[PHP-DB] Re: mysql results, arrays, and for loops
Hi, here is my solution (one of the possible) it is tested so it should work
fine:
?php
$recordset=mysql_query($query);
$num_righe=mysql_num_rows($recordset);
// here starts the dinamyc table
echo TABLE align=\left\ border=\1\ cellspacing=\2\ cellpadding=\2\
width=\80%\\ntr;
$record= mysql_fetch_array($recordset, MYSQL_ASSOC);
foreach ($record as $k =$val)
{
echo tdb$k/b/td\n; // print the field name in the first row of
the table
for ($i=0;$i$num_righe;$i++)
{
echo /tr;
mysql_data_seek($recordset,$i);
$record = mysql_fetch_array($recordset, MYSQL_ASSOC);
foreach ($record as $val){
echo td$val/td\n; // print the result of query, starting under
1st row
}
}
mysql_free_result($recordset);
mysql_close();
echo /tr/tablebr;
?
Let me know if it fits your problem.
Best regards
Francesco Basile
Philip Thompson [EMAIL PROTECTED] ha scritto nel messaggio
news:[EMAIL PROTECTED]
Hi all!
I am using a select statement to obtain all the dates whenever someone
submitted a problem in a database. Well, I want to get the result
(which could be multiple dates) and then print that in a table format
with some other information on a webpage. So I want to use a FOR loops
and go through each of the dates and dynamically create a table.
My question is: how do I store the results of the select query? Would I
want to store them in an array, and then just parse through each
element of the array, and what is the syntax for that? Or is there a
better way?
Thanks,
~Philip
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[PHP-DB] Re: how to reuse DB results
Sorry for mystakes, (shame on me, i said i was new) using Object array,
here is a working version (tested) with use of array.
Bye
$recordset=mysql_query('SELECT * FROM my_tabella'); // or whatever query is
/* define an array with fields to be shown on top of page. Attention
...values name must be the same of table field names you want to show on top
of the page*/
$my_key = array('key01','key02','key03'); /* u can use $_REQUEST variables
instead of
fixed values*/
$i=0;
while($record=mysql_fetch_array($recordset, MYSQL_ASSOC))
{
foreach ($record as $k=$val)
if ($k==$my_key[$i])
{
// echo the top of the page
echo p$k :$val/p;
/* if u need a top_array to store result shown on top of the
page, comment this line only
$top_array[$i]=$val; // loose the matching key-field!! */
$i++;
}
}
//reset the resource
mysql_data_seek($recordset,0);
// some middle-page code here
echo 'pthis is the bottom page and displays the whole result/p';
while ($record=mysql_fetch_array($recordset, MYSQL_ASSOC))
foreach ($record as $k=$val)
echo p$k : $val/p; // make your choice for html code
mysql_free_result($recordset);
// Bye
Aaron Wolski [EMAIL PROTECTED] ha scritto nel messaggio
news:[EMAIL PROTECTED]
Hi All,
Got this logic problem I don't know how to solve.
Is there any way I can make a call to the DB for some records.
Display some info from a column or two say at the top of the page and
then display the full result set in a while() loop?
Right now, I am making two calls to the Db to get the data I need to
display at the top of the page and then a second query to retrieve the
full result set.
I'm confused!
Thanks for any help!
Aaron
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[PHP-DB] Re: Retrieve data from a table, edit/add it and enter it in a new table
I see one bug in your code , that is you never rewind the pointer of
mysql_fetch_array($result), so at the end of the first cycle ...while
($r=mysq...)... the pointer is at the end of the query resource. You should
use mysql_data_seek($result,0) to rewind before doing another while cycle.
Hope it can help
Bye
Francesco Basile (PHP_newbee)
Justin [EMAIL PROTECTED] ha scritto nel messaggio
news:[EMAIL PROTECTED]
Hi,
I am trying to do the following:
Retrieve some information from a table, edit it by appending some further
information to it (a few more fields) and then enter the new data record
into a new table, and delete the old data in the original table. Sounds
confusing I know.
The code is below (I apologise for its poor style etc as I am very new to
php), but when I click on the 'Enter Information' button, nothing happens.
The existing data is retrieved without any problems and I can select it
using the radio button. But when I try and add data to the new fields
('option_close_price' and 'notes'), nothing happens.
