SV: [PHP-DB] Detailed Report

2005-10-27 Thread Henrik Hornemann
How about:

$sqlstmt1 = select category, name, code, city from table order by category;
$categry='';
while ($row1 = mysql_fetch_object ($rs1) {
  if ($old_category != $row1-category) {
$ category = $row1-category;
Print($category.br);
  } 
  Print($row1-name. '-' .$row1-code. '-' .$row1-city. \n');
}


Regards Henrik Hornemann



-Oprindelig meddelelse-
Fra: Shahmat Dahlan [mailto:[EMAIL PROTECTED] 
Sendt: 27. oktober 2005 12:29
Til: Danny
Cc: php-db@lists.php.net
Emne: Re: [PHP-DB] Detailed Report

Not sure whether this would be such a good idea. Does anybody else have 
a better one than this?

#main loop, gather distinct / unique category names
$sqlstmt = select distinct(category) as category from table;
$rs = mysql_query ($sqlstmt);
while ($row = mysql_fetch_object ($rs)) {
   print $row-category . \n; # output Customers / Providers
   $sqlstmt1 = select name, code, city from table;
   $rs1 = mysql_query ($sqlstmt);
   # inner loop, display those which category is from the main loop
   while ($row1 = mysql_fetch_object ($rs1) {
  print $row1-name . '-' . $row1-code . '-' . $row1-city . 
\n';# output John - A36 - City
   }
}


Danny wrote:

Hi All,
 I´ve got a connection, to a MySQL db, and get the following
ResultSet(Category | Name | Code | City)
 Customers | John | A36 | New York
Customers | Jason | B45 | Los Angeles
Customers | Max | A36 | Paris
Providers | John | A36 | London
Providers | Mark | B67 | Madrid
And I need the report in the following format:
 Customers
John - A36 - New York
Jason - B45 - Los Angeles
Max - A36 - Paris
 Providers
John - A36 - London
Mark - B67 - Madrid
 Any one can help? I´m a bit stalled
 thx
regards.

  



-- 
Best Regards,

Shahmat Dahlan
Research and Development
SAINS

Mobile: (60)16 882 6130
Office: (60)82 426 733 ext 5512

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SV: [PHP-DB] Detailed Report

2005-10-27 Thread Henrik Hornemann
Hi,

Sorry about the stupid mistakes, should ofcourse have been:

$sqlstmt1 = select category, name, code, city from table order by category;
$rs1 = mysql_query ($sqlstmt1);
$category='';

while ($row1 = mysql_fetch_object ($rs1) {
  if ($category != $row1-category) {
   $category = $row1-category;
   Print($category.br);
  } 
  Print($row1-name. '-' .$row1-code. '-' .$row1-city. \n');
}

Regards Henrik Hornemann


-Oprindelig meddelelse-
Fra: Henrik Hornemann [mailto:[EMAIL PROTECTED] 
Sendt: 27. oktober 2005 14:24
Til: php-db@lists.php.net
Emne: SV: [PHP-DB] Detailed Report

How about:

$sqlstmt1 = select category, name, code, city from table order by category;
$categry='';
while ($row1 = mysql_fetch_object ($rs1) {
  if ($old_category != $row1-category) {
$ category = $row1-category;
Print($category.br);
  } 
  Print($row1-name. '-' .$row1-code. '-' .$row1-city. \n');
}


Regards Henrik Hornemann



-Oprindelig meddelelse-
Fra: Shahmat Dahlan [mailto:[EMAIL PROTECTED] 
Sendt: 27. oktober 2005 12:29
Til: Danny
Cc: php-db@lists.php.net
Emne: Re: [PHP-DB] Detailed Report

Not sure whether this would be such a good idea. Does anybody else have 
a better one than this?

#main loop, gather distinct / unique category names
$sqlstmt = select distinct(category) as category from table;
$rs = mysql_query ($sqlstmt);
while ($row = mysql_fetch_object ($rs)) {
   print $row-category . \n; # output Customers / Providers
   $sqlstmt1 = select name, code, city from table;
   $rs1 = mysql_query ($sqlstmt);
   # inner loop, display those which category is from the main loop
   while ($row1 = mysql_fetch_object ($rs1) {
  print $row1-name . '-' . $row1-code . '-' . $row1-city . 
\n';# output John - A36 - City
   }
}


Danny wrote:

Hi All,
 I´ve got a connection, to a MySQL db, and get the following
ResultSet(Category | Name | Code | City)
 Customers | John | A36 | New York
Customers | Jason | B45 | Los Angeles
Customers | Max | A36 | Paris
Providers | John | A36 | London
Providers | Mark | B67 | Madrid
And I need the report in the following format:
 Customers
John - A36 - New York
Jason - B45 - Los Angeles
Max - A36 - Paris
 Providers
John - A36 - London
Mark - B67 - Madrid
 Any one can help? I´m a bit stalled
 thx
regards.

  



-- 
Best Regards,

Shahmat Dahlan
Research and Development
SAINS

Mobile: (60)16 882 6130
Office: (60)82 426 733 ext 5512

-- 
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php

-- 
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php

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PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php