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Posted by Petr Vileta on 10/18/92 11:55
Hi, I'm new here and excuse me if this question was be here earlier.
I have a simple code
<html><body>
<?php
<?php
$link = mysql_connect("localhost", "user", "password")
or die("Grr: " . mysql_error());
mysql_select_db("my_dbf") or die("Grr");
$query = "select id from my_table where id < 4 order by id";
$result = mysql_query($query) or die("Grr: " . mysql_error());
$rows = mysql_num_rows($result);
echo "Rows=", $rows, "<br>" ;
list_it($result);
function list_it($result)
{
while ($row = mysql_fetch_array($result, MYSQL_ASSOC))
{
echo $row["id"], "<br>";
}
}
?>
</body></html>
This code produce this output:
Rows=3
1
2
3
Warning: mysql_fetch_array(): 5 is not a valid MySQL result resource in
F:\webpub\php\test.php on line ...
Why warning?
This is a simple example. In my real code I must use a few different
functions for displaing mysql_fetch_array() result and $query variable I
must create before any html output because I use sessions.
How to pass resource type variable into function as parameter?
Oh, my PHP version is 4.3.7 on Windoze ;-)
--
Petr Vileta, Czech republic
(My server rejects all messages from Yahoo and Hotmail. Send me your mail
from another non-spammer site please.)
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