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PHP Discussion Thread
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Jan 3rd, 2009

urgent help...

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I m new in php.i want some help in my code..plz..
i have 15 image in mysql table .and i want to retreive atleast 3 images in first row.den other 3 in second row..how can i make it possible..
the code is given below..!

php Syntax (Toggle Plain Text)
<?php
include "dbconfig.php";
?>
 
<link href="main.css" rel="stylesheet" type="text/css" />
<table border="1" cellpadding="0" cellspacing="0" bordercolor="#BCC5C7">
 
<tr><?php
$results = "SELECT * FROM files WHERE menu_item_id='5' limit 5";
$Img_result = mysql_query ($results, $conn);
if (mysql_num_rows ($Img_result) >= 0){
$Img_menu = array();
while ($rows = mysql_fetch_array ($Img_result, MYSQL_ASSOC)){
 
$realname = $rows ['filename'];
$small_preview = 'download/wallpapers/'.$realname.'';
$large_preview = 'download/wallpapers/'.$realname.'';
 
?><td><?php echo "<a href='$large_preview' target='_blank'><img width='120' height='200' src='$small_preview' border='0'></a>";?>
<br /><div class="pic"><strong>Name :</strong> <?php echo $rows ['title']; ?></div></td><? } } ?></tr>
 
</table>
<?php
include "dbconfig.php";
?>

<link href="main.css" rel="stylesheet" type="text/css" />
<table border="1" cellpadding="0" cellspacing="0" bordercolor="#BCC5C7">

<tr><?php
$results = "SELECT * FROM files WHERE menu_item_id='5' limit 5";
$Img_result = mysql_query ($results, $conn);
if (mysql_num_rows ($Img_result) >= 0){
$Img_menu = array();
while ($rows = mysql_fetch_array ($Img_result, MYSQL_ASSOC)){

$realname = $rows ['filename'];
$small_preview = 'download/wallpapers/'.$realname.'';
$large_preview = 'download/wallpapers/'.$realname.'';

?><td><?php echo "<a href='$large_preview' target='_blank'><img width='120' height='200' src='$small_preview' border='0'></a>";?>
<br /><div class="pic"><strong>Name :</strong> <?php echo $rows ['title']; ?></div></td><? } } ?></tr>

</table>

Last edited by peter_budo; Jan 6th, 2009 at 2:04 pm. Reason: Keep It Organized - For easy readability, always wrap programming code within posts in [code] (code blocks) and [icode] (inline code) tags.

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SkyVValker

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Jan 3rd, 2009

Re: urgent help...

Are you saying you want 3 images per table row?
If this is what you want to do, you may also want to change your limit on the query results to 6.

Then you would put your table creation inside of a php loop:

PHP Syntax (Toggle Plain Text)
<?php
$totalResults=mysql_num_rows($yourQuery);
$numberOfImagesPerRow=3;
$numberOfRows=$totalResults/$numberOfImagesPerRow;
echo "<table">;
for($x=0;$x<$numberOfRows;$x++)
{
echo "<tr>";
  for($i=0;$i<$numberOfImagesPerRow;$i++)
   {
      echo "<td>";
      //picture display code here
      echo "</td>;
   }
echo"</tr>";
}
echo "</table>";
?>
 
<?php
$totalResults=mysql_num_rows($yourQuery);
$numberOfImagesPerRow=3;
$numberOfRows=$totalResults/$numberOfImagesPerRow;
echo "<table">;
for($x=0;$x<$numberOfRows;$x++)
{
echo "<tr>";
  for($i=0;$i<$numberOfImagesPerRow;$i++)
   {
      echo "<td>";
      //picture display code here
      echo "</td>;
   }
echo"</tr>";
}
echo "</table>";
?>

And if that isn't what you want to do, well now you know how to do that!

Last edited by DiGSGRL; Jan 3rd, 2009 at 7:16 pm. Reason: Forgot a bracket

DiGSGRL

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Jan 4th, 2009

Re: urgent help...

