Re: [PHP-DB] Resource id #2
Umm, he is putting them into an array, I quote:
while ($row = mysql_fetch_array($result)) {
$row['Books.Title'];
$row['Books.Author'];
$row['Books.ISBN'];
$row['BookList.dbase'];
$row['BookList.dbase_user'];
$row['BoxSet.BoxSet'];
$row['Category.Category'];
$row['Category.Sub_category'];
$row['Publisher.Publisher'];
$row['AuthUsers.email'];
}
See the while condition?
On Wed, 2002-11-27 at 06:03, Chris Barnes wrote:
You need to put your $result into an array. you can use:
$result_array = mysql_fetch_array($result);
then, if you know the field names in the array, print them like so:
echo $result_array[field1];
echo $result_array[field2];
or if you dont know their names you can refer to their position numbers
starting from 0 e.g.
echo $result_array[0];
echo $result_array[1];
using the position numbers you could put together a quick script to
crawl through the array and print all the fields with a few lines of
code.
On Wed, 2002-11-27 at 05:09, The Cossins Fam wrote:
Hello.
I am using MySQL as a database for a departmental library. I have written
a quick search script, but keep getting resource id #2 as a result to my
search. I have read the online documentation for the
mysql_fetch_array() function and must say, I don't see that I'm missing
anything. However, I've only started programming, much less working with
PHP, so perhaps someone can help me out. Here's my code:
?
$quickSearch = mcse;
$table1 = Books;
$table2 = BookList;
$table3 = BoxSet;
$table4 = Category;
$table5 = Publisher;
$table6 = AuthUsers;
$table7 = CD;
$connection = mysql_connect(localhost, root) or die(Couldn't connect
to the library database.);
$db_select = mysql_select_db(library, $connection) or die(Couldn't
select the library database.);
$search = SELECT * FROM $table1 LEFT JOIN $table2 ON Books.BookListID =
BookList.BookListID
LEFT JOIN $table3 ON Books.BoxSetID = BoxSet.BoxSetID
LEFT JOIN $table4 ON Books.CategoryID = Category.CategoryID
LEFT JOIN $table5 ON Books.PublisherID = Publisher.PublisherID
LEFT JOIN $table6 ON Books.auID = AuthUsers.auID
LEFT JOIN $table7 ON Books.CD = CD.CD_ID
WHERE Books.Title LIKE \%'$quickSearch'%\
OR Books.Author LIKE \%'$quickSearch'%\
OR Books.ISBN LIKE \%'$quickSearch'%\
OR BookList.dbase LIKE \%'$quickSearch'%\
OR BookList.dbase_user LIKE \%'$quickSearch'%\
OR BoxSet.BoxSet LIKE \%'$quickSearch'%\
OR Category.Category LIKE \%'$quickSearch'%\
OR Category.Sub_category LIKE \%'$quickSearch'%\
OR Publisher.Publisher LIKE \%'$quickSearch'%\;
$result = mysql_query($search, $connection) or die(Couldn't search the
library.);
while ($row = mysql_fetch_array($result)) {
$row['Books.Title'];
$row['Books.Author'];
$row['Books.ISBN'];
$row['BookList.dbase'];
$row['BookList.dbase_user'];
$row['BoxSet.BoxSet'];
$row['Category.Category'];
$row['Category.Sub_category'];
$row['Publisher.Publisher'];
$row['AuthUsers.email'];
}
?
I then have some HTML to display the result of the search. I don't
receive any error messages - I just see an empty table from the HTML
code I wrote. I added an echo of the $result to find the resouce id
#2.
Thanks for any help you can provide.
--joel
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Re: [PHP-DB] Resource id #2
But what do all those $row['fieldname'} rows do? Call me ignorant
(you wouldn't be the first), but a statement that simply has a
variable name doesn't DO anything. Should these have echos in front
of them?
--- Adam Voigt [EMAIL PROTECTED] wrote:
Umm, he is putting them into an array, I quote:
while ($row = mysql_fetch_array($result)) {
$row['Books.Title'];
$row['Books.Author'];
$row['Books.ISBN'];
$row['BookList.dbase'];
$row['BookList.dbase_user'];
$row['BoxSet.BoxSet'];
$row['Category.Category'];
$row['Category.Sub_category'];
$row['Publisher.Publisher'];
$row['AuthUsers.email'];
}
See the while condition?
On Wed, 2002-11-27 at 06:03, Chris Barnes wrote:
You need to put your $result into an array. you can use:
$result_array = mysql_fetch_array($result);
then, if you know the field names in the array, print them like
so:
echo $result_array[field1];
echo $result_array[field2];
or if you dont know their names you can refer to their position
numbers
starting from 0 e.g.
echo $result_array[0];
echo $result_array[1];
using the position numbers you could put together a quick script
to
crawl through the array and print all the fields with a few lines
of
code.
On Wed, 2002-11-27 at 05:09, The Cossins Fam wrote:
Hello.
I am using MySQL as a database for a departmental library. I
have written
a quick search script, but keep getting resource id #2 as a
result to my
search. I have read the online documentation for the
mysql_fetch_array() function and must say, I don't see that I'm
missing
anything. However, I've only started programming, much less
working with
PHP, so perhaps someone can help me out. Here's my code:
?
