wrote:
 Hi All
 
 I'm using PHP 4 and MySQL in W2K .
 
 When i submit a form and call a php file, i get this
 message :
 
 __
 
 Warning: Page has Expired


__
Do You Yahoo!?
Get email alerts  NEW webcam video instant messaging with Yahoo! Messenger
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[PHP-DB] newbie needs to format time field

2001-09-03 Thread Eric J Schwinder


This may be a dumb question but here goes:

I have a mySQL database that I am using PHP to interface with.  I have
fields in the database that are DATE and TIME types.  Can I format these
values so that the user sees September 15, 2001  or  2:00 PM  instead of
2001-09-15 and 14:00:00 when I show the values on the web page?

If so, can anyone suggest a reference which will help me do this?  I didn't
find anything in the mySQL or PHP manuals, but maybe I was looking in the
wrong places!  Thanks in advance from a relatively new PHP user!


Eric J Schwinder
eric.AT.bergencomputing.DOT.com



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RE: [PHP-DB] Searching database tables

2001-09-03 Thread Seb Frost


$searchCategory=Company;[insert html code for drop down box instead of
this line]
$searchKey=IBM;

$result = mysql_query(SELECT * from $table WHERE
($searchCategory='$searchKey'));

or something like that anyway


- seb


-Original Message-
From: Devon [mailto:[EMAIL PROTECTED]]
Sent: 04 September 2001 02:15
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Searching database tables


I have a database in mysql that only has one table, that table has approx 25
fields . I need to be able to search for data under each of these
fields.

Field examples:  Company, TechContact, AdminContact, Mobile etc etc

I need I guess a pull down menu that would select a field (c above) then be
able to search under each field, as in search for company name, or mobile
number etc etc.

As of yet I cant get my search working ... does any one have any sample code
or any suggestion?  Any help would be amazing . if you have an answer
post on here or write to me at work [EMAIL PROTECTED]



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[PHP-DB] unable to get more than one value to display in the array

2001-09-03 Thread Christopher


Hi all,

I am hoping someone can help me from my hair pulling.

I am using a class called graphite.php (which I have attached) to generate a
graph,
It pulls the last seven days of hits to a web site from the db and graphs
these as;
Hits (y-axis) vs Day (x-axis) as a line graph.

Unfortunately I am only able to graph the last element in the array. :(

Can anyone explain why ?

Thanks.

Chrisotpher.

Code is below...


?php
// Graph to display the Last Seven Days Hits...
//--
include(graphite.php);

//--
// set the db table variable
//--
$dbsql = stats;
$db = mysql_connect(localhost,test,test);
mysql_select_db(logs, $db);
//--
// SELECT statement to retrieve the count of site visitors
//--
$sql = SELECT count(http_x_forwarded_for) as cnt from $dbsql;
$result = mysql_query($sql);
while ($row = mysql_fetch_array($result)) {
$overall = $row[cnt];
}
//--
// Daily Averages queries
//--
echo Last Seven Daily Averages:BR;
$sql = SELECT count(*) as cnt, DAYNAME(recieved) as dnr,
TO_DAYS(recieved) as tdr
FROM $dbsql
WHERE recieved = DATE_SUB( CURRENT_DATE, INTERVAL 6 day )
GROUP BY dnr,tdr
ORDER BY tdr;
$result = mysql_query($sql);
while ($row = mysql_fetch_array($result)) {
echo $row[dnr] $row[cnt]BR;
$day = $row[dnr];
$count = $row[cnt];
}
//--

$test = new graphite_line();

$test-filename = mygif.gif;
$test-width = 400;
$test-height = 300;
$test-y_axis_label = Hits;
$test-y_lables = xrange(0, 20, 1);
$test-x_axis_label = Days of the Week;
$test-data = array($count);
$test-data_lables = array($day);

$test-init();

$test-prep();
$test-plot(20);

$test-data = array();
$test-graph_color = ImageColorAllocate($test-image,0,255,0);

$test-prep();
$test-plot(20);
$test-close();

echo img src=\mygif.gif\;

?



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SV: [PHP-DB] newbie needs to format time field

2001-09-03 Thread Torgil Zechel


Checkout:
http://www.mysql.com/doc/D/a/Date_and_time_functions.html

the function DATE_FORMAT(date,format) does what you want...

 -Ursprungligt meddelande-
 Från: Eric J Schwinder [mailto:[EMAIL PROTECTED]]
 Skickat: den 3 september 2001 21:39
 Till: [EMAIL PROTECTED]
 Ämne: [PHP-DB] newbie needs to format time field


 This may be a dumb question but here goes:

 I have a mySQL database that I am using PHP to interface with.  I have
 fields in the database that are DATE and TIME types.  Can I format these
 values so that the user sees September 15, 2001  or  2:00 PM
 instead of
 2001-09-15 and 14:00:00 when I show the values on the web page?

 If so, can anyone suggest a reference which will help me do this?
  I didn't
 find anything in the mySQL or PHP manuals, but maybe I was looking in the
 wrong places!  Thanks in advance from a relatively new PHP user!


 Eric J Schwinder
 eric.AT.bergencomputing.DOT.com



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 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]




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RE: [PHP-DB] Warning: Page has Expired

2001-09-03 Thread Dave Watkinson


look in php.ini (or whatever the Linux equivalent is) for
session.cache_limiter and set it to session.cache_limiter = nocache

HTH 8^D

Dave


-Original Message-
From: olinux o [mailto:[EMAIL PROTECTED]]
Sent: 03 September 2001 19:34
To: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Warning: Page has Expired


I have the same troubles. I believe that the only
solution is to use GET rather than POST as your FORM
METHOD. There may be another way, but this may work
fine, as long as you are not working with
passwords/sensitive info.

olinux


--- Mad Nas [EMAIL PROTECTED] wrote:
 Hi All
 
 I'm using PHP 4 and MySQL in W2K .
 
 When i submit a form and call a php file, i get this
 message :
 
 __
 
 Warning: Page has Expired


__
Do You Yahoo!?
Get email alerts  NEW webcam video instant messaging with Yahoo!
Messenger
http://im.yahoo.com

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PHP Database Mailing List (http://www.php.net/)
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To contact the list administrators, e-mail: [EMAIL PROTECTED]


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Re: RE: [PHP-DB] Random

[PHP-DB] newbie having problem

SV: [PHP-DB] newbie having problem

Re: [PHP-DB] newbie having problem

[PHP-DB] Re: Search Results - Next 10 records

Re: [PHP-DB] Searching sql database using php

[PHP-DB] Re: How can you hide database login passwd in your script?

[PHP-DB] Re: Dropping tables in MySQL

Re: [PHP-DB] Re: Search Results - Next 10 records

[PHP-DB] Re: Search Results - Next 10 records

Re: [PHP-DB] Use the User Input Critia as part of Query! (Mysql)

Re: [PHP-DB] newbie having problem

[PHP-DB] Use the User Input Critia as part of Query! (Mysql)

Re: [PHP-DB] Use the User Input Critia as part of Query! (Mysql)

[PHP-DB] $3000 Per Hour or $1 Million in 26 weeks? ? How Is This Possible??

Re: [PHP-DB] Warning: Page has Expired

[PHP-DB] newbie needs to format time field

RE: [PHP-DB] Searching database tables

[PHP-DB] unable to get more than one value to display in the array

SV: [PHP-DB] newbie needs to format time field

RE: [PHP-DB] Warning: Page has Expired

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