Any help appreciated.
form action=? echo $PHP_SELF ? method=post
?
mysql_pconnect(localhost,root,password);
mysql_select_db(options);
if(!$cmd)
{
$result = mysql_query(select * from open_trades);
while($r=mysql_fetch_array($result))
{
$open_date=$r[open_date];
$share=$r[share];
$code=$r[code];
$short_long_trade=$r[short_long_trade];
$id=$r[id];
$expiry=$r[expiry];
$exercise=$r[excercise];
$option_price=$r[option_price];
$no_purchased=$r[no_purchased];
$no_sold=$r[no_sold];
$income_in=$r[income_in];
$income_out=$r[income_out];
print table border=\1\ cellpadding=\3\ cellspacing=\0\\n;
print td/tdtdOpen Date/tdtdShare/tdtdCode/tdtdShort
orbr Long Trade/tdtdExpiry/tdtdExcercise/tdtdOption
Price/tdtdNumberbr Purchased/tdtdNumber Sold/tdtdIncome
In/tdtdIncome Out/tdtd
/tr;
while ($row = mysql_fetch_array($result))
{
print trtd;
print INPUT TYPE='RADIO' NAME='id' VALUE='echo $id';
print /tdtd;
print $row[open_date];
print /tdtd;
print $row[share];
print /tdtd;
print $row[code];
print /tdtd;
print $row[short_long_trade];
print /tdtd;
print $row[expiry];
print /tdtd;
print $row[excercise];
print /tdtd;
print $row[option_price];
print /tdtd;
print $row[no_purchased];
print /tdtd;
print $row[no_sold];
print /tdtd;
print $row[income_in];
print /tdtd;
print $row[income_out];
print /td/tr\n;
}
print /table\n;
?
? }?
input type=submit name=cmd value=Close/form
? }
?
?
if($cmd==Close)
{
if (!$submit)
{
$result = mysql_query(select * from open_trades);
while($myrow=mysql_fetch_array($result))
?
input type=hidden name=id value=?php echo $myrow[id] ?
Option Close PriceINPUT TYPE=TEXT NAME=option_close_price VALUE=?PHP
echo $myrow[option_close_price]? SIZE=7 br
Notes:INPUT TYPE=TEXT NAME=notes VALUE=?php echo $myrow[notes] ?
SIZE=60br
input type=hidden name=cmd value=edit
input type=Submit name=submit value=Enter information
/form
? } ?
?
if($submit)
{
$query = INSERT INTO closed_trades SET code='$code',
option_price='$option_price', option_close_price='$option_close_price',
no_sold='$no_sold', open_date='$open_date',
short_long_trade='$short_long_trade, share='$share', expiry='$expiry',
excercise='$excercise', no_purchased='$no_purchased',
income_in='$income_in', income_out='$income_out', notes='$notes', id=$id';
$result = mysql_query($ql);
$query = DELETE FROM open_trades WHERE id=$id;
$result = mysql_query($sql);
echo Thank you! Information updated.;
}
}
?
/td
/table
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Re: [PHP-DB] Re: HTTP header information
You may try with this:
?php
ob_start();/* buffer the output so no header is sent until the script is
complete*/
session_start();
if (!session_is_registered('_isLoggedIn')) {
header (Location:viewer.php?type=login);
} else {
// do the stuff to submit a problem if they have
// logged in and have pressed the submit button
// in the html
}
?
// then here is all the html for that page
?php
ob_end_flush();// send the output now
?
Bye
Basile Francesco
Philip Thompson [EMAIL PROTECTED] ha scritto nel messaggio
news:[EMAIL PROTECTED]
On Jun 10, 2004, at 9:44 AM, Torsten Roehr wrote:
Philip Thompson [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
Hi all.
I am running a website to where a user needs to login to authenticate
themselves to perform certain tasks. So a user logs in, and I start a
session (in PHP, of course). Well, the catch is, I am doing this all
from one page, 'viewer.php', and I just tack on the specific view/page
that I want them to see, depending on the link selected on that page.
Meaning, they click on the 'submit problem' link and it goes to
'viewer.php?type=submitproblem'.
The problem comes whenever I load the view 'submitproblem' and I start
a session with session_start(), which carries over the variable to
tell
whether or not the user is logged in. If they have not logged in
whenever they click on 'submitproblem' then it will redirect them to
'viewer.php?type=login'. So I log in, and then go to 'submitproblem'.
