RE: [PHP-DB] Stuck on db entry from select box...

2002-04-05 Thread Rick Emery

jas,

Ya didn't do what I told you to do.  Now, do this:

SELECT NAME=\files\;

$sql = UPDATE $table_name SET file_name=\$files\;

-Original Message-
From: Jas [mailto:[EMAIL PROTECTED]]
Sent: Friday, April 05, 2002 12:01 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Stuck on db entry from select box...


Ok here is my problem and where I am at this far, first thing I needed to do
is to read the contents of a directory and place the results into a select
box in a form (accomplished), now where I am stuck is passing the selection
from said select box to another script so it may be put into a mysql
database table... my code is as follows:
?php
$dir_name = /path/to/images/directory/;
$dir = opendir($dir_name);
$file_list .= pFORM METHOD=\post\ ACTION=\index_done.php3\
SELECT NAME=\files\$file_name;
 while ($file_name = readdir($dir)) {
  if (($file_name != .)  ($file_name !=..)) {
  $file_list .= OPTION VALUE=\$file_name\
NAME=\$file_name\$file_name/OPTION;
  }
 }
 $file_list .= /SELECTbrbrINPUT TYPE=\submit\ NAME=\submit\
VALUE=\select\/FORM/p;
 closedir($dir);
?
within the page... I echo the results like so:
? echo $file_list; ?
so far so good, now on the index_done.php3 my code is put into a require
statement and the required file code is as follows...
?php
$db_name = database_name;
$table_name = cm_index;
$connection = @mysql_connect(localhost, username, password) or die
(Could not connect to database.  Please try again later.);
$db = @mysql_select_db($db_name,$connection) or die (Could not select
database table. Please try again later.);
$sql = UPDATE $table_name SET file_name=\$file_name\;
$result = @mysql_query($sql, $connection) or die (Could not execute query.
Please try again later.);
?
I think I need a new pair of eyes to review this and let me know where I am
going wrong, also on another note I don't want to put the file in the
database table I want to put the path to the file in the database table, any
help would be greatly appriciated. =)
Jas



-- 
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




RE: [PHP-DB] Stuck on db entry from select box...

2002-04-05 Thread Rick Emery

I had a typo:

$sql = UPDATE $table_name SET file_name=\$files\;

-Original Message-
From: Rick Emery [mailto:[EMAIL PROTECTED]]
Sent: Friday, April 05, 2002 12:29 PM
To: 'Jas'; [EMAIL PROTECTED]
Subject: RE: [PHP-DB] Stuck on db entry from select box...


jas,

Ya didn't do what I told you to do.  Now, do this:

SELECT NAME=\files\;

$sql = UPDATE $table_name SET file_name=\$files\;

-Original Message-
From: Jas [mailto:[EMAIL PROTECTED]]
Sent: Friday, April 05, 2002 12:01 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Stuck on db entry from select box...


Ok here is my problem and where I am at this far, first thing I needed to do
is to read the contents of a directory and place the results into a select
box in a form (accomplished), now where I am stuck is passing the selection
from said select box to another script so it may be put into a mysql
database table... my code is as follows:
?php
$dir_name = /path/to/images/directory/;
$dir = opendir($dir_name);
$file_list .= pFORM METHOD=\post\ ACTION=\index_done.php3\
SELECT NAME=\files\$file_name;
 while ($file_name = readdir($dir)) {
  if (($file_name != .)  ($file_name !=..)) {
  $file_list .= OPTION VALUE=\$file_name\
NAME=\$file_name\$file_name/OPTION;
  }
 }
 $file_list .= /SELECTbrbrINPUT TYPE=\submit\ NAME=\submit\
VALUE=\select\/FORM/p;
 closedir($dir);
?
within the page... I echo the results like so:
? echo $file_list; ?
so far so good, now on the index_done.php3 my code is put into a require
statement and the required file code is as follows...
?php
$db_name = database_name;
$table_name = cm_index;
$connection = @mysql_connect(localhost, username, password) or die
(Could not connect to database.  Please try again later.);
$db = @mysql_select_db($db_name,$connection) or die (Could not select
database table. Please try again later.);
$sql = UPDATE $table_name SET file_name=\$file_name\;
$result = @mysql_query($sql, $connection) or die (Could not execute query.
Please try again later.);
?
I think I need a new pair of eyes to review this and let me know where I am
going wrong, also on another note I don't want to put the file in the
database table I want to put the path to the file in the database table, any
help would be greatly appriciated. =)
Jas



-- 
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

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




Re: [PHP-DB] Stuck on db entry from select box...

