Solved

Building a html table from the database?

Posted on 2007-03-28
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Last Modified: 2013-12-13
Hello,

I wondered if anyone could advise, I am trying to write some php code to access the data from the products table, and then display these products as 4 per row in the html table.

For example if there were 4 products there would be 1 html table row displaying the 4, and 10 products 3 rows.

I am not quite sure where to start with the display. I wondered anyone would be-able to offer any guidance to writing this type of nested loop?

Thank you

<?php
$Sproducts = "SELECT title, price, description, small_image, large_image FROM products";
$Srproducts = mysql_query($Sproducts);
                     
if($Srproducts && mysql_num_rows($Srproducts) > 0) {                                                    
?>        

<?php  
     while ($rproducts = mysql_fetch_assoc($Srproducts)) {                            
?>

<?php              
     }
}

?>
0
Comment
Question by:AntoniRyszard
  • 4
  • 3
7 Comments
 
LVL 12

Expert Comment

by:Jeff Darling
ID: 18814280

The html table looks like this:

<br />
<table border="0">
<tr>
<td><img src="1.jpg" /></td>
<td><img src="2.jpg" /></td>
<td><img src="3.jpg" /></td>
<td><img src="4.jpg" /></td>
</tr>
<tr>
<td><img src="5.jpg" /></td>
<td><img src="6.jpg" /></td>
<td><img src="7.jpg" /></td>
<td><img src="8.jpg" /></td>
</tr>
<tr>
<td><img src="9.jpg" /></td>
<td><img src="10.jpg" /></td>
</tr>
</table>


The PHP code that generates the html above forces the table to wrap at 4 columns.


$results = mysql_query($sql) or die ('problem with db query' . mysql_error());

echo "<table border=\"0\">\n";
echo "<tr>\n";

$rowcnt = mysql_num_rows($results);
$row = 0;
$col = 0;

for($u=0;$u<$rowcnt;$u++) {

$col = $col + 1;

$ProdPic=mysql_result($results,$u,'ProdPic');
$ProdTitle=mysql_result($results,$u,'ProdTitle');

// start new row
if($col > 4) {
 $row = row + 1;
 $col = 1;
 echo "</tr>\n";
 echo "<tr>\n";
}


echo "<td><img src=\"$ProdPic\" /></td>\n";

}

echo "</tr>\n";
echo "</table>\n";

You can see a demo at http://www.polysyncronism.com/CodeShare/PHPHTMLTableDemo/phpsql1.php
0
 

Author Comment

by:AntoniRyszard
ID: 18819660
Thank you, I will try the code with my database.

I was hoping to display the name and price of the product under each photo. Would adding another two rows under the image be simple to add to the existing loop?
0
 
LVL 12

Expert Comment

by:Jeff Darling
ID: 18824514

I think you are looking for something like this?

<table border="1">
<tr>
<td><img src="1.jpg" /></td>
<td><img src="2.jpg" /></td>
<td><img src="3.jpg" /></td>
<td><img src="4.jpg" /></td>
</tr>
<tr align="center">
<td>one $1</td>
<td>two $2</td>
<td>three $3</td>
<td>four $4</td>
</tr>
<tr>
<td><img src="5.jpg" /></td>
<td><img src="6.jpg" /></td>
<td><img src="7.jpg" /></td>
<td><img src="8.jpg" /></td>
</tr>
<tr align="center">
<td>five $5</td>
<td>six $6</td>
<td>seven $7</td>
<td>eight $8</td>
</tr>
</table>

I'll build the PHP to create this later.  
Here is a demo of the above html: http://www.polysyncronism.com/CodeShare/PHPHTMLTableDemo/Table1.html
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Author Comment

by:AntoniRyszard
ID: 18826002
Thank you, I would be very interested to see the how to add an extra row of prices under the image.
0
 

Author Comment

by:AntoniRyszard
ID: 18832560
Thank you, I would be very interested to look through the code to add a row under the image.

Could I ask if the code would be posted here, or on the site below?

www.polysyncronism.com/CodeShare/PHPHTMLTableDemo/Table1.html

Thank you
0
 
LVL 12

Expert Comment

by:Jeff Darling
ID: 18834443

Take a look at this.  It may be a bit messy but it appears to work.  

You should get the general idea.

http://www.polysyncronism.com/CodeShare/PHPHTMLTableDemo/phpsql2.php
0
 
LVL 12

Accepted Solution

by:
Jeff Darling earned 90 total points
ID: 18836130
$sql = "SELECT * FROM $DBName WHERE $DBField>0 ORDER BY $DBField";

$results = mysql_query($sql) or die ('problem with db query' . mysql_error());


echo "<table border=\"0\">\n";


$rowcnt = mysql_num_rows($results);
$row = 0;
$col = 0;

$ProdPicIdx=0;
$ProdPriceIdx=0;

for($u=0;$u<$rowcnt;$u++) {

$col = $col + 1;


 if($col == 1){
  echo "<tr>\n";
 }
 $ProdPic=mysql_result($results,$ProdPicIdx,'ProdPic');
 echo "<td><img src=\"$ProdPic\" /></td>\n";
 $ProdPicIdx = $ProdPicIdx + 1;

// start new row?

if($col > 3) {
  $row = row + 1;
  $col = 0;

  // Process Price and Title row

    echo "<tr>\n";
    for($x=0;$x<4;$x++){
      $ProdTitle=mysql_result($results,$ProdPriceIdx,'ProdTitle');
      $ProdPrice=mysql_result($results,$ProdPriceIdx,'ProdPrice');
      echo "<td align=\"CENTER\">".$ProdTitle." - $".$ProdPrice."</td>\n";
      $ProdPriceIdx = $ProdPriceIdx + 1;
    }
    echo "</tr>\n";

}

}

    // Process remaining Price and Titles for last row.
    echo "<tr>\n";
    for($x=0;$x<$col;$x++){
      $ProdTitle=mysql_result($results,$ProdPriceIdx,'ProdTitle');
      $ProdPrice=mysql_result($results,$ProdPriceIdx,'ProdPrice');
      echo "<td align=\"Center\">".$ProdTitle." - $".$ProdPrice."</td>\n";
      $ProdPriceIdx = $ProdPriceIdx + 1;
    }
      echo "</tr>\n";

echo "</table>\n";

mysql_close ($dbh);
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