This is where I get the error: Warning: session_start(): Cannot send
session cookie - headers already sent. Essentially, I understand why
this is occurring, but is there an easy way to get around it without
creating a new page, such as 'submitproblem.php' instead of
'viewer.php?type=submitproblem'???
Make sure that NO output is done before session_start() is called. Can
you
post some of your code?
Regards,
Torsten Roehr
See, that's the case. Because I'm essentially just changing the content
within the page, it never leaves the page 'viewer.php' - it just
changes the content by tacking on '?type=login, submitproblem, etc'.
But my code in the beginning of the file 'submitproblem.view' is:
?php
session_start();
if (!session_is_registered('_isLoggedIn')) {
header (Location:viewer.php?type=login);
} else {
// do the stuff to submit a problem if they have
// logged in and have pressed the submit button
// in the html
}
?
// then here is all the html for that page
So, does that help any?
~Philip
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[PHP-DB] Re: how to reuse DB results
Hi here is my logic, i haven't tested but it should work (i hope)!!!
i'm not sure if it works only on php5...!?! Let me know if it is a complete
crap i've just started learning php.
Bye
$recordset=mysql_query($query); // whatever query is
$obj_array=new ArrayObject($recordset);
// get the iterator now
$interator=$obj_array-getIterator();
// you can now easly navigate in the recordset by the iterator
/* define an array with n fields to be shown on top of page. Attention
...values name must be the same of table field names you want to show on top
of the page*/
$my_key = array(0='key1',1= 'key2',,n-1='keyn');
// echo the top of the page
$i=0;
while($iterator-valid())
{
if ($iterator-key()==$my_key[$i] )
{
echo p.$iterator-key(). : .$iterator-current()./p;
/* if u need a top_array to store result shown on top of the page,
uncomment this line only
$top_array[$i]=$iterator-current(); // loose the matching
key-field!! */
$i++;
$iterator-next();
}
else
$iterator-next();
}
/* to display the whole recordset resulting from the query don't forget
you still have
$recordset free to use with mysql_fetch_array() or other way to display */
echo 'pthis is the bottom page and displays the whole result/p';
while ($res=mysql_fetch_array($recordset))
for each ($res as $k=$val)
echo p$k : $val/p; // make your choice for html code
Aaron Wolski [EMAIL PROTECTED] ha scritto nel messaggio
news:[EMAIL PROTECTED]
Hi All,
Got this logic problem I don't know how to solve.
Is there any way I can make a call to the DB for some records.
Display some info from a column or two say at the top of the page and
then display the full result set in a while() loop?
Right now, I am making two calls to the Db to get the data I need to
display at the top of the page and then a second query to retrieve the
full result set.
I'm confused!
Thanks for any help!
Aaron
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[PHP-DB] Mysql not receiving the data
Good day list, I was running a Mysql/PHP DB/Webpage that was hosted on Redhat 8.0 It was a simple DB - only 1 table, and the php page connected and fed data to it. Last weekend I rebuilt the server to Fedora Core 2 - using the default PHP/Mysql/apache installs. I setup the databases, and imported the old data (taken from a MySQL dump) into the new DB, and setup Apache, so that I could view the pages that displayed the data, and also the page that I used to add the data. Online I could see everything, and the pages gave the appearance of working, however when I went into the DB using PHPMYADMIN to check the status of the new data entered, all I found was blank rows ( for the new data since the rebuild, all the old data was there) There were the correct number of new rows for the amount of records that I had entered, which tells me (unless I am nistaken) that the PHP is talking to the DB, and is atleast sending a insert command, but the rest of the data is not getting in. - I hope that I have made sense, Please note that the php/html pages used to input the data used to work on the old server and had been running for several years flawlessly. As I said I have a default conf for apache, obviously changing the bits that needed to be changed (servername, etc) and did not change anything on the PHP MySQL confs. Any ideas? Thank you Andrew Rothwell -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Mysql not receiving the data
Online I could see everything, and the pages gave the appearance of working, however when I went into the DB using PHPMYADMIN to check the status of the new data entered, all I found was blank rows ( for the new data since the rebuild, all the old data was there) There were the correct number of new rows for the amount of records that I had entered, which tells me (unless I am nistaken) that the PHP is talking to the DB, and is atleast sending a insert command, but the rest of the data is not getting in. - Without seeing any code whatsoever and since this worked before but no longer works on a new install, I can only assume that your code was written with the assumption that register_globals was turned on and it's not on in your current configuration. If that is the case, see the PHP manual or search the Web for the solution ($_POST, $_GET, etc.). Larry -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Mysql not receiving the data
Hi Larry, Thank you very much for the very quick response, I set my php.ini
file (located /etc/php.ini ) for the register_globals = On (it was off by
default)
Now however I get an error
Error adding entry: You have an error in your SQL syntax near 's spanish
driver is found shot dead, Inspector Jacques Clouseau is the first off' at
line 8
My Database is a movie database of my dvd's that I own (for insurance
reasons)
My addmovie.php is this
?
