[RESOLVED] Displaying results in table

[trehy2006]

Hi,

I haven't started with this one myself, as I have know idea how to start it!!

Basically I want to run a SELECT query, to be displayed in a table.

The table has to show 3 records acrros and unltd. going down. What I need to know is how do I make it so that every third record it shows a </tr><tr> amongst the table code??

TIA

Robert

[alimadzi]

Try this...

<?php

$i = 1;
echo "<table><tr>";
$result = mysql_query( "SELECT foo FROM bar" );
while ( $row = mysql_fetch_assoc( $result ) ) {
    echo "<td>$row[bar]</td>";
    if ( $i % 3 == 0 ) { echo "</tr><tr>"; }
    $i++;
}
mysql_free_result( $result );
echo "</tr></table>";

?>

That should work for you. You will need to add extra code to handle situations where the number of records is not evenly divisible by three. This will get you started at least...

[trehy2006]

Thanks for that...

alimadzi wrote:

You will need to add extra code to handle situations where the number of records is not evenly divisible by three. This will get you started at least...

How would I go about doing that?

TIA

Robert

[alimadzi]

I'll give you a hint. I think it's useful to try it on your own first...

With three columns, there are three possibilities when you reach the end of your cells:

1. Just added 3rd cell to row: do nothing
2. Just added 2nd cell to row: add one more empty cell to complete table
3. Just added 1st cell to row: add two more empty cells to complete table

Or with #3 you could add a single blank cell with a rowspan of 2. The modulus operator (%) will be your friend in doing all of this. Good luck. We'll help if you get stuck.

[trehy2006]

Okay,

Would it be something along the lines of expanding the if statement a couple of times...like so...

if ( $i % 3 == 0 ) { echo "</tr><tr>"; }
else if ( $i % 2 == 0 ) { echo "<td>&nbsp;</td>"; }
else if( $i % 1 == 0 ) { echo "<td colspan='2'>&nbsp;</td>";}

Does that look okay, Im happy with (as in to me it looks alright) the % 2... (if the row has two items in it??)
But I'm not happy with $i % 1 == 0 (if only 1 item)... Is this correct tho... Unfortunately my server is down atm so I can't even run it to check it.. I was hoping you would be able to tell me by looking at it if this is the right direction!

Thanks

Robert

[laserlight]

But I'm not happy with $i % 1 == 0 (if only 1 item)... Is this correct tho...

That will be true for all integers $i, thus it would be better as just an else instead of an elseif.

However, it is wrong, as the logic cannot be within the loop. You only want to handle the 'missing' records after the main loop has ended.

[trehy2006]

so... how would I go about that? Sorry, I'm a bit confused/lost now!! 🙁

[alimadzi]

trehy2006 wrote:

so... how would I go about that? Sorry, I'm a bit confused/lost now!! 🙁

OK, I'll help. You were on the right track. The following should go after the foreach loop and before your closing table tag...

if ( $i % 3 == 2 ) {
    echo "<td>&nbsp;</td><td>&nbsp;</td>\n";
} elseif ( $i % 3 == 0 ) {
    echo "<td>&nbsp;</td>\n";
}

Using the counter as I did in my first code sample, the remainder will be 1 after a complete row. Add one more cell and a new row starts, so you need two empty cells. Add two more cells and you're one short of another row.

Hope this helps...

[trehy2006]

Thanks,

So I would end up with something like this??

<?php 

$i = 1; 
echo "<table><tr>"; 
$result = mysql_query( "SELECT foo FROM bar" ); 
while ( $row = mysql_fetch_assoc( $result ) ) { 
    echo "<td>$row[bar]</td>"; 
    if ( $i % 3 == 0 ) { echo "</tr><tr>"; } 
    $i++; 
} 
mysql_free_result( $result ); 
if ( $i % 3 == 2 ) { 
    echo "<td>&nbsp;</td><td>&nbsp;</td>\n"; 
} elseif ( $i % 3 == 0 ) { 
    echo "<td>&nbsp;</td>\n"; 
} 
echo "</tr></table>"; 

?>

I think this is what you mean??

Thanks
Robert

[alimadzi]

Yeah, exactly. Have you tried it yet?

Sorry about leading you on and trying to get you to do some of this. I used to teach high school and old habits die hard. Speaking of which...

Your next exercise (should you choose to accept it) is to store the number of columns in a variable that you can change at will. All the table generating code should adapt to whatever that value is. Have fun! 😉

[laserlight]

There is actually a more general solution by which you can replace:

if ( $i % 3 == 2 ) {
    echo "<td>&nbsp;</td><td>&nbsp;</td>\n";
} elseif ( $i % 3 == 0 ) {
    echo "<td>&nbsp;</td>\n";
}
echo "</tr></table>";

with:

for ($i %= 3; $i < 3 && $i != 0; ++$i) {
    echo "<td>&nbsp;</td>";
}
echo "\n</tr></table>";

Sorry about leading you on and trying to get you to do some of this.

It is usually better to give hints or give an algorithm outline than to spoonfeed with a code example solution right away, which unfortunately I am prone to do.

EDIT:
Sorry, I noticed that your count begins from 1 instead of 0, thus what I suggested does not work. If you count from 0, the only change needed should be changing:

    if ( $i % 3 == 0 ) { echo "</tr><tr>"; }
    $i++;

in the first loop to:

    $i++;
    if ( $i % 3 == 0 ) { echo "</tr><tr>"; }

Basically, you just increment before you test whether the end of the table row has been reached.

Incidentally, you can also use a switch with fall-through to handle the 'missing' columns. When counting from 1:

switch ($i % 3) {
case 2: echo "<td>&nbsp;</td>";
case 0: echo "<td>&nbsp;</td>";
}

[alimadzi]

Yeah, I realized after the fact that I probably should have initialized the counter variable to zero instead of one, but didn't want to confuse the poor guy... 😉