2002-04-05 Thread Jas

Ok I made the changes you suggested and unfortunately I am getting the same
results... Could not execute query, I manually put entries into the tables
so that the update command would just overwrite the current contents.  Here
is my code this far, I think my problem (correct me if I am wrong) lies in
the fact that each of select boxes does not contain a unique name, needless
to say I am stuck:
?php
$dir_name = /path/to/images/directory/;
$dir = opendir($dir_name);
$file_list .= pFORM METHOD=\post\ ACTION=\index_done.php3\
SELECT NAME=\files\;
 while ($file_name = readdir($dir)) {
  if (($file_name != .)  ($file_name !=..)) {
  $file_list .= OPTION VALUE=\$file_name\
NAME=\$file_name\$file_name/OPTION;
  }
 }
 $file_list .= /SELECTbrbrINPUT TYPE=\submit\ NAME=\submit\
VALUE=\select\/FORM/p;
 closedir($dir);
?
Select boxes are filled with files from specified folder... then echoed to
screen
? echo $file_list; ?
Working fine... when user click on image using said select box it links to
this file, code is as follows:
?php
$db_name = db_name;
$table_name = cm_index;
$connection = mysql_connect(localhost, user_name, password) or die
(Could not connect to database.  Please try again later.);
$db = mysql_select_db($db_name,$connection) or die (Could not select
database table. Please try again later.);
$sql = UPDATE $table_name SET file_name=\$files\;
$result = mysql_query($sql, $connection) or die (Could not execute query.
Please try again later.);
?
Still getting could not execute query... sorry about not knowing where I am
going wrong but still new to these advanced functions.  Thanks in advance,
Jas



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




Re: [PHP-DB] Stuck on db entry from select box...

2002-04-05 Thread Jas

Upon doing a print $sql before the mysql_querie I get this output;
UPDATE cm_index SET ad01_t=$filesCould not execute query. Please try again
later



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




RE: [PHP-DB] Stuck on db entry from select box...

2002-04-05 Thread Bob

I got it to work like this.. I don't know if you didn't want to use INSERT
or something but it sticks the filename in the database table for me..

?php
$dir_name = /home/httpd/html/database/images;
$dir = opendir($dir_name);
$file_list .= pFORM METHOD=\post\ ACTION=\$PHP_SELF\
SELECT NAME=\files\$file_name;
 while ($file_name = readdir($dir)) {
  if (($file_name != .)  ($file_name !=..)) {
  $file_list .= OPTION VALUE=\$file_name\
NAME=\$file_name\$file_name/OPTION;
  }
 }
 $file_list .= /SELECTbrbrINPUT TYPE=\submit\ NAME=\submit\
VALUE=\select\/FORM/p;
 closedir($dir);
?
within the page... I echo the results like so:
? echo $file_list; ?
so far so good, now on the index_done.php3 my code is put into a require
statement and the required file code is as follows...
?php
$db_name = phpTemp;
$table_name = fileList;
$connection = mysql_connect(localhost, notme, getyourown) or die
(Could not connect to database.  Please try again later.);
$db = mysql_select_db($db_name,$connection) or die (Could not select
database table. Database said:  . mysql_error() );
$sql = INSERT INTO $table_name( files ) VALUES( \$files\ );
//$sql = UPDATE $table_name SET files = \$files\;
$result = mysql_query($sql, $connection) or die (Could not execute query.
Database said:  .  mysql_error() );
?

I changed the names of the database and table, stuff like that..