$quickSearch = mcse;
$table1 = Books;
$table2 = BookList;
$table3 = BoxSet;
$table4 = Category;
$table5 = Publisher;
$table6 = AuthUsers;
$table7 = CD;
$connection = mysql_connect(localhost, root) or
die(Couldn't connect
to the library database.);
$db_select = mysql_select_db(library, $connection) or
die(Couldn't
select the library database.);
$search = SELECT * FROM $table1 LEFT JOIN $table2 ON
Books.BookListID =
BookList.BookListID
LEFT JOIN $table3 ON Books.BoxSetID = BoxSet.BoxSetID
LEFT JOIN $table4 ON Books.CategoryID =
Category.CategoryID
LEFT JOIN $table5 ON Books.PublisherID =
Publisher.PublisherID
LEFT JOIN $table6 ON Books.auID = AuthUsers.auID
LEFT JOIN $table7 ON Books.CD = CD.CD_ID
WHERE Books.Title LIKE \%'$quickSearch'%\
OR Books.Author LIKE \%'$quickSearch'%\
OR Books.ISBN LIKE \%'$quickSearch'%\
OR BookList.dbase LIKE \%'$quickSearch'%\
OR BookList.dbase_user LIKE \%'$quickSearch'%\
OR BoxSet.BoxSet LIKE \%'$quickSearch'%\
OR Category.Category LIKE \%'$quickSearch'%\
OR Category.Sub_category LIKE \%'$quickSearch'%\
OR Publisher.Publisher LIKE \%'$quickSearch'%\;
$result = mysql_query($search, $connection) or die(Couldn't
search the
library.);
while ($row = mysql_fetch_array($result)) {
$row['Books.Title'];
$row['Books.Author'];
$row['Books.ISBN'];
$row['BookList.dbase'];
$row['BookList.dbase_user'];
$row['BoxSet.BoxSet'];
$row['Category.Category'];
$row['Category.Sub_category'];
$row['Publisher.Publisher'];
$row['AuthUsers.email'];
}
?
I then have some HTML to display the result of the search. I
don't
receive any error messages - I just see an empty table from the
HTML
code I wrote. I added an echo of the $result to find the
resouce id
#2.
Thanks for any help you can provide.
--joel
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Re: [PHP-DB] Resource id #2
They don't do anything, but my point was, he said that what he pulled
from the DB needed to be put into an array, and I was pointing out, it
already was.
On Wed, 2002-11-27 at 12:23, Mark wrote:
But what do all those $row['fieldname'} rows do? Call me ignorant
(you wouldn't be the first), but a statement that simply has a
variable name doesn't DO anything. Should these have echos in front
of them?
--- Adam Voigt [EMAIL PROTECTED] wrote:
Umm, he is putting them into an array, I quote:
while ($row = mysql_fetch_array($result)) {
$row['Books.Title'];
$row['Books.Author'];
$row['Books.ISBN'];
$row['BookList.dbase'];
$row['BookList.dbase_user'];
$row['BoxSet.BoxSet'];
$row['Category.Category'];
$row['Category.Sub_category'];
$row['Publisher.Publisher'];
$row['AuthUsers.email'];
}
See the while condition?
On Wed, 2002-11-27 at 06:03, Chris Barnes wrote:
You need to put your $result into an array. you can use:
$result_array = mysql_fetch_array($result);
then, if you know the field names in the array, print them like
so:
echo $result_array[field1];
echo $result_array[field2];
or if you dont know their names you can refer to their position
numbers
starting from 0 e.g.
echo $result_array[0];
echo $result_array[1];
using the position numbers you could put together a quick script
to
crawl through the array and print all the fields with a few lines
of
code.
On Wed, 2002-11-27 at 05:09, The Cossins Fam wrote:
Hello.
I am using MySQL as a database for a departmental library. I
have written
a quick search script, but keep getting resource id #2 as a
result to my
search. I have read the online documentation for the
mysql_fetch_array() function and must say, I don't see that I'm
missing
anything. However, I've only started programming, much less
working with
PHP, so perhaps someone can help me out. Here's my code:
?
$quickSearch = mcse;
$table1 = Books;
$table2 = BookList;
$table3 = BoxSet;
$table4 = Category;
$table5 = Publisher;
$table6 = AuthUsers;
$table7 = CD;
$connection = mysql_connect(localhost, root) or
die(Couldn't connect
to the library database.);
$db_select = mysql_select_db(library, $connection) or
die(Couldn't
select the library database.);
$search = SELECT * FROM $table1 LEFT JOIN $table2 ON
Books.BookListID =
BookList.BookListID
LEFT JOIN $table3 ON Books.BoxSetID = BoxSet.BoxSetID
LEFT JOIN $table4 ON Books.CategoryID =
Category.CategoryID
LEFT JOIN $table5 ON Books.PublisherID =
Publisher.PublisherID
LEFT JOIN $table6 ON Books.auID = AuthUsers.auID
LEFT JOIN $table7 ON Books.CD = CD.CD_ID
WHERE Books.Title LIKE \%'$quickSearch'%\
OR Books.Author LIKE \%'$quickSearch'%\
OR Books.ISBN LIKE \%'$quickSearch'%\
OR BookList.dbase LIKE \%'$quickSearch'%\
OR BookList.dbase_user LIKE \%'$quickSearch'%\
OR BoxSet.BoxSet LIKE \%'$quickSearch'%\
OR Category.Category LIKE \%'$quickSearch'%\
OR Category.Sub_category LIKE \%'$quickSearch'%\
OR Publisher.Publisher LIKE \%'$quickSearch'%\;
$result = mysql_query($search, $connection) or die(Couldn't
search the
library.);
while ($row = mysql_fetch_array($result)) {
$row['Books.Title'];
$row['Books.Author'];
$row['Books.ISBN'];
$row['BookList.dbase'];
$row['BookList.dbase_user'];
$row['BoxSet.BoxSet'];
$row['Category.Category'];
$row['Category.Sub_category'];
$row['Publisher.Publisher'];
$row['AuthUsers.email'];
}
?
I then have some HTML to display the result of the search. I
don't
receive any error messages - I just see an empty table from the
HTML
code I wrote. I added an echo of the $result to find the
resouce id
#2.
Thanks for any help you can provide.
