Re: [PHP-DB] Cookies

2001-04-25 Thread johndmiller

I believe that the cookie part needs to be sent to the browser before the
html.  You can put the set cookies any where in the php portion as long as
the first thing you send is the cookie stuff

setcookie..
html header info...
html

not html header info.
setcookie
html..

Hopes this helps
John

 On Wed, 25 Apr 2001, Tobbe wrote:

 Hi !!

 I'm trying to but down 2 variables ($iCount and $sPhone) into a cookie and I
 used the code example from the php manual from php.net put I only get the
 error messages (down below)

 And how do i get back the var in to the page

 or is there another beter way to save the variables when reloading the page


 ___

setcookie (cookie[two], $iCount);
setcookie (cookie[one], $sPhone);
if (isset ($cookie)) {
while (list ($name, $value) = each ($cookie)) {
echo $name == $valuebr\n; } }

 __
 Warning: Cannot add header information - headers already sent by (output
 started at c:\program files\apache group\apache\htdocs\masscot\delete.php:5)
 in c:\program files\apache group\apache\htdocs\masscot\delete.php on line 51

 Warning: Cannot add header information - headers already sent by (output
 started at c:\program files\apache group\apache\htdocs\masscot\delete.php:5)
 in c:\program files\apache group\apache\htdocs\masscot\delete.php on line 52


 Thanks / Tobbe






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[PHP-DB] MySql select and insert

2001-04-25 Thread johndmiller

Ok here is my problem.  I have this database and the key is comprised of
two fields, filename and path to file.  When I read based on this key, I
get 0 records found. When I insert the same record, it will say that the
key already exists.(which id does).  Why can't MySql findthe record.

Also when I log into MySql and type the search in manually, it doesnot
find it, then when I put a wild card at the end of the path and use like
MySql finds the record.  The File name and Path are derived from the
system so it is not atyping error.  Any know what is up with it

Thanks

John


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Re: [PHP-DB] Variable passing

2001-04-24 Thread johndmiller

Use sessions.  This will store the values on the server rather than
passing them back and forth.  Lots less typing too.


On Tue, 24 Apr 2001, Johannes Janson wrote:

 You can output the passed varibles from the previous page
 in an input type=hidden  in the form of the following page.

 Johannes


 Brinzoi Constantin Aurelian [EMAIL PROTECTED] schrieb im Newsbeitrag
 [EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
  Hi !
 
  Here is the problem: I got 8 pages, shown one after another. Each have 2
  buttons (one for continue and one for abort). When I click to Continue I
  go to the next page. On the last page I have to submit ALL variables from
  these 8 pages to the database.
 
  My test shown that only the variables from the last page are filled-in and
  otherones are empty.
 
  I have tried with another php page that includes all 8 -
  include(page...). Worthless!
 
  What can I do?
 
  TIA,
 
  Constantin Brinzoi
  Dept. Sondaje
  IRECSON
 
 
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[PHP-DB] trouble reading a database (fwd)

2001-04-23 Thread johndmiller


Below is the output and code for my problem.  When the database gets read,
it returns 0 rows.  When it writes, it says it can't because of duplicate
entry.  FLD_FileName and FDL_Path are the Primary key.

Two questions, what I am doing wrong and is there a better way to create a
string of text then the one used below.

TIA
John

Read_sql SELECT * FROM TBL_Picture WHERE FLD_FileName = SailBoat.jpgand
FLD_Path = /var/www/html/slidecollection/WinterCarnival
Result of read Resource id #2
Error Results 0
Number or Rows 0
Insert Sql INSERT INTO TBL_Picture VALUES 
(SailBoat.jpg,/var/www/html/slidecollection/WinterCarnival,
1234,gfgfdgfddg,gfdgfd)
Error Results Duplicate entry 
'/var/www/html/slidecollection/WinterCarnival-SailBoat.jpg'
for key 1Result is of 2 is

//reading the database to see if record already exists
  $result = mysql_select_db (Picture_DB, $link) or die (Could not get the 
database);
  $read_sql  = SELECT * FROM TBL_Picture WHERE FLD_FileName = \;
  $read_sql .= $fn_FileName;
  $read_sql .= \and FLD_Path =\ ;
  $read_sql .= $fn_Path;
  $read_sql .= \;
  echo  Read_sql ;
  echo $read_sql;
  $result = mysql_query($read_sql, $link) or die(could not read the table);
  echo Result of read ;
  echo $result;
  echo  Error Results ;
  echo mysql_errno ($link);
  $number_of_rows = mysql_num_rows($result);
  echo  Number or Rows ;
  echo $number_of_rows;
//if return no rows then we need to add the record
  if  ($number_of_rows == 0){
$insert_sql  = INSERT INTO TBL_Picture  VALUES (\;
$insert_sql .= $fn_FileName;
$insert_sql .= \,\;
$insert_sql .= $fn_Path;
$insert_sql .= \,\;
$insert_sql .= $fn_Year;
$insert_sql .= \,\;
$insert_sql .= $fn_DescriptiveText;
$insert_sql .= \,\;
$insert_sql .= $fn_PictureName;
$insert_sql .= \);
echo  Insert Sql ;
echo $insert_sql;
$result2 = mysql_query($insert_sql, $link);
 // or die(could not write to the table );
  echo  Error Results ;
  echo mysql_error ($link);
echo Result is of 2 is;
echo $result2;




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[PHP-DB] RE: [PHP] trouble reading a database (fwd)

2001-04-23 Thread johndmiller



-- Forwarded message --
Date: Mon, 23 Apr 2001 20:19:57 -0500 (CDT)
From: johndmiller [EMAIL PROTECTED]
To: Jason Murray [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
Subject: RE: [PHP]  trouble reading a database (fwd)

Thanks for the pointer about  ' vs . I have changed my code to used
these.  The code is a lot cleaner.  I am still getting a not found error.
Below is the new cleaner messages that I display.  One point that may be
important, FLD_Name and FLD_Path are the primary key.  Couple of
questions, can I recreate the same key and use just one name (looks like
I could) and would this help with the problem.