mysql_connect(localhost,username,password);
mysql_select_db(movies);
$add = INSERT INTO movies SET
movie_name='$movie_name',
genre='$genre',
director='$director',
star1='$star1',
star2='$star2',
star3='$star3',
brief_synopsis='$brief_synopsis',
imdb_link='$imdb_link';
if (@mysql_query($add))
{
echo(pYour entry has been added. br
$movie_name/p);
}
else
{
echo(pError adding entry: .
mysql_error() . /p);
}
?
And the addmovie.htm page (atleast the form action is this)
body bgcolor=#FF
form method=post action=addmovie.php name=addmovies
table width=300 border=0 cellspacing=2 cellpadding=2
bordercolordark=#FF0033 bordercolorlight=#66
tr
td width=41% bgcolor=#99Movie Name /td
td width=59% bgcolor=#99FFCC
input type=text name=movie_name
/td
/tr
Andrew
-Original Message-
From: Larry E. Ullman [mailto:[EMAIL PROTECTED]
Sent: Sunday, June 13, 2004 11:22 AM
To: Andrew Rothwell
Cc: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Mysql not receiving the data
Online I could see everything, and the pages gave the appearance of
working, however when I went into the DB using PHPMYADMIN to check the
status of the new data entered, all I found was blank rows ( for the
new data since the rebuild, all the old data was there) There were the
correct number of new rows for the amount of records that I had
entered, which tells me (unless I am nistaken) that the PHP is talking
to the DB, and is atleast sending a insert command, but the rest of
the data is not getting in. -
Without seeing any code whatsoever and since this worked before but no
longer works on a new install, I can only assume that your code was written
with the assumption that register_globals was turned on and it's not on in
your current configuration.
If that is the case, see the PHP manual or search the Web for the solution
($_POST, $_GET, etc.).
Larry
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Re: [PHP-DB] Mysql not receiving the data
Andrew Rothwell wrote: Hi Larry, Thank you very much for the very quick response, I set my php.ini file (located /etc/php.ini ) for the register_globals = On (it was off by default) Now however I get an error Error adding entry: You have an error in your SQL syntax near 's spanish driver is found shot dead, Inspector Jacques Clouseau is the first off' at line 8 I would think it fairly obvious that the extra ' in the text you are loading is causing a problem, use an ` instead, or escape the single ' Not sure if MySQL uses '' or \' :) -- Lester Caine - L.S.Caine Electronic Services -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Mysql not receiving the data
The apostrophe (') in your data is, most likely, killing the SQL statement
when it is sent to the server. Use addslashes() around all of your form data
to prevent this and also to help guard against SQL injection attacks.
Ex:
$add = INSERT INTO movies SET
movie_name='.addslashes($movie_name).',
genre='.addslashes($genre).',
director='.addslashes($director).',
star1='.addslashes($star1).',
star2='.addslashes($star2).',
star3='.addslashes($star3).',
brief_synopsis='.addslashes($brief_synopsis).',
imdb_link='$imdb_link';
Hope this helped.
Rich
-Original Message-
From: Andrew Rothwell [mailto:[EMAIL PROTECTED]
Sent: Sunday, June 13, 2004 1:48 PM
To: [EMAIL PROTECTED]
Subject: RE: [PHP-DB] Mysql not receiving the data
Hi Larry, Thank you very much for the very quick response, I set my php.ini
file (located /etc/php.ini ) for the register_globals = On (it was off by
default)
Now however I get an error
Error adding entry: You have an error in your SQL syntax near 's spanish
driver is found shot dead, Inspector Jacques Clouseau is the first off' at
line 8
My Database is a movie database of my dvd's that I own (for insurance
reasons)
My addmovie.php is this
?
mysql_connect(localhost,username,password);
mysql_select_db(movies);
$add = INSERT INTO movies SET
movie_name='$movie_name',
genre='$genre',
director='$director',
star1='$star1',
star2='$star2',
star3='$star3',
brief_synopsis='$brief_synopsis',
imdb_link='$imdb_link';
if (@mysql_query($add))
{
echo(pYour entry has been added. br
$movie_name/p);
}
else
{
echo(pError adding entry: .
mysql_error() . /p);
}
?
