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Posted by Jerry Stuckle on 12/17/93 11:55
Petr Vileta wrote:
> 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 ;-)
>
The code you showed should not display this error. Unfortunately, you
left out what's probably the most important piece - the line number of
the failing code.
--
==================
Remove the "x" from my email address
Jerry Stuckle
JDS Computer Training Corp.
jstucklex@attglobal.net
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