Re: [PHP] data from database
it works,
thanks,
Alain
John Nichel [EMAIL PROTECTED] schreef in bericht
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alain dhaene wrote:
Hi,
I will write a function that returns the result of a recordset.
I tried this:
Function getPersonen()
{
openDB(); //function that I implements in another file
$query = SELECT * FROM Cursisten;
$result = mysql_query($query)
or die(Fout bij uitvoeren query);
$resultrow = mysql_fetch_array( $result );
Get rid of the above line, unless you plan on passing the array. From
the looks below, you want to pass the identifier though. If you do want
to pass the result array instead of the identifier, leave as is.
closeDB(); //function that I implements in another file
mysql_free_result($result);
If you are going to pass the identifier, you just killed it here. If
you are going to pass the data array, swap these two lines, ie free your
result before you close your connections.
return $resultrow;
Change this line to...
return $result
if you are not planning on passing the array.
}
I call this function:
$resultaat = getPersonen();
// Printen resultaten in HTML
print table\n;
while ($line = mysql_fetch_array($resultaat, MYSQL_ASSOC)) {
What you do with this, all depends on how you rewrite your function. If
you're passing the identifier, you can leave it as is, if you're passing
the result data array, you need to get rid of the while loop.
print \ttr\n;
foreach ($line as $col_value) {
print \t\ttd$col_value/td\n;
}
print \t/tr\n;
}
print /table\n;
In runtime I have the following error
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL
result resource in
/home/schoolre/public_html/Hitek/Online/Registratie/showPerson.php on
line 8
What is wrong?
Alot, I suggest you check out the MySQL functions
http://us4.php.net/manual/en/ref.mysql.php
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[PHP] data from database
Hi,
I will write a function that returns the result of a recordset.
I tried this:
Function getPersonen()
{
openDB(); //function that I implements in another file
$query = SELECT * FROM Cursisten;
$result = mysql_query($query)
or die(Fout bij uitvoeren query);
$resultrow = mysql_fetch_array( $result );
closeDB(); //function that I implements in another file
mysql_free_result($result);
return $resultrow;
}
I call this function:
$resultaat = getPersonen();
// Printen resultaten in HTML
print table\n;
while ($line = mysql_fetch_array($resultaat, MYSQL_ASSOC)) {
print \ttr\n;
foreach ($line as $col_value) {
print \t\ttd$col_value/td\n;
}
print \t/tr\n;
}
print /table\n;
In runtime I have the following error
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL
result resource in
/home/schoolre/public_html/Hitek/Online/Registratie/showPerson.php on line 8
What is wrong?
Alain
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RE: [PHP] data from database
$result = mysql_query($query) You need to pass the connection resource returned from your mysql_connect call as a second parameter. Regards. -Gregory -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] data from database
Gregory Kornblum wrote: $result = mysql_query($query) You need to pass the connection resource returned from your mysql_connect call as a second parameter. Regards. -Gregory While it is a good practice to do this, it is not necessary... If link_identifier isn't specified, the last opened link is assumed. If no link is open, the function tries to establish a link as if mysql_connect() was called with no arguments, and use it. The result of the query is buffered. -- By-Tor.com It's all about the Rush http://www.by-tor.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] data from database
alain dhaene wrote:
Hi,
I will write a function that returns the result of a recordset.
I tried this:
Function getPersonen()
{
openDB(); //function that I implements in another file
$query = SELECT * FROM Cursisten;
$result = mysql_query($query)
or die(Fout bij uitvoeren query);
$resultrow = mysql_fetch_array( $result );
Get rid of the above line, unless you plan on passing the array. From
the looks below, you want to pass the identifier though. If you do want
to pass the result array instead of the identifier, leave as is.
closeDB(); //function that I implements in another file
mysql_free_result($result);
If you are going to pass the identifier, you just killed it here. If
you are going to pass the data array, swap these two lines, ie free your
result before you close your connections.
return $resultrow;
Change this line to...
return $result
if you are not planning on passing the array.
}
I call this function:
$resultaat = getPersonen();
// Printen resultaten in HTML
print table\n;
while ($line = mysql_fetch_array($resultaat, MYSQL_ASSOC)) {
What you do with this, all depends on how you rewrite your function. If
you're passing the identifier, you can leave it as is, if you're passing
the result data array, you need to get rid of the while loop.
print \ttr\n;
foreach ($line as $col_value) {
print \t\ttd$col_value/td\n;
}
print \t/tr\n;
}
print /table\n;
In runtime I have the following error
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL
result resource in
/home/schoolre/public_html/Hitek/Online/Registratie/showPerson.php on line 8
What is wrong?
Alot, I suggest you check out the MySQL functions
http://us4.php.net/manual/en/ref.mysql.php
--
By-Tor.com
It's all about the Rush
http://www.by-tor.com
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