And the addmovie.htm page (atleast the form action is this)
body bgcolor=#FF
form method=post action=addmovie.php name=addmovies
table width=300 border=0 cellspacing=2 cellpadding=2
bordercolordark=#FF0033 bordercolorlight=#66
tr
td width=41% bgcolor=#99Movie Name /td
td width=59% bgcolor=#99FFCC
input type=text name=movie_name
/td
/tr
Andrew
-Original Message-
From: Larry E. Ullman [mailto:[EMAIL PROTECTED]
Sent: Sunday, June 13, 2004 11:22 AM
To: Andrew Rothwell
Cc: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Mysql not receiving the data
Online I could see everything, and the pages gave the appearance of
working, however when I went into the DB using PHPMYADMIN to check the
status of the new data entered, all I found was blank rows ( for the
new data since the rebuild, all the old data was there) There were the
correct number of new rows for the amount of records that I had
entered, which tells me (unless I am nistaken) that the PHP is talking
to the DB, and is atleast sending a insert command, but the rest of
the data is not getting in. -
Without seeing any code whatsoever and since this worked before but no
longer works on a new install, I can only assume that your code was written
with the assumption that register_globals was turned on and it's not on in
your current configuration.
If that is the case, see the PHP manual or search the Web for the solution
($_POST, $_GET, etc.).
Larry
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Re: [PHP-DB] Mysql not receiving the data
I agree, the slashes are killing the query. I would suggets doing this:
$add = INSERT INTO movies SET
movie_name=\$movie_name\,
genre=\$genre\,
director=\$director\,
star1=\$star1\,
star2=\$star2\,
star3=\$star3\,
brief_synopsis=\$brief_synopsis\,
imdb_link=\$imdb_link\;
Rich Hutchins [EMAIL PROTECTED] ha scritto nel messaggio
news:[EMAIL PROTECTED]
The apostrophe (') in your data is, most likely, killing the SQL statement
when it is sent to the server. Use addslashes() around all of your form
data
to prevent this and also to help guard against SQL injection attacks.
Ex:
$add = INSERT INTO movies SET
movie_name='.addslashes($movie_name).',
genre='.addslashes($genre).',
director='.addslashes($director).',
star1='.addslashes($star1).',
star2='.addslashes($star2).',
star3='.addslashes($star3).',
brief_synopsis='.addslashes($brief_synopsis).',
imdb_link='$imdb_link';
Hope this helped.
Rich
-Original Message-
From: Andrew Rothwell [mailto:[EMAIL PROTECTED]
Sent: Sunday, June 13, 2004 1:48 PM
To: [EMAIL PROTECTED]
Subject: RE: [PHP-DB] Mysql not receiving the data
Hi Larry, Thank you very much for the very quick response, I set my
php.ini
file (located /etc/php.ini ) for the register_globals = On (it was off by
default)
Now however I get an error
Error adding entry: You have an error in your SQL syntax near 's spanish
driver is found shot dead, Inspector Jacques Clouseau is the first off' at
line 8
My Database is a movie database of my dvd's that I own (for insurance
reasons)
My addmovie.php is this
?
mysql_connect(localhost,username,password);
mysql_select_db(movies);
$add = INSERT INTO movies SET
movie_name='$movie_name',
genre='$genre',
director='$director',
star1='$star1',
star2='$star2',
star3='$star3',
brief_synopsis='$brief_synopsis',
imdb_link='$imdb_link';
if (@mysql_query($add))
{
echo(pYour entry has been added. br
$movie_name/p);
}
else
{
echo(pError adding entry: .
mysql_error() . /p);
}
?