Personally, I would do it like this:

?php

if( !$command ) main();
if( $command == goForIt ) doTheDeed( $files );
exit;

function main()
{
$dir_name = /home/httpd/html/database/images;
$dir = opendir($dir_name);
$file_list .= pFORM METHOD=\post\
ACTION=\$PHP_SELF?command=goForIt\
SELECT NAME=\files\$file_name;
 while ($file_name = readdir($dir)) {
  if (($file_name != .)  ($file_name !=..)) {
  $file_list .= OPTION VALUE=\$file_name\
NAME=\$file_name\$file_name/OPTION;
  }
 }
 $file_list .= /SELECTbrbrINPUT TYPE=\submit\ NAME=\submit\
VALUE=\select\/FORM/p;
 closedir($dir);

echo $file_list;
}

function doTheDeed( $files )
{
$db_name = phpTemp;
$table_name = fileList;
$connection = mysql_connect(localhost, notme, getyourown) or die
(Could not connect to database.  Please try again later.);
$db = mysql_select_db($db_name,$connection) or die (Could not select
database table. Database said:  . mysql_error() );
$sql = INSERT INTO $table_name( files ) VALUES( \$files\ );
//$sql = UPDATE $table_name SET files = \$files\;
$result = mysql_query($sql, $connection);
$error = mysql_error();
if( eregi( uplicate, $error ) ) echo( Sorry, $files is already in the
database.brbr );
else echo( $files added to database successfully.brbr );
echo( a href=\javascript:history.back();\Add another file?/a );
}

?

But it is your project...
Later,

Bob Weaver


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




Re: [PHP-DB] Stuck on db entry from select box...

2002-04-05 Thread Bob

Okay, so I figured out why you might want to use UPDATE instead of SELECT.
And I also found that no matter what I tried, UPDATE doesn't seem to be able
to UPDATE a table if there are no rows in it.. Otherwise it works fine.. So
I found that I had to put an if test in that would check to see if the table
was empty and assign a different value to $sql than if the table wasn't
empty. This is what I ended up with:

?php

if( !$command ) main();
if( $command == goForIt ) doTheDeed( $files );
exit;

function main()
{
$dir_name = /home/httpd/html/database/images;
$dir = opendir($dir_name);
$file_list .= pFORM METHOD=\post\
ACTION=\$PHP_SELF?command=goForIt\
SELECT NAME=\files\$file_name;
 while ($file_name = readdir($dir)) {
  if (($file_name != .)  ($file_name !=..)) {
  $file_list .= OPTION VALUE=\$file_name\
NAME=\$file_name\$file_name/OPTION;
  }
 }
 $file_list .= /SELECTbrbrINPUT TYPE=\submit\ NAME=\submit\
VALUE=\select\/FORM/p;
 closedir($dir);

echo $file_list;
}

function doTheDeed( $files )
{
$db_name = phpTemp;
$table_name = fileList;
$connection = mysql_connect(localhost, root, sinister) or die
(Could not connect to database.  Please try again later.);
$db = mysql_select_db($db_name,$connection) or die (Could not select
database table. Database said:  . mysql_error() );
echo( Executing Query. You submitted font color=maroon$files/font to
the database.br\n );
$getdata = mysql_query( SELECT * FROM fileList );
$number = mysql_num_rows( $getdata );
if( $number = '0' ) $sql = INSERT INTO $table_name( files ) VALUES(
\$files\ );
else $sql = UPDATE $table_name SET files = \$files\;
$result = mysql_query($sql, $connection);
$error = mysql_error();
if( eregi( uplicate, $error ) ) echo( Sorry, $files is already in the
database.brbr\n );
else echo( $files added to database successfully.brbr\n );
echo( a href=\javascript:history.back();\Add another file?/a );
}

?

Later,

Bob

- Original Message -
From: Bob [EMAIL PROTECTED]
To: jas [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
Sent: Friday, April 05, 2002 5:43 PM
Subject: RE: [PHP-DB] Stuck on db entry from select box...


 I got it to work like this.. I don't know if you didn't want to use INSERT
 or something but it sticks the filename in the database table for me..

