PHP Colums Error

[shahbaz13]

Hi,
I am trying to display images from my db with the following code..

<?php
			include_once("scripts/connect.php");
			if ($q == "All Images") {
			$sql = mysql_query("SELECT * FROM gallery");
			$q = "All Images";
			} else {
			$sql = mysql_query("SELECT * FROM gallery WHERE caption LIKE '%$q%'");
			}
			$limit = 7;
			$count = 0;
			while($row = mysql_fetch_array($sql)) {
			$caption = $row['caption'];
			$id = $row['id'];
			$num_results = mysql_num_rows($sql);
			if($num_results == 0) {
			echo "<tr><td><font face='Arial'><b>Sorry, no results found...</b></font></td></tr>";
			} else {
			if($count < $limit) {
			if ($count ==0) {
			echo "<tr>";
			}
			echo "<td style='border-style:none; border-width:medium;'>";
			echo "<p align='center'>";
			echo "<a href='cms/gallery/$id.jpg' title='$caption' rel='lightbox[PhotoGallery1]'>";
			echo "<img id='$id' style='opacity:0.65' onmouseover=\"Animate('$id', '', '', '', '', '100', 250);return 

false;\" onmouseout=\"Animate('$id', '', '', '', '', '65', 250);return false;\" src='cms/gallery/$id.jpg' width='100' >";
			echo "</a>";
			echo "</td>";
} 
else {
$count == 0;
			echo "</tr><td style='border-style:none; border-width:medium;'>";
			echo "<p align='center'>";
			echo "<a href='cms/gallery/$id.jpg' title='$caption' rel='lightbox[PhotoGallery1]'>";
			echo "<img id='$id' style='opacity:0.65' onmouseover=\"Animate('$id', '', '', '', '', '100', 250);return 

false;\" onmouseout=\"Animate('$id', '', '', '', '', '65', 250);return false;\" src='cms/gallery/$id.jpg' width='100' >";
			echo "</a>";
			echo "</td>";
}
$count++;
}}
echo "</tr>";
?>

But when I execute it, it display first seven results in columns and the others 1 in 1 row... PLZ HELp...

[darkagn]

The line:

echo "</tr><td style='border-style:none; border-width:medium;'>";

is at fault I think. You close the row with the </tr> followed immediately with a <td> node. Are you able to view the source in the browser? I think you will find that you will have cells (<td>) outside of rows (<tr>) because of the line mentioned above, which will confuse the browser when it renders the table.

Also, you have some <p> tags that aren't closed, and your <img> tags aren't closed either.

[shahbaz13]

No, i tried everything but it didn't work... PLZ HELP...!!!

[rpv_sen]

hi

I think line no 9 $limit = 7;

and also you are checking if($count < $limit)

you have given the limit as 7 by default so it is not going beyond 7 so please try to change and test it

[shahbaz13]

no, that's not the problem, i have changed it many times... i don't know what else..

[ddymacek]

please post your last iteration of code.
you said you've 'changed it many times' so what are you looking at now, what issues are you still having?

[shahbaz13]

plzzzzzzzzzzz help
But when I execute it, it display first seven results in columns and the others 1 in 1 row... PLZ HELp...

[karthik_ppts]

in what format do you want to display your images?

[shahbaz13]

.jpg

[karthik_ppts]

in what format do you want to display your images?

Sorry.. I think my question is not understandable to you. format means 4x4 or 3x3 grid format.

[shahbaz13]

it can be any format....

[shahbaz13]

any format

[ddymacek]

I would not exactly recommend doing it this way, but this will solve your issue.

<?php

    include_once("scripts/connect.php");
    if ($q == "All Images") {
		$sql = mysql_query("SELECT * FROM gallery");
		$q = "All Images";
    } else {
		$sql = mysql_query("SELECT * FROM gallery WHERE caption LIKE '%$q%'");
    }
    
    $count = 0;
	$limit = 7;
	$num_results = mysql_num_rows($sql);
	/******
	// fake the data
	$cap=array();
	$ids = array();
	for ($s=0;$s<21;$s++) {
		array_push($cap,'caption-'.$s);
		array_push($ids,'id-'.$s);
		//$caption[$s] = 'caption-'$s;
		//$id[$s] = 'id-'$s;
	}
	*/
	// fake data from array above, this will be in the numrows variable.
	// $cntr = count($cap);
	//echo "cntr = " . $cntr . "<BR>";
	// want to loop on 7,
	echo "<table>";
	//foreach($cap as $c) {
	//echo "c = " . $c . "<BR>";
	//echo "id = " . $ids[$cnt] . "<BR>";
	while($row = mysql_fetch_array($sql)) {

		//echo "caption = " . $caption . "<BR />";
		//echo "id = " . $id . "<BR>";
		$caption = $c;
		$id = $ids[$cnt];
		//$caption = $row['caption'];
		//$id = $row['id'];		
		if($num_results == 0) {
			echo "<tr><td><font face='Arial'><b>Sorry, no results found...</b></font></td></tr>";
		} else {
			if($count < $limit) {
				if ($count == 0) {
					//echo "CNT was 0<br>";
					echo "<tr>";
				}
				
				echo "<td style='border-style:none; border-width:medium;'>";
				echo "<p align='center'>";
				echo "<a href='cms/gallery/$id.jpg' title='$caption' rel='lightbox[PhotoGallery1]'>";
				echo "<img id='$id' style='opacity:0.65' onmouseover=\"Animate('$id', '', '', '', '', '100', 250);return
				 
				false;\" onmouseout=\"Animate('$id', '', '', '', '', '65', 250);return false;\" src='cms/gallery/$id.jpg' width='100' >";
				echo "</a>";
				echo "</td>";
			} else {
			    // this is not how I would recommend solving your problem but this code will do the action you are requesting.
				// based on limit of 7 you want to start a new row for every 8th total record coming across so:
			    if ($count == 8 || $count == 16 || $count == 24 || $count == 32 || $count == 40 || $count == 48) {
					echo "</tr><tr>";
				}			
				//$cnt == 0;
				echo "<td style='border-style:none; border-width:medium;'>";
				echo "<p align='center'>";
				echo "<a href='cms/gallery/$id.jpg' title='$caption' rel='lightbox[PhotoGallery1]'>";
				echo "<img id='$id' style='opacity:0.65' onmouseover=\"Animate('$id', '', '', '', '', '100', 250);return
				false;\" onmouseout=\"Animate('$id', '', '', '', '', '65', 250);return false;\" src='cms/gallery/$id.jpg' width='100' >";
				echo "</a>";
				echo "</td>";
			}
		
		//$count++;  <--- this was in the wrong place.
		}		
	$count++;
	}
echo "</tr>";
echo "</table>";
?>

[shahbaz13]

ddymacek Thankyou very much...!!! It Worked...