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=
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
$recordset=mysql_query('SELECT * FROM my_tabella'); // or whatever query is
/* define an array with fields to be shown on top of page. Attention
...values name
I see one bug in your code , that is you never rewind the pointer of
mysql_fetch_array($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
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
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
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\,
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
send by GET, COOKIE method.
Bye
Hans Lellelid [EMAIL PROTECTED] ha scritto nel messaggio
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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
the code by find replace
tools.
Hope can help
ps. FORZA ITALIA AGLI EUROPEI
Franciccio
Alessandro Folghera [EMAIL PROTECTED] ha scritto nel messaggio
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I hope that phpers will be not angry if I'm posting that maybe stupid
question.
I have the following script, an URL Directory
Here is the lean, corrected code:
?php
$link = mysql_connect( $site, $id, $pass ) or die (error connecting);
echo 'connection okbr';
mysql_select_db( $dbname, $link) or die (eror selecting db);
echo 'selected db ok'br;
$result = mysql_query( 'Select * From newsletter_subscribers',$link) or die
Have you tried with a insted of A for am/pm indication?
Hope can help
Bye
Franciccio
Philip Thompson [EMAIL PROTECTED] ha scritto nel messaggio
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Hi all.
Maybe there's something I'm doing incorrectly, but I cannot get the
date() function to return the appropriate time
I don't think you can have a relation (table) without a unique key. Either a
single field or a unique relation between 2 or more fields is necessary for
the table.
Franciccio
Bob Lockie [EMAIL PROTECTED] ha scritto nel messaggio
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On 06/16/04 09:53 John Nichel spoke:
Bob
I suggest to first analize the problem and then go through the query, php
coding ...etc etc
In the table u have only one superkey which is also a key and it is made of
the three fields (attributes) 'category', 'language' and ' name'. You should
consider to look for a prymary key randomly,
Jochem Maas ha scritto:
Han wrote:
Hmm,
still no luck. Thanks for the help. I think I'll have to break the
you mean that it still times out? crashes mysql?
maybe the table (tc_countries) is corrupt? try doing a repair.
select up into 2 selects and throw the results of the first into arrays.
Eve Atley ha scritto:
Platform: Redhat Linux Enterprise WS 3
PHP installed: 4.3.2
MySQL installed: 4.0.21
Apache installed: 2.0.46
When setting up PhpMyAdmin today, I got the error:
Cannot load mysql extension,
Please check PHP configuration
My phpinfo() shows:
'with-mysql=shared,/usr' (yes, the
Squeakypants ha scritto:
From a forum that recommended this to me, I changed the query execution to this:
$query = SELECT credits FROM krypto WHERE user=$user AND pass=$pass;
echo query = . $query .\n;
$result = mysql_query($query);
echo result = . $result .\n;
So now it outputs this:
query =
You may try to create a VIEW. In this case the tables will be dynamically
linked, that is when some values change in both tales the views
automatically update itself each time you call it.
CPT John W. Holmes [EMAIL PROTECTED] ha scritto nel messaggio
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From: Roger
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