--joel
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You can't demand something as a
Re: [PHP-DB] Resource id #2
Add: print_r($row)
In your while loop, that will show you everything that is being returned
with both it's numeric and text based position.
On Tue, 2002-11-26 at 13:09, The Cossins Fam wrote:
Hello.
I am using MySQL as a database for a departmental library. I have written
a quick search script, but keep getting resource id #2 as a result to my
search. I have read the online documentation for the
mysql_fetch_array() function and must say, I don't see that I'm missing
anything. However, I've only started programming, much less working with
PHP, so perhaps someone can help me out. Here's my code:
?
$quickSearch = mcse;
$table1 = Books;
$table2 = BookList;
$table3 = BoxSet;
$table4 = Category;
$table5 = Publisher;
$table6 = AuthUsers;
$table7 = CD;
$connection = mysql_connect(localhost, root) or die(Couldn't connect
to the library database.);
$db_select = mysql_select_db(library, $connection) or die(Couldn't
select the library database.);
$search = SELECT * FROM $table1 LEFT JOIN $table2 ON Books.BookListID =
BookList.BookListID
LEFT JOIN $table3 ON Books.BoxSetID = BoxSet.BoxSetID
LEFT JOIN $table4 ON Books.CategoryID = Category.CategoryID
LEFT JOIN $table5 ON Books.PublisherID = Publisher.PublisherID
LEFT JOIN $table6 ON Books.auID = AuthUsers.auID
LEFT JOIN $table7 ON Books.CD = CD.CD_ID
WHERE Books.Title LIKE \%'$quickSearch'%\
OR Books.Author LIKE \%'$quickSearch'%\
OR Books.ISBN LIKE \%'$quickSearch'%\
OR BookList.dbase LIKE \%'$quickSearch'%\
OR BookList.dbase_user LIKE \%'$quickSearch'%\
OR BoxSet.BoxSet LIKE \%'$quickSearch'%\
OR Category.Category LIKE \%'$quickSearch'%\
OR Category.Sub_category LIKE \%'$quickSearch'%\
OR Publisher.Publisher LIKE \%'$quickSearch'%\;
$result = mysql_query($search, $connection) or die(Couldn't search the
library.);
while ($row = mysql_fetch_array($result)) {
$row['Books.Title'];
$row['Books.Author'];
$row['Books.ISBN'];
$row['BookList.dbase'];
$row['BookList.dbase_user'];
$row['BoxSet.BoxSet'];
$row['Category.Category'];
$row['Category.Sub_category'];
$row['Publisher.Publisher'];
$row['AuthUsers.email'];
}
?
I then have some HTML to display the result of the search. I don't
receive any error messages - I just see an empty table from the HTML
code I wrote. I added an echo of the $result to find the resouce id
#2.
Thanks for any help you can provide.
--joel
_
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RE: [PHP-DB] Resource id #2
Try this...
either...
while ($row = mysql_fetch_array($result)) {
$title = $row['Books.Title'];
$author = $row['Books.Author'];
...
print $title;
}
or...
while ($row = mysql_fetch_array($result)) {
print $row['Title'];
...
}
-Original Message-
From: The Cossins Fam [mailto:[EMAIL PROTECTED]]
Sent: Tuesday, November 26, 2002 1:10 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Resource id #2
Hello.
I am using MySQL as a database for a departmental library. I have written a
quick search script, but keep getting resource id #2 as a result to my
search. I have read the online documentation for the
mysql_fetch_array() function and must say, I don't see that I'm missing
anything. However, I've only started programming, much less working with
PHP, so perhaps someone can help me out. Here's my code:
?
$quickSearch = mcse;
$table1 = Books;
$table2 = BookList;
$table3 = BoxSet;
$table4 = Category;
$table5 = Publisher;
$table6 = AuthUsers;
$table7 = CD;
$connection = mysql_connect(localhost, root) or die(Couldn't connect to
the library database.);
$db_select = mysql_select_db(library, $connection) or die(Couldn't select
the library database.);
$search = SELECT * FROM $table1 LEFT JOIN $table2 ON Books.BookListID =
BookList.BookListID
LEFT JOIN $table3 ON Books.BoxSetID = BoxSet.BoxSetID
LEFT JOIN $table4 ON Books.CategoryID = Category.CategoryID
LEFT JOIN $table5 ON Books.PublisherID = Publisher.PublisherID
LEFT JOIN $table6 ON Books.auID = AuthUsers.auID
LEFT JOIN $table7 ON Books.CD = CD.CD_ID
WHERE Books.Title LIKE \%'$quickSearch'%\
OR Books.Author LIKE \%'$quickSearch'%\
OR Books.ISBN LIKE \%'$quickSearch'%\
OR BookList.dbase LIKE \%'$quickSearch'%\
OR BookList.dbase_user LIKE \%'$quickSearch'%\
OR BoxSet.BoxSet LIKE \%'$quickSearch'%\
OR Category.Category LIKE \%'$quickSearch'%\
OR Category.Sub_category LIKE \%'$quickSearch'%\
OR Publisher.Publisher LIKE \%'$quickSearch'%\;
$result = mysql_query($search, $connection) or die(Couldn't search the
library.);
while ($row = mysql_fetch_array($result)) {
$row['Books.Title'];
$row['Books.Author'];
$row['Books.ISBN'];
$row['BookList.dbase'];
$row['BookList.dbase_user'];
$row['BoxSet.BoxSet'];
$row['Category.Category'];
$row['Category.Sub_category'];
$row['Publisher.Publisher'];
$row['AuthUsers.email'];
}
?
I then have some HTML to display the result of the search. I don't receive
any error messages - I just see an empty table from the HTML code I wrote.
I added an echo of the $result to find the resouce id #2.
Thanks for any help you can provide.
