Question

Running a db query, outputting as HTML table, but need to add 2 additional columns. How?

Asked by: eddo9696

Hello,
I am running a query against a mysql db and the output is sent to a function that creates a html table. It works, but now I need to be able to add two additional columns. One of the columns needs to contain a submit button(no functionality necessary), the other a link to an image that is stored locally. The file name of the image is stored in the database and I've written a function to change the file name into an img tag in html.  I believe that the key to solving this is editing my function for the html table but am not quite sure how to do it. I've attached my img tag function and html table loop. Any suggestions are appreciated. Thanks.

/***********************************************************************
* Function to print query as table
*
* Argument(s): 
* $result 		- Result from SQL query
***********************************************************************/ 
function dbtable($result) 
{
  ?><table border="1" width="80%"><tr><?php
  	if(! $result) 
		{ 
		?><th>result not valid</th><?php 
		}
  	else 
		{
		$i = 0;
		while ($i < mysql_num_fields($result)) 
		{
		  $meta = mysql_fetch_field($result, $i);
		  ?><th style="white-space:nowrap"><?=$meta->name?></th><?php
		  $i++;
		}
    ?></tr><?php
    if(mysql_num_rows($result) == 0) 
		{ 
		?><tr><td colspan="<?=mysql_num_fields($result)?>">
		<strong><center>no result</center></strong>
		</td></tr><?php 
		} 
	else
		while($row=mysql_fetch_assoc($result)) 
		{
		?><tr style="white-space:nowrap"><?php
			foreach($row as $key=>$value) 
			{ 
			?><td><?=$value?></td><?php 
			}
			?></tr><?php
		 }
  }
  ?></table><?php
} 

/***********************************************************************
* Function to create HTML image tag
*
* Argument(s): 
*	$imagename	string	- Name of image file
*
* Returns: Image link
***********************************************************************/ 
function nametoimage($imagename)
{
  if ($imagename != null)  
  {
    $result = "<img src=\"images/{$imagename} \" />" ;
  }
echo $result;  
}

                                  
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Asked On
2009-11-08 at 04:16:56ID24881488
Tags

database query table columns add image

Topics

PHP Scripting Language

,

PHP and Databases

,

PHP for Windows

Participating Experts
4
Points
500
Comments
7

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Answers

 

by: mmarthPosted on 2009-11-08 at 05:12:45ID: 25770357

try something like this. change the column names to fit your columns.


?>
<table>
      <tr>
         <td align="center" width="88" valign='bottom'><span class='black12B'>firstname</span></td>
         <td align="center" width="88" valign='bottom'><span class='black12B'>lastname</span></td>
         <td align="center" width="88" valign='bottom'><span class='black12B'>email</span></td>
      </tr>
      <?php

function openRS($query)
{
   mysql_close;
      mysql_connect(classSERVER,classUSERNAME,classPASSWORD)  or die();
      mysql_select_db(classDATABASE)  or (die() and print "1");
      $result=mysql_query($query)   or (die() and print "2");
      return $result;
}


$query = "SELECT * FROM Table ";
$rsClass=openRS($query);
$rowcount = mysql_num_rows($rsClass);
if ($rowcount > 0)
{
   while($tempClass = mysql_fetch_array($rsClass,MYSQL_ASSOC))
   {
      $username      = $tempClass["username"];
      $firstname     = $tempClass["firstname"];
      $lastname      = $tempClass["lastname"];
      $imagename     = $tempClass["imagename"];
   
      ?>
           
            <tr>
            <form METHOD="POST" action="updateClassMate.php" name="UpdateForm"  >
               <td align="left" width="88" valign='bottom'><input NAME="firstname" SIZE="22"   value="<?=$firstname?>" maxlength="22"></td>
               <td align="left" width="88" valign='bottom'><input NAME="lastname" SIZE="22"   value="<?=$lastname?>" maxlength="22"> </td>
               <td align="left" width="88" valign='bottom'><input NAME="email"  SIZE="40"   value="<?=$email?>" maxlength="55"></td>
               <td align="left" width="88" valign='bottom'>
                  <?php
                  if ($imagename != null)  
                  {
                     ?>
                        <img src="images/$imagename " />
                     <?php
                  }
                  ?>
               </td>
               <td align="left" width="88" valign='bottom'>
                  <input type='submit' NAME="ipClassMate"  SIZE="20"   value="update <?=$firstname?>">
               </td>
            </form>
            </tr>
   <?php
   }
}

?>

 

by: fiboPosted on 2009-11-08 at 10:33:30ID: 25771311

JUst a small comment...