 ?php
 $dir_name = /home/httpd/html/database/images;
 $dir = opendir($dir_name);
 $file_list .= pFORM METHOD=\post\ ACTION=\$PHP_SELF\
 SELECT NAME=\files\$file_name;
  while ($file_name = readdir($dir)) {
   if (($file_name != .)  ($file_name !=..)) {
   $file_list .= OPTION VALUE=\$file_name\
 NAME=\$file_name\$file_name/OPTION;
   }
  }
  $file_list .= /SELECTbrbrINPUT TYPE=\submit\ NAME=\submit\
 VALUE=\select\/FORM/p;
  closedir($dir);
 ?
 within the page... I echo the results like so:
 ? echo $file_list; ?
 so far so good, now on the index_done.php3 my code is put into a require
 statement and the required file code is as follows...
 ?php
 $db_name = phpTemp;
 $table_name = fileList;
 $connection = mysql_connect(localhost, notme, getyourown) or die
 (Could not connect to database.  Please try again later.);
 $db = mysql_select_db($db_name,$connection) or die (Could not select
 database table. Database said:  . mysql_error() );
 $sql = INSERT INTO $table_name( files ) VALUES( \$files\ );
 //$sql = UPDATE $table_name SET files = \$files\;
 $result = mysql_query($sql, $connection) or die (Could not execute query.
 Database said:  .  mysql_error() );
 ?

 I changed the names of the database and table, stuff like that..


 Personally, I would do it like this:

 ?php

 if( !$command ) main();
 if( $command == goForIt ) doTheDeed( $files );
 exit;

 function main()
 {
 $dir_name = /home/httpd/html/database/images;
 $dir = opendir($dir_name);
 $file_list .= pFORM METHOD=\post\
 ACTION=\$PHP_SELF?command=goForIt\
 SELECT NAME=\files\$file_name;
  while ($file_name = readdir($dir)) {
   if (($file_name != .)  ($file_name !=..)) {
   $file_list .= OPTION VALUE=\$file_name\
 NAME=\$file_name\$file_name/OPTION;
   }
  }
  $file_list .= /SELECTbrbrINPUT TYPE=\submit\ NAME=\submit\
 VALUE=\select\/FORM/p;
  closedir($dir);

 echo $file_list;
 }

 function doTheDeed( $files )
 {
 $db_name = phpTemp;
 $table_name = fileList;
 $connection = mysql_connect(localhost, notme, getyourown) or die
 (Could not connect to database.  Please try again later.);
 $db = mysql_select_db($db_name,$connection) or die (Could not select
 database table. Database said:  . mysql_error() );
 $sql = INSERT INTO $table_name( files ) VALUES( \$files\ );
 //$sql = UPDATE $table_name SET files = \$files\;
 $result = mysql_query($sql, $connection);
 $error = mysql_error();
 if( eregi( uplicate, $error ) ) echo( Sorry, $files is already in the
 database.brbr );
 else echo( $files added to database successfully.brbr );
 echo( a href=\javascript:history.back();\Add another file?/a );
 }

 ?

 But it is your project...
 Later,

 Bob Weaver



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

Re: [PHP-DB] Stuck on db entry from select box...

2002-04-05 Thread Bob

Bug report: You'll need to add another = to the if test..

if( $number == '0' ) $sql = INSERT INTO $table_name( files ) VALUES(
 \$files\ );

Dumb mistake..

Bob

- Original Message -
From: Bob [EMAIL PROTECTED]
To: jas [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
Sent: Friday, April 05, 2002 7:44 PM
Subject: Re: [PHP-DB] Stuck on db entry from select box...


 Okay, so I figured out why you might want to use UPDATE instead of SELECT.
 And I also found that no matter what I tried, UPDATE doesn't seem to be
able
 to UPDATE a table if there are no rows in it.. Otherwise it works fine..
So
 I found that I had to put an if test in that would check to see if the
table
 was empty and assign a different value to $sql than if the table wasn't
 empty. This is what I ended up with:

 ?php

 if( !$command ) main();
 if( $command == goForIt ) doTheDeed( $files );
 exit;

 function main()
 {
 $dir_name = /home/httpd/html/database/images;
 $dir = opendir($dir_name);
 $file_list .= pFORM METHOD=\post\
 ACTION=\$PHP_SELF?command=goForIt\
 SELECT NAME=\files\$file_name;
  while ($file_name = readdir($dir)) {
   if (($file_name != .)  ($file_name !=..)) {
   $file_list .= OPTION VALUE=\$file_name\
 NAME=\$file_name\$file_name/OPTION;
   }
  }
  $file_list .= /SELECTbrbrINPUT TYPE=\submit\ NAME=\submit\
 VALUE=\select\/FORM/p;
  closedir($dir);