And the addmovie.htm page (atleast the form action is this)
body bgcolor=#FF
form method=post action=addmovie.php name=addmovies
table width=300 border=0 cellspacing=2 cellpadding=2
bordercolordark=#FF0033 bordercolorlight=#66
tr
td width=41% bgcolor=#99Movie Name /td
td width=59% bgcolor=#99FFCC
input type=text name=movie_name
/td
/tr
Andrew
-Original Message-
From: Larry E. Ullman [mailto:[EMAIL PROTECTED]
Sent: Sunday, June 13, 2004 11:22 AM
To: Andrew Rothwell
Cc: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Mysql not receiving the data
Online I could see everything, and the pages gave the appearance of
working, however when I went into the DB using PHPMYADMIN to check the
status of the new data entered, all I found was blank rows ( for the
new data since the rebuild, all the old data was there) There were the
correct number of new rows for the amount of records that I had
entered, which tells me (unless I am nistaken) that the PHP is talking
to the DB, and is atleast sending a insert command, but the rest of
the data is not getting in. -
Without seeing any code whatsoever and since this worked before but no
longer works on a new install, I can only assume that your code was
written
with the assumption that register_globals was turned on and it's not on in
your current configuration.
If that is the case, see the PHP manual or search the Web for the solution
($_POST, $_GET, etc.).
Larry
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[PHP-DB] Re: Problem in passing the necessary variable.
Ciao..hi, i would suggest to embed the html code in php like this:
?php
$_your_variable=$row[imgid];
echo
a href=\../images/image.php?img=$_your_variable\
target=\_blank\
onClick=\window.open(this.href, this.target,
\'width=200,height=250\');
return false;\img SRC=\../preview.jpg\ ALT=\Click to look at
the picture
you are charging\ border=\0\/a;
?
If it works let me know , bye
Francesco Basile
__
Alessandro Folghera [EMAIL PROTECTED] ha scritto nel messaggio
news:[EMAIL PROTECTED]
Hi,
I have developed a news system .. I can select to insert an image into the
articles.
The function to extract the image calling them by the imgid of images
table is the following:
function getImm() {
?
select name=image
?
connect();
$sql = select * from images;
$result = mysql_query($sql);
while($row = mysql_fetch_array($result)) {
printf(option value=\%s\%s/option, $row[id_img], $row[imgid]);
}
?
/select
?
I'd like to be able to see the image before to publish the piece of news.
I think about a window pop up with java in order to create a preview of
the
image. The code follows:
a href=../images/image.php?img=XX target=_blank
onClick=window.open(this.href, this.target, 'width=200,height=250');
return false;img SRC=../preview.jpg ALT=Click to look at the picture
you are charging border=0/a
The problem is that I can't pass the imgid to the href inspite of the
May someone tell me how to pass the id number in order to show me the
image
or how may I solve the trouble?
I hope someone will help me ...
Sincerely,
Alexander
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RE: [PHP-DB] Mysql not receiving the data
Thank you everybody that responded so quickly - I used the suggestion of Franciccio - and the data is now gow into the db Thank you very much - I really appreciate the help. Another question - with this fix in place - do I still need the register_globals = On ? Or can I now turn it off? Thank you all again Andrew -Original Message- From: franciccio [mailto:[EMAIL PROTECTED] Sent: Sunday, June 13, 2004 12:26 PM To: [EMAIL PROTECTED] Subject: Re: [PHP-DB] Mysql not receiving the data I agree, the slashes are killing the query. I would suggets doing this: $add = INSERT INTO movies SET movie_name=\$movie_name\, genre=\$genre\, director=\$director\, star1=\$star1\, star2=\$star2\, star3=\$star3\, brief_synopsis=\$brief_synopsis\, imdb_link=\$imdb_link\; -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Mysql not receiving the data
Hi Andrew, Andrew Rothwell wrote: Thank you everybody that responded so quickly - I used the suggestion of Franciccio - and the data is now gow into the db Thank you very much - I really appreciate the help. Another question - with this fix in place - do I still need the register_globals = On ? Or can I now turn it off? It seems like you should have kept your old php.ini file, as this other error you encountered was probably due to your old php.ini file having this setting: magic_quotes_gpc = 1 That INI var instructs PHP to automatically addslashes() to any GET/POST/COOKIE data. I would suggest turning this back on, unless you've thoroughly redesigned your code to not need it. This is unrelated to register_globals, which you will need to leave on unless you redesign your application. Hans -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] creating a nested multidimensional array from a query
Hi :)
I am on my second week of php so be gentle
I am creating an XML file out of a mysql query with nested arrays.