Thanx DisGSGurl...!i got it..thanx a billion

SkyVValker

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Jan 4th, 2009

Re: urgent help...

there is an other prob..it repeat 1 image various times..n i want to retri all images from database.this code showing 3 images per row..
plz chk ma array n loop..
==========

php Syntax (Toggle Plain Text)
<?php
$results = "SELECT * FROM files WHERE menu_item_id='5' order by rand()";
$Img_result = mysql_query ($results, $conn);
$totalResults=mysql_num_rows($Img_result);
$numberOfImagesPerRow=3;
$numberOfRows=$totalResults/$numberOfImagesPerRow;
 
if (mysql_num_rows ($Img_result) >= 0){
$Img_menu = array();
while ($rows = mysql_fetch_array ($Img_result, MYSQL_ASSOC)){
 
 
echo "<table>";
for($x=0;$x<$numberOfRows;$x++)
{
echo "<tr>";
  for($i=0;$i<$numberOfImagesPerRow;$i++)
   {
      echo "<td>";
      //PREVIEW IMAGE IF EXISTS
$realname = $rows ['filename'];
$small_preview = 'download/wallpapers/'.$realname.'';
 
	echo "<div align='center'><img src='$small_preview' border='0'></div><BR>";
 
 
      echo "</td>"; 
     }
 
	}  
echo "</tr>";
}
}
echo "</table>";
 
?>
<?php
$results = "SELECT * FROM files WHERE menu_item_id='5' order by rand()";
$Img_result = mysql_query ($results, $conn);
$totalResults=mysql_num_rows($Img_result);
$numberOfImagesPerRow=3;
$numberOfRows=$totalResults/$numberOfImagesPerRow;

if (mysql_num_rows ($Img_result) >= 0){
$Img_menu = array();
while ($rows = mysql_fetch_array ($Img_result, MYSQL_ASSOC)){


echo "<table>";
for($x=0;$x<$numberOfRows;$x++)
{
echo "<tr>";
  for($i=0;$i<$numberOfImagesPerRow;$i++)
   {
      echo "<td>";
      //PREVIEW IMAGE IF EXISTS
$realname = $rows ['filename'];
$small_preview = 'download/wallpapers/'.$realname.'';

	echo "<div align='center'><img src='$small_preview' border='0'></div><BR>";

 
      echo "</td>"; 
     }
	 
	}  
echo "</tr>";
}
}
echo "</table>";

?>

===============

SkyVValker

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Jan 4th, 2009

Re: urgent help...

The reason you are only showing one file over and over is due to this line:

PHP Syntax (Toggle Plain Text)
$realname = $rows ['filename'];
$realname = $rows ['filename'];

You need to tell it to iterate through the results array. You should keep a counter and append the number onto that line:

PHP Syntax (Toggle Plain Text)
$i=0;(outside of your while loop)
$realname = $rows ['filename'][$i];
$i++;
 
WHERE menu_item_id='5'
$i=0;(outside of your while loop)
$realname = $rows ['filename'][$i];
$i++;

WHERE menu_item_id='5'

Your where clause determines how many records are going to be returned. If you are having trouble after trying the first suggestion of moving through your array, then make sure the where is what you need it to be.

DiGSGRL

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Jan 4th, 2009

Re: urgent help...

all recordes..
let me explain..!
if i have 12 images in my db..den
1 image repeat 12 time.den 2nd image repeat 12 times,..

SkyVValker

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Jan 4th, 2009

Re: urgent help...

Take the table out of the while loop.
Use the while loop to create your image array.

PHP Syntax (Toggle Plain Text)
$i=0;
while ($rows = mysql_fetch_array ($Img_result, MYSQL_ASSOC))
{
$imageArray=$rows['filename'][$i};
$i++;
}
$i=0;
while ($rows = mysql_fetch_array ($Img_result, MYSQL_ASSOC))
{
$imageArray=$rows['filename'][$i};
$i++;
}

Then inside your table where you display your filename:

PHP Syntax (Toggle Plain Text)
$c=0;(outside of the for loop)
$realname = $rows ['filename'][$c];
$c++;
$c=0;(outside of the for loop)
$realname = $rows ['filename'][$c];
$c++;

DiGSGRL

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Jan 4th, 2009

Re: urgent help...

DiGSGurl...would u like to eidt ur suggestion in ma code plz ..

SkyVValker

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