--joel
_
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Re: [PHP-DB] Resource id #2
On Wednesday 27 November 2002 02:09, The Cossins Fam wrote:
Hello.
I am using MySQL as a database for a departmental library. I have written
a quick search script, but keep getting resource id #2 as a result to my
search. I have read the online documentation for the
mysql_fetch_array() function and must say, I don't see that I'm missing
anything. However, I've only started programming, much less working with
PHP, so perhaps someone can help me out. Here's my code:
?
$quickSearch = mcse;
$table1 = Books;
$table2 = BookList;
$table3 = BoxSet;
$table4 = Category;
$table5 = Publisher;
$table6 = AuthUsers;
$table7 = CD;
$connection = mysql_connect(localhost, root) or die(Couldn't connect
to the library database.);
$db_select = mysql_select_db(library, $connection) or die(Couldn't
select the library database.);
$search = SELECT * FROM $table1 LEFT JOIN $table2 ON Books.BookListID =
BookList.BookListID
LEFT JOIN $table3 ON Books.BoxSetID = BoxSet.BoxSetID
LEFT JOIN $table4 ON Books.CategoryID = Category.CategoryID
LEFT JOIN $table5 ON Books.PublisherID = Publisher.PublisherID
LEFT JOIN $table6 ON Books.auID = AuthUsers.auID
LEFT JOIN $table7 ON Books.CD = CD.CD_ID
WHERE Books.Title LIKE \%'$quickSearch'%\
OR Books.Author LIKE \%'$quickSearch'%\
OR Books.ISBN LIKE \%'$quickSearch'%\
OR BookList.dbase LIKE \%'$quickSearch'%\
OR BookList.dbase_user LIKE \%'$quickSearch'%\
OR BoxSet.BoxSet LIKE \%'$quickSearch'%\
OR Category.Category LIKE \%'$quickSearch'%\
OR Category.Sub_category LIKE \%'$quickSearch'%\
OR Publisher.Publisher LIKE \%'$quickSearch'%\;
It's always a good idea to print out your query so you can visually check
whether it looks OK. So:
print $search;
$result = mysql_query($search, $connection) or die(Couldn't search the
library.);
It's also a good idea to see what really went wrong by using mysql_error().
IE:
if (($result = mysql_query($search, $connection) === FALSE)) {
print mysql_error();
die(Couldn't search the library.);
}
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RE: [PHP-DB] resource id#2 - ????
The quick answer is that that is what it's supposed to return... that's all
the result is. A nicer, longer answer is to give you some of my code so you
can see one way that you actually get the data out (I use sybase_fetch_row,
but you can also use db_fetch_array which returns an associated array or
something - check the manual):
$dataResult = sybase_query($sql);
while ($array_ref = sybase_fetch_row($dataResult)) {
if ($array_ref[4] == 1) {
if ($array_ref[0] == $male ) {
$left[$i] = $array_ref[2];
$names[$i] = $array_ref[3];
$ages[$i] = $array_ref[1];
$i++;
}
elseif ($array_ref[0] == $female ) {
$right[$j] = $array_ref[2];
$j++;
}
}
}//while
Last note - the index of the array ref refers to something in your SQL
starting with 0 (so my sql started with select sex...)
-Natalie
-Original Message-
From: Jas [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, June 12, 2002 1:42 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] resource id#2 -
Not sure how to over come this, the result of a database query keeps giving
me this: ?php
/* Get Ip address, where they came from, and stamp the time */ if
(getenv(HTTP_X_FORWARDED_FOR)){
$ipaddy = getenv(HTTP_X_FORWARDED_FOR);
} else {
$ipaddy = getenv(REMOTE_ADDR); }
$referrer = $HTTP_REFERER;
$date_stamp = date(Y-m-d H:i:s);
/* Start session, and check registered variables */ session_start(); if
(isset($HTTP_SESSION_VARS['ipaddy']) ||
isset($HTTP_SESSION_VARS['referrer']) ||
isset($HTTP_SESSION_VARS['date_stamp'])) {
$main = Video clips stored in Database;
/* Insert client info to database table */
require('/path/to/connection/class/con.inc');
$table_sessions = dev_sessions;
$sql_insert = INSERT INTO $table_sessions
(ipaddy,referrer,date_stamp) VALUES ('$ipaddy','$referrer','$date_stamp');
$result = @mysql_query($sql_insert,$dbh) or die(Couldn't execute
insert to database!);
/* Pull video info from database table into an array */
$table_content = dev_videos;
$sql_content = @mysql_query(SELECT * FROM $table_content,$dbh);
while ($row = @mysql_fetch_array($record)) {
$id = $row['id'];
$file_name = $row['file_name'];
$file_size = $row['file_size'];
$file_properties = $row['file_properties'];
$file_codec = $row['file_codec'];
$file_author = $row['file_author'];
$result = @mysql_query($sql_content,$dbh) or die(Couldn't execute
query on database!);
/* loop through records and print results of array into table */
$current .=
trtd$id/tdtd$file_name/tdtd$file_size/tdtd$file_properties
/tdtd$file_codec/tdtd$file_author/td/tr; }
} else {
/* Start a new session and register variables */
session_start();
session_register('ipaddy');
session_register('referrer');
session_register('date_stamp'); }
?
Just to make sure everything is working correctly I have echoed every
statement to the screen so I may see what's going on like so: ?php echo
$main; ? table width=75% border=1 ?php echo $current; ? // my
results should be here but instead I get Resource ID #2 printed wtf?
/tablebr ?php echo $ipaddy; ?br ?php echo $referrer; ?br ?php
echo $date_stamp; ?br ?php echo $sql_insert; ?br ?php echo
$sql_content; ?br ?php echo $id; ?br ?php echo $file_name; ?br
?php echo $file_size; ?br
Any help would be great!
Jas
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Re: [PHP-DB] resource id#2 - ????