$query = "SELECT * FROM Table ";
...
      $username      = $tempClass["username"];
      $firstname     = $tempClass["firstname"];
      $lastname      = $tempClass["lastname"];
      $imagename     = $tempClass["imagename"];

You should avoid SELECT * as much as possible.
In that case, it should be

$query = "SELECT username, firstname, lastname, imagename FROM Table ";

SELECT * can really be a magnific and hidden performance trap...

 

by: silver_surfer_44Posted on 2009-11-09 at 01:07:37ID: 25774343

I changed your existing code to add a submit button and image.

/***********************************************************************
* Function to print query as table
*
* Argument(s):
* $result               - Result from SQL query
***********************************************************************/
function dbtable($result)
{
  ?><table border="1" width="80%"><tr><?php
        if(! $result)
                {
                ?><th>result not valid</th><?php
                }
        else
                {
                $i = 0;
                while ($i < mysql_num_fields($result))
                {
                  $meta = mysql_fetch_field($result, $i);
                  ?><th style="white-space:nowrap"><?=$meta->name?></th><?php
                  $i++;
                }
<th style="white-space:nowrap">Submit</th>
    ?></tr><?php
    if(mysql_num_rows($result) == 0)
                {
                ?><tr><td colspan="<?=mysql_num_fields($result)?>">
                <strong><center>no result</center></strong>
                </td></tr><?php
                }
        else
                while($row=mysql_fetch_assoc($result))
                {
                ?>
<tr style="white-space:nowrap">
<?php
                        foreach($row as $key=>$value)
                        {
                       if ($key == 'image'){
                        ?><td><?php echo nametoimage($value)?></td><?php
                       }
                       else{
                       ?><td><?=$value?></td><?php
                       }
                        }
                        ?>
<td><input type=submit value=submit /></td>

</tr><?php
                 }
  }
  ?></table><?php
}

 

by: pg-expertPosted on 2009-11-09 at 01:21:00ID: 25774417

Try this code,

you may need to include form tags around, if you dont have one.

HTH,
~PG

<?php
/***********************************************************************
* Function to print query as table
*
* Argument(s): 
* $result 		- Result from SQL query
***********************************************************************/ 
function dbtable($result) 
{
  ?><table border="1" width="80%"><tr><?php
  	if(! $result) 
		{ 
		?><th>result not valid</th><?php 
		}
  	else 
		{
		$i = 0;
		// considering imagename is fetched in the fourth column
		while ($i < mysql_num_fields($result)) 
		{
		  $meta = mysql_fetch_field($result, $i);
		  ?><th style="white-space:nowrap"><?=$meta->name?></th><?php
		  $i++;
		}
		//Fifth column, submit button.
    ?>
          <th style="white-space:nowrap">Submit</th>
    	</tr><?php
    if(mysql_num_rows($result) == 0) 
		{ 
		?><tr><td colspan="<?=mysql_num_fields($result)?>">
		<strong><center>no result</center></strong>
		</td></tr><?php 
		} 
	else
		while($row=mysql_fetch_assoc($result)) 
		{
		?><tr style="white-space:nowrap"><?php
			foreach($row as $key=>$value) 
			{
				// considering the image field as "imagename", replace as required	
				if ( $key == "imagename" ) $value = nametoimage($imagename);
				?><td><?=$value?></td><?php 
			}
			?>
			<td><input type="submit" value="Submit" name="submit"></td></tr><?php
		 }
  }
  ?></table><?php
} 
 
/***********************************************************************
* Function to create HTML image tag
*
* Argument(s): 
*	$imagename	string	- Name of image file
*
* Returns: Image link
***********************************************************************/ 
function nametoimage($imagename)
{
  if ($imagename != null)  
  {
    $result = "<img src=\"images/{$imagename} \" />" ;
  }
echo $result;  
}
 
?>

                                              
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by: fiboPosted on 2009-12-05 at 05:39:50ID: 25978987

AngelIII,
Mine excepted, the (3) contribution made do represent work which should somehow get some points!

Thx for your hard work.
B.

20120131-EE-VQP-002

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