Currently I can get 1 element and 1 child with a properly formatted XML
file with the below script .
My question is: How do I add 3 to 4 more child elements to the below
'playlist' array ?
Currently ,I have one parent 'Artist' and one child 'english' ...
I need to add more child elements to the 'Artist' array like urlPath,
spanish, biography, etc
Do I have to add another dimension to the 'playlist' array? Do I need
another foreach loop ?
Is there an easier more efficient way to do this?
Be nice to spell out the schema in some way in the script...in case you
need to add more levels...like a 'subCategory'
I am a bit new to this so any help would be greatly appretiated
head is spinning a bit
$sql = 'SELECT artist.artist_name, media.english, media.path ';
$sql .= 'FROM media, artist ';
$sql .= 'WHERE artist.artist_id = media.artist_id LIMIT 0, 30 ';
$result = mysql_query($sql);
while ($row = mysql_fetch_array($result))
{
$playlist[$row['artist_name']] [] = $row['english'];
//would like to add more children here...
}
$xml = sirenreels\n;
foreach ($playlist as $artist = $media)
{
$num_media = count($media);
$xml .= artist\n;
$xml .= \tmeta\n;
$xml .= \t\ttitle.$artist./title\n;
$xml .= \t/meta\n;
$xml .= \tcontent\n;
foreach ($media as $mediaVal)
{
$xml .= \t\tmedia\n;
$xml .= \t\t\tenglish_name.$mediaVal./english_name\n;
///add more children
///add more children
$xml .= \t\t/media\n;
}
$xml .= \t/content\n;
$xml .= /artist\n;
}
$xml .= /sirenreels\n;
print $xml
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Re: [PHP-DB] Mysql not receiving the data
Some of the reason to have register_global on is to easly use variables sent by post, get, cookie method of a form. Ex. form name=form1 method=post action=anypage.php?get_var=1 input type=text name=text_post value=Hello World input type=submit name=Submit value=Submit /form in your code at the page anypage.php you will have available the variables: $get_var==1 and $text_post==Hello World if register_global=TRUE in php.ini file - in this case u still have available $_POST;GET ecc... $_GET['get_var']==1 and $_POST['text_post']==Hello World if register_global=FALSE in php.ini file - in this case you don't have the $get_var e $text_post available, so code is safer I would suggest to leave register global=FALSE as to have safer code unless u have to rewrite the whole code. Think about having hacked variables value send by GET, COOKIE method. Bye Hans Lellelid [EMAIL PROTECTED] ha scritto nel messaggio news:[EMAIL PROTECTED] Hi Andrew, Andrew Rothwell wrote: Thank you everybody that responded so quickly - I used the suggestion of Franciccio - and the data is now gow into the db Thank you very much - I really appreciate the help. Another question - with this fix in place - do I still need the register_globals = On ? Or can I now turn it off? It seems like you should have kept your old php.ini file, as this other error you encountered was probably due to your old php.ini file having this setting: magic_quotes_gpc = 1 That INI var instructs PHP to automatically addslashes() to any GET/POST/COOKIE data. I would suggest turning this back on, unless you've thoroughly redesigned your code to not need it. This is unrelated to register_globals, which you will need to leave on unless you redesign your application. Hans -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] MySQL persistent connections
Hello, I am no expert but I believe more information is required, like what would be the average return of a mysql operation ? How many different DB's are you accessing ? How many different usernames and passwords for mysql are you using ?? I believe persistent connections are only useful when the amount of DB connections being made is inversely proportional to the amount of data being returned. That is if you have 50 connections being created to check the same one field in one table. Then persistent connections would be worth it. But is you are returning 4-5MB of data per-query then persistent connections would not be worth it. Plus with persistent connections I have heard most people have issues with hitting the too many open connections limit. Because the connections are not being closed or reused. Michael. On Fri, 11 Jun 2004 11:38:37 -0400 Radek Zajkowski [EMAIL PROTECTED] wrote: Hey there PHP fiends, I have a bit of cookie here. We're designing a PHP based app that uses MySQL as the data storage. Scalability is an issue, we want to be able to handle up to 1 people utilizing the system (not at once of course). Question is, should I be connecting and disconnecting from the DB on each operation or increase the maximum number of allowed connections with MySQL and connect just once. TIA R - -- 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