 echo $file_list;
 }

 function doTheDeed( $files )
 {
 $db_name = phpTemp;
 $table_name = fileList;
 $connection = mysql_connect(localhost, oops, didanyonesee) or die
 (Could not connect to database.  Please try again later.);
 $db = mysql_select_db($db_name,$connection) or die (Could not select
 database table. Database said:  . mysql_error() );
 echo( Executing Query. You submitted font color=maroon$files/font to
 the database.br\n );
 $getdata = mysql_query( SELECT * FROM fileList );
 $number = mysql_num_rows( $getdata );
 if( $number = '0' ) $sql = INSERT INTO $table_name( files ) VALUES(
 \$files\ );
 else $sql = UPDATE $table_name SET files = \$files\;
 $result = mysql_query($sql, $connection);
 $error = mysql_error();
 if( eregi( uplicate, $error ) ) echo( Sorry, $files is already in the
 database.brbr\n );
 else echo( $files added to database successfully.brbr\n );
 echo( a href=\javascript:history.back();\Add another file?/a );
 }

 ?

 Later,

 Bob

 - Original Message -
 From: Bob [EMAIL PROTECTED]
 To: jas [EMAIL PROTECTED]
 Cc: [EMAIL PROTECTED]
 Sent: Friday, April 05, 2002 5:43 PM
 Subject: RE: [PHP-DB] Stuck on db entry from select box...


  I got it to work like this.. I don't know if you didn't want to use
INSERT
  or something but it sticks the filename in the database table for me..
 
  ?php
  $dir_name = /home/httpd/html/database/images;
  $dir = opendir($dir_name);
  $file_list .= pFORM METHOD=\post\ ACTION=\$PHP_SELF\
  SELECT NAME=\files\$file_name;
   while ($file_name = readdir($dir)) {
if (($file_name != .)  ($file_name !=..)) {
$file_list .= OPTION VALUE=\$file_name\
  NAME=\$file_name\$file_name/OPTION;
}
   }
   $file_list .= /SELECTbrbrINPUT TYPE=\submit\ NAME=\submit\
  VALUE=\select\/FORM/p;
   closedir($dir);
  ?
  within the page... I echo the results like so:
  ? echo $file_list; ?
  so far so good, now on the index_done.php3 my code is put into a require
  statement and the required file code is as follows...
  ?php
  $db_name = phpTemp;
  $table_name = fileList;
  $connection = mysql_connect(localhost, notme, getyourown) or die
  (Could not connect to database.  Please try again later.);
  $db = mysql_select_db($db_name,$connection) or die (Could not select
  database table. Database said:  . mysql_error() );
  $sql = INSERT INTO $table_name( files ) VALUES( \$files\ );
  //$sql = UPDATE $table_name SET files = \$files\;
  $result = mysql_query($sql, $connection) or die (Could not execute
query.
  Database said:  .  mysql_error() );
  ?
 
  I changed the names of the database and table, stuff like that..
 
 
  Personally, I would do it like this:
 
  ?php
 
  if( !$command ) main();
  if( $command == goForIt ) doTheDeed( $files );
  exit;
 
  function main()
  {
  $dir_name = /home/httpd/html/database/images;
  $dir = opendir($dir_name);
  $file_list .= pFORM METHOD=\post\
  ACTION=\$PHP_SELF?command=goForIt\
  SELECT NAME=\files\$file_name;
   while ($file_name = readdir($dir)) {
if (($file_name != .)  ($file_name !=..)) {
$file_list .= OPTION VALUE=\$file_name\
  NAME=\$file_name\$file_name/OPTION;
}
   }
   $file_list .= /SELECTbrbrINPUT TYPE=\submit\ NAME=\submit\
  VALUE=\select\/FORM/p;
   closedir($dir);
 
  echo $file_list;
  }
 
  function doTheDeed( $files )
  {
  $db_name = phpTemp;
  $table_name = fileList;
  $connection = mysql_connect(localhost, notme, getyourown) or die
  (Could not connect to database.  Please try again later.);
  $db = mysql_select_db($db_name,$connection) or die (Could not select
  database table. Database said:  . mysql_error() );
  $sql = INSERT INTO $table_name( files