At 11:42 -0600 6/12/02, Jas wrote:
Not sure how to over come this, the result of a database query keeps giving
me this:
?php
/* Get Ip address, where they came from, and stamp the time */
if (getenv(HTTP_X_FORWARDED_FOR)){
$ipaddy = getenv(HTTP_X_FORWARDED_FOR);
} else {
$ipaddy = getenv(REMOTE_ADDR); }
$referrer = $HTTP_REFERER;
$date_stamp = date(Y-m-d H:i:s);
/* Start session, and check registered variables */
session_start();
if (isset($HTTP_SESSION_VARS['ipaddy']) ||
isset($HTTP_SESSION_VARS['referrer']) ||
isset($HTTP_SESSION_VARS['date_stamp'])) {
$main = Video clips stored in Database;
/* Insert client info to database table */
require('/path/to/connection/class/con.inc');
$table_sessions = dev_sessions;
$sql_insert = INSERT INTO $table_sessions
(ipaddy,referrer,date_stamp) VALUES ('$ipaddy','$referrer','$date_stamp');
$result = mysql_query($sql_insert,$dbh) or die(Couldn't execute
insert to database!);
/* Pull video info from database table into an array */
$table_content = dev_videos;
$sql_content = mysql_query(SELECT * FROM $table_content,$dbh);
while ($row = mysql_fetch_array($record)) {
$record isn't defined anywhere. Do you mean $sql_content?
$id = $row['id'];
$file_name = $row['file_name'];
$file_size = $row['file_size'];
$file_properties = $row['file_properties'];
$file_codec = $row['file_codec'];
$file_author = $row['file_author'];
$result = mysql_query($sql_content,$dbh) or die(Couldn't execute
query on database!);
/* loop through records and print results of array into table */
$current .=
trtd$id/tdtd$file_name/tdtd$file_size/tdtd$file_properties
/tdtd$file_codec/tdtd$file_author/td/tr; }
} else {
/* Start a new session and register variables */
session_start();
session_register('ipaddy');
session_register('referrer');
session_register('date_stamp'); }
?
Just to make sure everything is working correctly I have echoed every
statement to the screen so I may see what's going on like so:
?php echo $main; ?
table width=75% border=1
?php echo $current; ? // my results should be here but instead I get
Resource ID #2 printed wtf?
/tablebr
?php echo $ipaddy; ?br
?php echo $referrer; ?br
?php echo $date_stamp; ?br
?php echo $sql_insert; ?br
?php echo $sql_content; ?br
?php echo $id; ?br
?php echo $file_name; ?br
?php echo $file_size; ?br
Any help would be great!
Jas
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RE: [PHP-DB] resource id#2 - ????
I totally didn't even see that section of code. I am sorry I wrote back
like you were an idiot! I think we know who the real idiot is today... :-)
Sorry again!
-Natalie
-Original Message-
From: Paul DuBois [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, June 12, 2002 1:53 PM
To: Jas; [EMAIL PROTECTED]
Subject: Re: [PHP-DB] resource id#2 -
At 11:42 -0600 6/12/02, Jas wrote:
Not sure how to over come this, the result of a database query keeps
giving me this: ?php
/* Get Ip address, where they came from, and stamp the time */
if (getenv(HTTP_X_FORWARDED_FOR)){
$ipaddy = getenv(HTTP_X_FORWARDED_FOR);
} else {
$ipaddy = getenv(REMOTE_ADDR); }
$referrer = $HTTP_REFERER;
$date_stamp = date(Y-m-d H:i:s);
/* Start session, and check registered variables */ session_start();
if (isset($HTTP_SESSION_VARS['ipaddy']) ||
isset($HTTP_SESSION_VARS['referrer']) ||
isset($HTTP_SESSION_VARS['date_stamp'])) {
$main = Video clips stored in Database;
/* Insert client info to database table */
require('/path/to/connection/class/con.inc');
$table_sessions = dev_sessions;
$sql_insert = INSERT INTO $table_sessions
(ipaddy,referrer,date_stamp) VALUES ('$ipaddy','$referrer','$date_stamp');
$result = @mysql_query($sql_insert,$dbh) or die(Couldn't
execute insert to database!);
/* Pull video info from database table into an array */
$table_content = dev_videos;
$sql_content = @mysql_query(SELECT * FROM $table_content,$dbh);
while ($row = @mysql_fetch_array($record)) {
$record isn't defined anywhere. Do you mean $sql_content?
$id = $row['id'];
$file_name = $row['file_name'];
$file_size = $row['file_size'];
$file_properties = $row['file_properties'];
$file_codec = $row['file_codec'];
$file_author = $row['file_author'];
$result = @mysql_query($sql_content,$dbh) or die(Couldn't
execute query on database!);
/* loop through records and print results of array into table */
$current .=
trtd$id/tdtd$file_name/tdtd$file_size/tdtd$file_proper
ties
/tdtd$file_codec/tdtd$file_author/td/tr; }
} else {
/* Start a new session and register variables */
session_start();
session_register('ipaddy');
session_register('referrer');
session_register('date_stamp'); }
?
Just to make sure everything is working correctly I have echoed every
statement to the screen so I may see what's going on like so:
?php echo $main; ?
table width=75% border=1
?php echo $current; ? // my results should be here but instead I get
Resource ID #2 printed wtf?
/tablebr
?php echo $ipaddy; ?br
?php echo $referrer; ?br
?php echo $date_stamp; ?br
?php echo $sql_insert; ?br
?php echo $sql_content; ?br
?php echo $id; ?br
?php echo $file_name; ?br
?php echo $file_size; ?br
Any help would be great!
Jas
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Re: [PHP-DB] resource id#2 - ????
Thanks a ton, I was really looking right past that variable.
