sql-query works alone but not in function...

[Dennis]

Hy,

My english is not the best, so if you don´t understand something,
please ask!

I want to print the content of a mysql-table where a field is a
specified value.

For that i have written some code that works fine. But i don´t want to
rewrite the code everytime the specified value changes, so i tried to
make a function out of my code.
But now, i get a mysql error:
Warning: Supplied argument is not a valid MySQL result resource in
..........\conf ig.php on line 35

I have only added one line before the code:
function selLinkKategori e() {

and i want to call that function with:
selLinkKategori e();

It would be very nice if someone can help me a bit, cause i´m still
learning php, and i don´t see the problem...

Here is the code (line 35 from the error is line 6 here):



function selLinkKategori e() {
$verbindung = mysql_connect($ baseurl, $dbuser, $dbpass);
mysql_select_db ($dbase, $verbindung);
$sql = "SELECT * FROM $links WHERE kategorie = 'tuts'";
$result = mysql_query($sq l, $verbindung);

$num = mysql_num_rows( $result);
$cur = 1;

echo '<table width="727" border="0" align="center" cellspacing="2"
cellpadding="1" class="content" >
<tr>
<td nowrap><font color="#003366" ><b><u>Link</u></b></font></td>
<td nowrap><font color="#003366" ><b><u>Beschrei bung</u></b></font></td>
</tr>';

while ($num >= $cur) {
$row = mysql_fetch_arr ay($result);

$link = $row["link"];
$beschr = $row["beschreibu ng"];

printf ("<tr><td nowrap bgcolor='%s'>$l ink</td>", bgcolor(3));
printf ("<td nowrap bgcolor='%s'>$b eschr</td>", bgcolor(3));

$cur++;
}
echo "</table>";
}

[Tom Thackrey]

Re: sql-query works alone but not in function...


On 27-Sep-2003, dennis@schilde. org (Dennis) wrote:
[color=blue]
> My english is not the best, so if you don´t understand something,
> please ask!
>
> I want to print the content of a mysql-table where a field is a
> specified value.
>
> For that i have written some code that works fine. But i don´t want to
> rewrite the code everytime the specified value changes, so i tried to
> make a function out of my code.
> But now, i get a mysql error:
> Warning: Supplied argument is not a valid MySQL result resource in
> .........\confi g.php on line 35
>
> I have only added one line before the code:
> function selLinkKategori e() {
>
> and i want to call that function with:
> selLinkKategori e();
>
> It would be very nice if someone can help me a bit, cause i´m still
> learning php, and i don´t see the problem...
>
> Here is the code (line 35 from the error is line 6 here):
>
>
>
> function selLinkKategori e() {
> $verbindung = mysql_connect($ baseurl, $dbuser, $dbpass);
> mysql_select_db ($dbase, $verbindung);
> $sql = "SELECT * FROM $links WHERE kategorie = 'tuts'";
> $result = mysql_query($sq l, $verbindung);
>
> $num = mysql_num_rows( $result);
> $cur = 1;
>
> echo '<table width="727" border="0" align="center" cellspacing="2"
> cellpadding="1" class="content" >
> <tr>
> <td nowrap><font color="#003366" ><b><u>Link</u></b></font></td>
> <td nowrap><font
> color="#003366" ><b><u>Beschrei bung</u></b></font></td>
> </tr>';
>
> while ($num >= $cur) {
> $row = mysql_fetch_arr ay($result);
>
> $link = $row["link"];
> $beschr = $row["beschreibu ng"];
>
> printf ("<tr><td nowrap bgcolor='%s'>$l ink</td>", bgcolor(3));
> printf ("<td nowrap bgcolor='%s'>$b eschr</td>", bgcolor(3));
>
> $cur++;
> }
> echo "</table>";
> }[/color]

You either need to pass parameters for $baseurl, $dbuser, $dbpass, and
$dbase or declare them as globals.

--
Tom Thackrey

https://www.creative-light.com

[Jan Pieter Kunst]

Re: sql-query works alone but not in function...

In article <de9e51d.030927 0720.1c472c04@p osting.google.c om>,
dennis@schilde. org (Dennis) wrote:
[color=blue]
> It would be very nice if someone can help me a bit, cause i´m still
> learning php, and i don´t see the problem...
>
> Here is the code (line 35 from the error is line 6 here):
>
>
>
> function selLinkKategori e() {
> $verbindung = mysql_connect($ baseurl, $dbuser, $dbpass);
> mysql_select_db ($dbase, $verbindung);
> $sql = "SELECT * FROM $links WHERE kategorie = 'tuts'";
> $result = mysql_query($sq l, $verbindung);
>
> $num = mysql_num_rows( $result);
> $cur = 1;
>
> echo '<table width="727" border="0" align="center" cellspacing="2"
> cellpadding="1" class="content" >
> <tr>
> <td nowrap><font color="#003366" ><b><u>Link</u></b></font></td>
> <td nowrap><font color="#003366" ><b><u>Beschrei bung</u></b></font></td>
> </tr>';
>
> while ($num >= $cur) {
> $row = mysql_fetch_arr ay($result);
>
> $link = $row["link"];
> $beschr = $row["beschreibu ng"];
>
> printf ("<tr><td nowrap bgcolor='%s'>$l ink</td>", bgcolor(3));
> printf ("<td nowrap bgcolor='%s'>$b eschr</td>", bgcolor(3));
>
> $cur++;
> }[/color]


You need to explicitly pass the variables $baseurl, $dbuser, $dbpass and
$dbase into your function. Otherwise those variables are undefined
inside the function.

function selLinkKategori e($baseurl, $dbuser, $dbpass, $dbase) {
...
}

By the way, the mysql_fetch_arr ay loop can be done with fewer steps:

while($row = mysql_fetch_arr ay($result)) {
...
}

JP

--
Sorry, <devnull@cauce. org> is een "spam trap".
E-mail adres is <jpk"at"akamail .com>, waarbij "at" = @.

[Geoff Berrow]

Re: sql-query works alone but not in function...

I noticed that Message-ID:
<MZhdb.9927$TW. 7975@newssvr25. news.prodigy.co m> from Tom Thackrey
contained the following:
[color=blue]
>You either need to pass parameters for $baseurl, $dbuser, $dbpass, and
>$dbase or declare them as globals.[/color]

And also to the where clause.

--
Geoff Berrow
It's only Usenet, no one dies.
My opinions, not the committee's, mine.
Simple RFDs http://www.ckdog.co.uk/rfdmaker/