Ask script to echo n images per row

Hi all,

i have some problem with "show images". This is the field from table 'gallery',

id int(25)
imagePath varchar(250)
imageTitle varchar(250)

which is the image path taken from field 'imagePath', so i could be like this:

<img src="imgGallery/<?PHP $row["imagePath"]; ?>">

PHP:

The result that i want is like this, all images displayed in each row is 4 images:

<TABLE>
	<TR>
		<TD>
			<TABLE>
				<TR>
					<TD><img src="imgGallery/1.jpg"></TD>
				</TR>
				<TR>
					<TD>tile: image number 1</TD>
				</TR>
			</TABLE>
		</TD>
		<TD>
			<TABLE>
				<TR>
					<TD><img src="imgGallery/2.jpg"></TD>
				</TR>
				<TR>
					<TD>tile: image number 2</TD>
				</TR>
			</TABLE>
		</TD>
		<TD>
			<TABLE>
				<TR>
					<TD><img src="imgGallery/3.jpg"></TD>
				</TR>
				<TR>
					<TD>tile: image number 3</TD>
				</TR>
			</TABLE>
		</TD>
		<TD>
			<TABLE>
				<TR>
					<TD><img src="imgGallery/4.jpg"></TD>
				</TR>
				<TR>
					<TD>tile: image number 4</TD>
				</TR>
			</TABLE>
		</TD>
	</TR>
	<TR>
		<TD>
			<TABLE>
				<TR>
					<TD><img src="imgGallery/5.jpg"></TD>
				</TR>
				<TR>
					<TD>tile: image number 5</TD>
				</TR>
			</TABLE>
		</TD>
		<TD>
			<TABLE>
				<TR>
					<TD><img src="imgGallery/6.jpg"></TD>
				</TR>
				<TR>
					<TD>tile: image number 6</TD>
				</TR>
			</TABLE>
		</TD>
		<TD>
			<TABLE>
				<TR>
					<TD><img src="imgGallery/7.jpg"></TD>
				</TR>
				<TR>
					<TD>tile: image number 7</TD>
				</TR>
			</TABLE>
		</TD>
		<TD>
			<TABLE>
				<TR>
					<TD><img src="imgGallery/8.jpg"></TD>
				</TR>
				<TR>
					<TD>tile: image number 8</TD>
				</TR>
			</TABLE>
		</TD>
	</TR>
</TABLE>

PHP:

My question is how is the complete php script (also the query), so i can display the images as 4 images in a row, and 4 images again in next row...

Thank you

 

I will code this for $25.

 

Personally, I think your nested tables will bloat your html. I would try something like this:

<?php

$result = mysql_query( "SELECT * FROM gallery" );

$count = 0;

while ( $Row = mysql_fetch_assoc( $result ) )
{
	$count++;
	
	echo '<TABLE style="float:left;">
<TR>
<TD><img src="imgGallery/' . $Row["imagePath"] . '"></TD>
</TR>
<TR>
<TD>tile: ' . $Row["imageTitle"] . '</TD>
</TR>
</TABLE>';

	if ( $count % 4 ) echo '<br style="clear:both;" />';

}

?>

PHP:

The table in the loop could be replaced by a div if need be. Have a play and see what you think.

Brew

 

@phpl33t : thanx for your offering, but i'm really sorry this i really dont have enough money for this code, thanx anyway.

@Brewster : thanx it's working properly, but i have to change

if ( $count % 4 )

PHP:

to be like this

if ( $count % 4 == 0 )

PHP:

Best Regards

 

A fix for that code:

$result = mysql_query( "SELECT * FROM `gallery`") or die(mysql_error());
if (mysql_num_rows($result) >= 1) {
	echo '<table border="0">';
	$column = 1;
	while($items = mysql_fetch_array($result)) {
		$imagepath = stripslashes($items['imagepath']);
		$imageTitle = stripslashes($items['imageTitle']);
		if ($column == 1) {
			echo '<tr>';
		}
      		echo '
      			<td valign="top" width="50%" align="center">
      				<div><img src="./imgGallery/'.$imagePath.'"></div>
                                                   <div><strong>Title: '.$imageTitle.'</div>
      			</td>
      		';
      		$column++;
      		if ($column == 3) { 
      			echo '</tr>';
      			$column = 1;
      		}
	}
	echo '</table>';
} else {
	echo '<div>No images found in the database!</div>';
}

PHP: