[Resolved] [Resolved] functions & tables

Sera

i'm unable to declare the table elements with the other arguments of my function, what can i do?

<?php

function cat_query ($g_viewc_query,$g_viewc_result,$num,$CatId,$CatName,$bg,$row,....){
.....
<td align="left">', $row[1], '</td>
<td align="left">', $row[2], '</td>
<td align="left">', $row[3], '</td>
<td align="left">', $row[4], '</td>
<td align="left">', $row[5], '</td>
</tr>';
}

	echo '</table>';

	.....		}

what can i do about the function not accepting $row[ * ] in the part indicated by the dots within the first set of brackets?

Weedpacket

Why do you need to? You're passing $row already, aren't you?

Sera

yeah, that's what i thought, but then when i tried to run the code & selected an item from a drop down list i got the following message...

Warning: Missing argument 1 for cat_query() in c:\apache\htdocs\cat_query.inc on line 3

Warning: Missing argument 2 for cat_query() in c:\apache\htdocs\cat_query.inc on line 3

Warning: Missing argument 3 for cat_query() in c:\apache\htdocs\cat_query.inc on line 3

Warning: Missing argument 4 for cat_query() in c:\apache\htdocs\cat_query.inc on line 3

Warning: Missing argument 5 for cat_query() in c:\apache\htdocs\cat_query.inc on line 3

Warning: Missing argument 6 for cat_query() in c:\apache\htdocs\cat_query.inc on line 3

Warning: Missing argument 7 for cat_query() in c:\apache\htdocs\cat_query.inc on line 3

cat_query is the include file that contains the function. line 3 is where the function code begins. the whole cat_query code:

<?php

function cat_query ($g_viewc_query,$g_viewc_result,$num,$CatId,$CatName,$bg,$row){

//Make the query
$g_viewc_query = "SELECT GameTitle, PubId, MaxNumPlay, GameCube, PS2, Xbox FROM Game WHERE CatId=$CatId ORDER BY GameTitle ASC";

//Run the query
$g_viewc_result = @ ($g_viewc_query);

//How many records exist?
$num = mysql_num_rows ($g_viewc_result);

if ($num > 0) { //If it ran ok, display the records.				
echo "<p><big><b>There are currently $num games under the category $CatName in the database.</b></big></p>";

//Table header
echo '<table align="left" cellspacing="2" cellpadding="2">
<tr>
<td align="left"><b>Game Title</b></td>
<td align="left"><b>Publisher</b></td>
<td align="left"><b>No. Of Players</b></td>
<td align="left"><b>GameCube</b></td>
<td align="left"><b>PlayStation 2</b></td>
<td align="left"><b>Xbox</b></td></tr>';

//Fetch and print all the records.
$bg = '#eeeeee'; //Set the background color.
	while ($row = mysql_fetch_array($g_viewc_result, MYSQL_NUM)) {
	$bg = ($bg == '#eeeeee' ? '#ffffff' : '#eeeeee'); //Switch the background color.
	echo '<tr bgcolor="', $bg, '"><td align="left">', $row[0], '</td>
	<td align="left">', $row[1], '</td>
	<td align="left">', $row[2], '</td>
	<td align="left">', $row[3], '</td>
	<td align="left">', $row[4], '</td>
	<td align="left">', $row[5], '</td>
	</tr>';
	}

	echo '</table>';

	mysql_free_result ($g_viewc_result); //Free up the resources.

	}else { //If it did not run Ok.
	echo '<p>The games could not be displayed due to a system error.</p><p>' . mysql_error() . '</p>';
	}

mysql_close(); //Close the database connection.
}

?>[/color]

leatherback

Hi,

when you are using cat_query() you forget to put the arguments in.
You need to call the function, and place the proper values in the statement:

cat_query ($g_viewc_query,$g_viewc_result,$num,$CatId,$CatNa
me,$bg,$row)

Where the arguments are the values you want to use in the function.

Sera

ahhhhhh.... Leatherback, thank u so much!!!😃 😃 😃

Sera

now i'm getting this error msg:

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in c:\apache\htdocs\cat_query.inc on line 12

The games could not be displayed due to a system error.

Unknown column 'C3' in 'where clause'

why is mysql_num_rows() not valid?

leatherback

Hi,

Is the variable $CatId : c3?

you should place quotes around the values in the where statement.

The error you get bascially that you have an invalid sql query. Therefore there is no result, and then you cannot use mysql_num_rows.

You can very easily check for this kind of errors yourself, by echo-ing the query just before you actually run it. That will show you the query.

Also, in your script you should use error checkers:

if(!$g_viewc_result)
{
die(mysql_error());
}

Sera

Hey Leatherback,

once again, thanx!!!
this thing's actually working now!

Sera

the footer is appearing beside the table, instead of underneath it or at the bottom of the page.
i've taken it away for now.... anyway i can fix it pls? 🙂

leatherback

Impossible to tell without knowing HOW you position the footer, wether it is nested in tables, classes etcetc.

I suggest that if you cannot get it fixed, you posta new thread reg. this problem.

Since you seem to have the previous trouble under control, could you please makr this thread resolved/ (See link at bottom of page).

Sera

ok leatherback, thanx 4 ur time & help!🙂