Read_sql SELECT * FROM TBL_Picture WHERE FLD_FileName = 'IceCastle.jpg' and FLD_Path = 
'/var/www/html/slidecollection/WinterCarnival'
Error Results
Result of read Resource id #2
Number or Rows 0
Insert Sql INSERT INTO TBL_Picture  VALUES ('IceCastle.jpg', 
'/var/www/html/slidecollection/WinterCarnival', '1984', 'This is a picture of an Ice 
Castle at night.', 'Ice Castle')
Error Results Duplicate entry 
'/var/www/html/slidecollection/WinterCarnival-IceCastle.jpg' for key 1

Can I set up my primary key On Tue, 24 Apr 2001, Jason Murray wrote:

  Read_sql SELECT * FROM TBL_Picture WHERE FLD_FileName =
  SailBoat.jpgand FLD_Path =
 /var/www/html/slidecollection/WinterCarnival

 This is a little hideous.

 For one thing, you should probably use ' and not . It may not
 make a difference to MySQL, but it does to other databases you
 may use in the future.

 For the FLD_Path comparison, you are looking for something that
 specifically STARTS with a space. I'd bet this is why the script
 is getting 0 rows back - probably none of your FLD_Path's start
 with a space.

 Jason




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[PHP-DB] db read

2001-04-23 Thread johndmiller

I am new a database stuff.  I am using mysql 4.0.1 with php on RH7.0.
I am setting up a db with three fields with no-one field being unique. Do
I have to have a unique field?

The purpose of this db is to store a word, name of a file, and path to the
file.  My plan is to use a where clause, but I am not sure if it would be
more efficent to do
WHERE field1=word and field2=filename and field3=path;
or
WHERE field1 = word
and then read through the rows using something like a for each and check
the other two fields.

Thank in Advance
John


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[PHP-DB] trouble reading a database

2001-04-22 Thread johndmiller


Below is the output and code for my problem.  When the database gets read,
it returns 0 rows.  When it writes, it says it can't because of duplicate
entry.  FLD_FileName and FDL_Path are the Primary key.

Two questions, what I am doing wrong and is there a better way to create a
string of text then the one used below.

TIA
John

Read_sql SELECT * FROM TBL_Picture WHERE FLD_FileName = "SailBoat.jpg"and
FLD_Path =" /var/www/html/slidecollection/WinterCarnival"
Result of read Resource id #2
Error Results 0
Number or Rows 0
Insert Sql INSERT INTO TBL_Picture VALUES 
("SailBoat.jpg","/var/www/html/slidecollection/WinterCarnival",
"1234","gfgfdgfddg","gfdgfd")
Error Results Duplicate entry 
'/var/www/html/slidecollection/WinterCarnival-SailBoat.jpg'
for key 1Result is of 2 is

//reading the database to see if record already exists
  $result = mysql_select_db ("Picture_DB", $link) or die ("Could not get the 
database");
  $read_sql  = "SELECT * FROM TBL_Picture WHERE FLD_FileName = \"";
  $read_sql .= $fn_FileName;
  $read_sql .= "\"and FLD_Path =\" ";
  $read_sql .= $fn_Path;
  $read_sql .= "\"";
  echo " Read_sql ";
  echo $read_sql;
  $result = mysql_query($read_sql, $link) or die("could not read the table");
  echo "Result of read ";
  echo $result;
  echo " Error Results ";
  echo mysql_errno ($link);
  $number_of_rows = mysql_num_rows($result);
  echo " Number or Rows ";
  echo $number_of_rows;
//if return no rows then we need to add the record
  if  ($number_of_rows == 0){
$insert_sql  = "INSERT INTO TBL_Picture  VALUES (\"";
$insert_sql .= $fn_FileName;
$insert_sql .= "\",\"";
$insert_sql .= $fn_Path;
$insert_sql .= "\",\"";
$insert_sql .= $fn_Year;
$insert_sql .= "\",\"";
$insert_sql .= $fn_DescriptiveText;
$insert_sql .= "\",\"";
$insert_sql .= $fn_PictureName;
$insert_sql .= "\")";
echo " Insert Sql ";
echo $insert_sql;
$result2 = mysql_query($insert_sql, $link);
 // or die("could not write to the table ");
  echo " Error Results ";
  echo mysql_error ($link);
echo "Result is of 2 is";
echo $result2;




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[PHP-DB] can't connect to mysql

2001-04-19 Thread johndmiller

I just looked at my mysql.sock file and it has a length of zero.  Is this
the right size, I don't think it is.  If not, do I have to reinstall MySQL
to get the file back or can I get it from somewhere else.

FYI:
running RH7.0, Apache

Thanks,
John


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[PHP-DB] MySql won't connect

2001-04-18 Thread johndmiller

I know that the pages that I am using worked.  I created them, then had to
reload my OS.  I saved everything before erasing the HD.  I have restored
the files but now I get a warning message that says MySQL Connection
Failed: Can't connect to local MySQL server through socket 

The mysql.sock does exist, there is also a pid file there pointing to the
current running occurance of MySQL.

I have root set up with a password and I am able to connect to MySQL from
a promopt.  Below is the offending code.  I think there is something that
I missed when I set things up.

FYI: I am running RH7.0 and Apache

you are welcome to check out this page:

http://24.163.168.40/phpinfo.php


  $link = mysql_connect("localhost", "root", "secret") or
die("can't connect");


Last, I did look for the php-mysql module and it is on the system.

Thanks

John


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