Jas
Paul Dubois [EMAIL PROTECTED] wrote in message
news:p05111763b92d3addd3b4@[192.168.0.33]...
At 11:42 -0600 6/12/02, Jas wrote:
Not sure how to over come this, the result of a database query keeps
giving
me this:
?php
/* Get Ip address, where they came from, and stamp the time */
if (getenv(HTTP_X_FORWARDED_FOR)){
$ipaddy = getenv(HTTP_X_FORWARDED_FOR);
} else {
$ipaddy = getenv(REMOTE_ADDR); }
$referrer = $HTTP_REFERER;
$date_stamp = date(Y-m-d H:i:s);
/* Start session, and check registered variables */
session_start();
if (isset($HTTP_SESSION_VARS['ipaddy']) ||
isset($HTTP_SESSION_VARS['referrer']) ||
isset($HTTP_SESSION_VARS['date_stamp'])) {
$main = Video clips stored in Database;
/* Insert client info to database table */
require('/path/to/connection/class/con.inc');
$table_sessions = dev_sessions;
$sql_insert = INSERT INTO $table_sessions
(ipaddy,referrer,date_stamp) VALUES
('$ipaddy','$referrer','$date_stamp');
$result = @mysql_query($sql_insert,$dbh) or die(Couldn't execute
insert to database!);
/* Pull video info from database table into an array */
$table_content = dev_videos;
$sql_content = @mysql_query(SELECT * FROM $table_content,$dbh);
while ($row = @mysql_fetch_array($record)) {
$record isn't defined anywhere. Do you mean $sql_content?
$id = $row['id'];
$file_name = $row['file_name'];
$file_size = $row['file_size'];
$file_properties = $row['file_properties'];
$file_codec = $row['file_codec'];
$file_author = $row['file_author'];
$result = @mysql_query($sql_content,$dbh) or die(Couldn't execute
query on database!);
/* loop through records and print results of array into table */
$current .=
trtd$id/tdtd$file_name/tdtd$file_size/tdtd$file_properties
/tdtd$file_codec/tdtd$file_author/td/tr; }
} else {
/* Start a new session and register variables */
session_start();
session_register('ipaddy');
session_register('referrer');
session_register('date_stamp'); }
?
Just to make sure everything is working correctly I have echoed every
statement to the screen so I may see what's going on like so:
?php echo $main; ?
table width=75% border=1
?php echo $current; ? // my results should be here but instead I get
Resource ID #2 printed wtf?
/tablebr
?php echo $ipaddy; ?br
?php echo $referrer; ?br
?php echo $date_stamp; ?br
?php echo $sql_insert; ?br
?php echo $sql_content; ?br
?php echo $id; ?br
?php echo $file_name; ?br
?php echo $file_size; ?br
Any help would be great!
Jas
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Re: [PHP-DB] Resource ID#2 ????
On Thursday 30 May 2002 00:17, Jas wrote: Ok here is my problem, I set this up so a user selects a name form a select box and that name or $user_id is then passed to this page so the user can edit the contact info etc. However it does not pull the selected $user_id and place each field into my form boxes, all I get is a Resource ID #2. Not sure how I can over come this... $table = auth_users; $record = mysql_query(SELECT * FROM $table WHERE user_id = '$user_id',$dbh); Make liberal use of error checking (see manual) and mysql_error(). -- Jason Wong - Gremlins Associates - www.gremlins.com.hk Open Source Software Systems Integrators * Web Design Hosting * Internet Intranet Applications Development * /* grep me no patterns and I'll tell you no lines. */ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Resource ID#2 ????
If you look at the previously posted code at the bottom of the form there is a echo for the sql select statement that is echoing Resource id #2 on the page. Now that error is the correct field id number in the database, I am just not sure how to itemize the data from that table, at least I think. Any help would be great! Jas Jason Wong [EMAIL PROTECTED] wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... On Thursday 30 May 2002 00:17, Jas wrote: Ok here is my problem, I set this up so a user selects a name form a select box and that name or $user_id is then passed to this page so the user can edit the contact info etc. However it does not pull the selected $user_id and place each field into my form boxes, all I get is a Resource ID #2. Not sure how I can over come this... $table = auth_users; $record = @mysql_query(SELECT * FROM $table WHERE user_id = '$user_id',$dbh); Make liberal use of error checking (see manual) and mysql_error(). -- Jason Wong - Gremlins Associates - www.gremlins.com.hk Open Source Software Systems Integrators * Web Design Hosting * Internet Intranet Applications Development * /* grep me no patterns and I'll tell you no lines. */ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Resource ID#2 ????
I could be missing something, but it looks like you are using the result of
the mysql_query as the actual result. It actually returns some weird
identifier. To access the real info you'd have to use something like
mysql_fetch_array to get it.
Check out this and see if it helps:
http://www.php.net/manual/en/function.mysql-query.php
-Natalie
-Original Message-
From: Jas [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, May 29, 2002 12:18 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Resource ID#2
Ok here is my problem, I set this up so a user selects a name form a select
box and that name or $user_id is then passed to this page so the user can
edit the contact info etc. However it does not pull the selected $user_id
and place each field into my form boxes, all I get is a Resource ID #2. Not
sure how I can over come this...
$table = auth_users;
$record = @mysql_query(SELECT * FROM $table WHERE user_id =
'$user_id',$dbh);
$var_form .= table width=\100%\ border=\0\ cellpadding=\7\form
name=\$user_id\ method=\post\ action=\del_account.php\
trtd width=\20%\ colspan=\2\bEdit Account
$user_id/b/td/tr
trtd width=\20%\First Name:/tdtd width=\80%\input
type=\text\ name=\$f_name\ size=\30\ maxlength=\30\
value=\$f_name\font class=\copyright\i.e. John/font/td/tr
trtd width=\20%\Last Name:/tdtd width=\80%\input
type=\text\ name=\$l_name\ size=\30\ maxlength=\30\
value=\$l_name\font class=\copyright\i.e. Doe/font/td/tr
trtd width=\20%\Email:/tdtd width=\80%\input
type=\text\ name=\$email_addy\ size=\30\ maxlength=\30\
value=\$email_addy\font class=\copyright\i.e.
