Hi, here is my solution (one of the possible) it is tested so it should work
fine:
\n";
$record= mysql_fetch_array($recordset, MYSQL_ASSOC);
foreach ($record as $k =>$val)
{
echo "$k\n"; // print the field name in the first row of
the table
for ($i=0;$i<$num_righe;$i++)
{
echo
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
$recordset=mysql_query('SELECT * FROM my_tabella'); // or whatever query is
/* define an array with fields to be shown on top of page. Attention
...values name mu
I see one bug in your code , that is you never "rewind" the pointer of
mysql_fetch_array($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:
// then here is all the html for that page
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
>
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 iterato
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:
";
?>
If it works let me know , bye
Francesco Basile
__
"Alessandro Folghera" <[EMAIL PROTECTED]> ha scritto nel messaggio
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> Hi,
>
> I ha
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 m
Wathever you are ging to do with meta tag they still are html tag, so as
long as you embend the tag in php code it shuold just work fine.
";
?>
Any text editor should help you modifying the code by "find & replace"
tools.
Hope can help
ps. FORZA ITALIA AG
Here is the lean, corrected code:
';
mysql_select_db( $dbname, $link) or die ("eror selecting db");
echo 'selected db ok';
$result = mysql_query( 'Select * From newsletter_subscribers',$link) or die
("error query");
// some here code to disply result
?>
It is easier to evaluate each time the return
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
news:[EMAIL PROTECTED]
> Hi all.
>
> Maybe there's something I'm doing incorrectl
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
news:[EMAIL PROTECTED]
> On 06/16/04
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, togheth
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.
Didn'
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 co
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:
que
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
news:[EMAIL PROTECTED]
> From: "Roger
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