[EMAIL PROTECTED]/font/td/tr
trtd width=\20%\User Name:/tdtd width=\80%\input
type=\text\ name=\$un\ size=\30\ maxlength=\30\ value=\$un\font
class=\copyright\i.e. j-doe/font/td/tr
trtd width=\20%\Password:/tdtd width=\80%\input
type=\password\ name=\$pw\ size=\30\ maxlength=\30\font
class=\copyright\(password must be alpha-numeric, i.e.
pAs5w0rd)/font/td/tr
trtd width=\20%\Confirm Password:/tdtd
width=\80%\input type=\password\ name=\$pw\ size=\30\
maxlength=\30\font class=\copyright\please confirm password
entered/font/td/tr
trtd width=\20%\nbsp;/tdtd width=\80%\input
type=\submit\ name=\add\ value=\edit user\nbsp;nbsp;input
type=\reset\ name=\reset\ value=\reset\/td/tr
/form/table;
echo $record;
} else {
blah blah
}
Thanks in advance,
Jas
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Re: [PHP-DB] Resource ID#2 ????
On Thursday 30 May 2002 00:41, Jas wrote: If you look at the previously posted code at the bottom of the form there is a echo for the sql select statement that is echoing Resource id #2 on the page. Now that error is the correct field id number in the database, I am just not sure how to itemize the data from that table, at least I think. Any help would be great! You are confused -- #2 just happens to coincide with your field id. You're not RFTM. mysql_query() returns a result_id (Resource id #2). You feed this into a mysql_fetch_row() (or any of the other similar functions) to actually retrieve records. -- Jason Wong - Gremlins Associates - www.gremlins.com.hk Open Source Software Systems Integrators * Web Design Hosting * Internet Intranet Applications Development * /* 1: No code table for op: ++post */ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Resource ID#2 ????
How can I get the form below to list the single record in a db that matches
the request of $user_id?
$table = auth_users;
$record = @mysql_query(SELECT * FROM $table WHERE user_id =
'$user_id',$dbh);
while ($row = mysql_fetch_row($record)) {
$user_id = $row['user_id'];
$f_name = $row['f_name'];
$l_name = $row['l_name'];
$email_addy = $row['email_addy'];
$un = $row['un'];
$pw = $row['pw']; }
$var_form .= table width=\100%\ border=\0\ cellpadding=\7\form
name=\$user_id\ method=\post\ action=\del_account.php\
trtd width=\20%\ colspan=\2\bEdit Account
$user_id/b/td/tr
trtd width=\20%\First Name:/tdtd width=\80%\input
type=\text\ name=\f_name\ size=\30\ maxlength=\30\
value=\$f_name\font class=\copyright\i.e. John/font/td/tr
trtd width=\20%\Last Name:/tdtd width=\80%\input
type=\text\ name=\l_name\ size=\30\ maxlength=\30\
value=\$l_name\font class=\copyright\i.e. Doe/font/td/tr
trtd width=\20%\Email:/tdtd width=\80%\input
type=\text\ name=\email_addy\ size=\30\ maxlength=\30\
value=\$email_addy\font class=\copyright\i.e.
[EMAIL PROTECTED]/font/td/tr
trtd width=\20%\User Name:/tdtd width=\80%\input
type=\text\ name=\un\ size=\30\ maxlength=\30\ value=\$un\font
class=\copyright\i.e. j-doe/font/td/tr
trtd width=\20%\Password:/tdtd width=\80%\input
type=\password\ name=\pw\ size=\30\ maxlength=\30\font
class=\copyright\(password must be alpha-numeric, i.e.
pAs5w0rd)/font/td/tr
trtd width=\20%\Confirm Password:/tdtd
width=\80%\input type=\password\ name=\pw\ size=\30\
maxlength=\30\font class=\copyright\please confirm password
entered/font/td/tr
trtd width=\20%\nbsp;/tdtd width=\80%\input
type=\submit\ name=\add\ value=\edit user\nbsp;nbsp;input
type=\reset\ name=\reset\ value=\reset\/td/tr
/form/table;
echo $record;
Thanks in advance,
Jas
Natalie Leotta [EMAIL PROTECTED] wrote in message
7546CB15C0A1D311BF0D0004AC4C4B0C024AC420@SSIMSEXCHNG">news:7546CB15C0A1D311BF0D0004AC4C4B0C024AC420@SSIMSEXCHNG...
I could be missing something, but it looks like you are using the result
of
the mysql_query as the actual result. It actually returns some weird
identifier. To access the real info you'd have to use something like
mysql_fetch_array to get it.
Check out this and see if it helps:
http://www.php.net/manual/en/function.mysql-query.php
-Natalie
-Original Message-
From: Jas [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, May 29, 2002 12:18 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Resource ID#2
Ok here is my problem, I set this up so a user selects a name form a
select
box and that name or $user_id is then passed to this page so the user can
edit the contact info etc. However it does not pull the selected $user_id
and place each field into my form boxes, all I get is a Resource ID #2.
Not
sure how I can over come this...
$table = auth_users;
$record = @mysql_query(SELECT * FROM $table WHERE user_id =
'$user_id',$dbh);
$var_form .= table width=\100%\ border=\0\
cellpadding=\7\form
name=\$user_id\ method=\post\ action=\del_account.php\
trtd width=\20%\ colspan=\2\bEdit Account
$user_id/b/td/tr
trtd width=\20%\First Name:/tdtd width=\80%\input
type=\text\ name=\$f_name\ size=\30\ maxlength=\30\
value=\$f_name\font class=\copyright\i.e. John/font/td/tr
trtd width=\20%\Last Name:/tdtd width=\80%\input
type=\text\ name=\$l_name\ size=\30\ maxlength=\30\
value=\$l_name\font class=\copyright\i.e. Doe/font/td/tr
trtd width=\20%\Email:/tdtd width=\80%\input
type=\text\ name=\$email_addy\ size=\30\ maxlength=\30\
value=\$email_addy\font class=\copyright\i.e.
[EMAIL PROTECTED]/font/td/tr
trtd width=\20%\User Name:/tdtd width=\80%\input
type=\text\ name=\$un\ size=\30\ maxlength=\30\
value=\$un\font
class=\copyright\i.e. j-doe/font/td/tr
trtd width=\20%\Password:/tdtd width=\80%\input
type=\password\ name=\$pw\ size=\30\ maxlength=\30\font
class=\copyright\(password must be alpha-numeric, i.e.
pAs5w0rd)/font/td/tr
trtd width=\20%\Confirm Password:/tdtd
width=\80%\input type=\password\ name=\$pw\ size=\30\
maxlength=\30\font class=\copyright\please confirm password
entered/font/td/tr
trtd width=\20%\nbsp;/tdtd width=\80%\input
type=\submit\ name=\add\ value=\edit user\nbsp;nbsp;input
type=\reset\ name=\reset\ value=\reset\/td/tr
/form/table;
echo $record;
} else {
blah blah
}
Thanks in advance,
Jas
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Re: [PHP-DB] Resource Id #2
Good, I'm glad it worked!!!
I see a lot of people using mysql_fetch_array().
Which is fine...But I'm more into Object Oriented programming...and
that's why I always use mysql_fetch_object()
Good luck!!!
Dan
On Wednesday, January 23, 2002, at 11:45 AM, [EMAIL PROTECTED] wrote:
My thanks for your quick response!
Here's the problem I'm dealing with:
I am inserting info with a unique identifier that auto-increments, then
take
that ID and place it into another table for later reference.
I was trying to use something like:
$query=select var1 from table1 where var2='$var3'
and var4='$var5';
$somevar=mysql_query($query) or die.. yada yada
echo $somevar;
I tried your code, and it works wonderfully! Thanks so much. :)
J. Wharton
[EMAIL PROTECTED]
- Original Message -
From: Dan Brunner [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
Sent: Wednesday, January 23, 2002 12:21 PM
Subject: Re: [PHP-DB] Resource Id #2
Use something like this...
...
...
...
$rows = mysql_num_rows($result);
for ($y = 0; $y $rows; $y++){
$data = mysql_fetch_object($result);
echo $data-Field_Name;
...
...
...
}
A Little more code would Help!!!
On Wednesday, January 23, 2002, at 10:56 AM, [EMAIL PROTECTED] wrote:
I'm having a problem with retrieving data from MySQL databases. I can
input
with no problems, but when I try to pull the data back out (single
field)
and echo it I get something showing up saying Resource Id #2. Any
ideas?
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Re: [PHP-DB] Resource Id #2
AHAHAHHAHA AHHAHAHAH Dan On Wednesday, January 23, 2002, at 02:19 PM, [EMAIL PROTECTED] wrote: Dan, Good, I'm glad it worked!!! I see a lot of people using mysql_fetch_array(). Which is fine...But I'm more into Object Oriented programming...and that's why I always use mysql_fetch_object() hey this is a family show - but I guess 'what you do in the privacy of' ... - and I'm just plain object-ionable, so I favor mysql_fetch_assoc() -- 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]
Re: [PHP-DB] resource id #2, #3, #4......
On Thursday 29 November 2001 01:29, Kevin Ruiz wrote:
I've come across yet another problem.
[snip]
$ci = select contactid from users where username='$username' and
password='$password';
$cir = mysql_query($ci)
or die(Couldn't execute);
$query = select * from my_contacts where contactid='$cir';
$result = mysql_query($query) or
die( mysql_error() );
When I try to print $cir to see if anything's getting passed I keep getting
something that reads resource id #2.
[snip]
Does anyone know what this means and how I can work around it.
$cir does NOT contain the actual results of your query. It is only a
*pointer* to your results. To get at the actual result(s):
print table\n;
while ($line = mysql_fetch_array($cir)) {
print \ttr\n;
while(list($col_name, $col_value) = each($line)) {
print \t\ttd$col_value/td\n;
}
print \t/tr\n;
}
print /table\n;
Read the manual for full details.
hth
--
Jason Wong - Gremlins Associates - www.gremlins.com.hk
/*
Total strangers need love, too; and I'm stranger than most.
*/
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Re: [PHP-DB] Resource id #2
I am trying to select a message from a table in a database. the message will will be in column 'msg' and will be in same row as id='0'. I need that msg put to the screen. With the code below I get the error: Resource id #2 Can someone please explain how I can fix this.. Thankyou in advance. ? $msgcre = SELECT msg FROM creators WHERE id='0'; $msg = mysql_query($msgcre); echo FONT color='#00' style='font-size: 10pt; font-weight: bold'$msg/font; ? Andrew, Although the assignment statement looks as though the LHS (left-hand side) should contain the result of the query, it does not (as you have discovered). Instead of being an integer or string, it is something called a 'resource'. You have already 'met' a resource in connection with the db connect process. After issuing a query, you must then use one of the mysql_fetch_... functions to extract the data, one field, or one row at a time, from the 'resource'. Please check out the manual at http://uk2.php.net/manual/en/ref.mysql.php. Also you might consider working through one of the many excellent texts and/or online articles discussing the PHP/MySQL combination. Regards, =dn -- 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]
