PHP - Conditional Format (cell Colour) On Table

hello,

lets say i want 3 of the following columns to only display on 1 line always, is there a way to do that?

Code: [Select]
<th align='center'>Date</th>
<th align='center'>Supplier</th>
<th align='center'>Requested By</th>
<th align='center'>Address of Job</th>
<th align='center'>Job Number</th>
<th align='center'>P.O. Number</th>
<th align='center'>Description</th>
<th align='center'>Amount</th>
<th align='center'>Notes</th>
<th align='center'></th>

how to format only one specific row based on the value of  one cell in a row?

$result = mysql_query(select bal,  etc. etc ..query )

while($row = mysql_fetch_assoc($result))

echo "<tr>";

echo "<td>" . $row['Account'] . "</td>";

 echo "<td>" . $row['licence'] . "</td>";

 echo "<td>" . $row['own'] . "</td>";

 echo "<td>" . $row['bal'] . "</td>";

  echo "</tr>";

Would appreciate help on how to format only the $row['bal'] output to red colour IF bal >5000

Thanks

Trying to run this function so it displays the generated image in a table cell -

Code: [Select]
<?php 
function imgsecuregen($size = 6){ 

$width = 11*$size; 
$height = 25; 

$string = ""; 
for($i = 1; $i <= $size; $i++){ 
$string .= rand (0,9).""; 
} // for 

$im = ImageCreate($width, $height); 
$bg = imagecolorallocate($im, 102, 102, 102);  // background
$black = imagecolorallocate($im, 0, 255, 0);   // text
$grey = imagecolorallocate($im, 102, 102, 102);    // border 
imagerectangle($im,0, 0, $width-1, $height-1, $grey); 
imagestring($im, 5, $size, 5, $string, $black); 
imagepng($im); 
imagedestroy($im); 
} 

imgsecuregen(8);  // string length

?>

Works if I load it directly, but not in a <td>.
 

This topic has been moved to Third Party PHP Scripts.

http://www.phpfreaks.com/forums/index.php?topic=314271.0

I am trying to insert data from a mysql in a table. I want the data to appear with the image on the left, item name on the top right followed by description underneath.


<?php
$sql = 'SELECT * FROM tbl_products ORDER BY id';
$result = $db->query($sql);
$output[] = '<ul>';
$output[] = '<table border="3" bordercolor="#000000">';
while ($row = $result->fetch()) {

         $output[] = '<tr>';
         $output[] = '<td><img src="images/'.$row['pic'].'" width="67" height="100" /></td>';
         $output[] = '<td class="style10"><strong>'.$row['item_name'].'</strong></td>';
         $output[] = '<td>'.$row['item_description'].'</td>';
         $output[] = '</tr>';

}
$output[] = '</table>';
echo join('',$output);
?>

I would like the items to appear the same way as this page

http://aafcollection.info/items/list.php

This topic has been moved to HTML Help.

http://www.phpfreaks.com/forums/index.php?topic=307147.0

Hi guys i have another question about my shopping cart.   I have in my sql database table called products. It consist of 4 columns (product_id, description, quantity, price). Now on my index page, I want to call just one product from cell, which would have special price. But i don't know how to call one single product.   So my code looks like that:

 <?php
			
			$sql="SELECT product_id FROM Products";
			$query=mysql_query($sql);
			$row=mysql_fetch_array($query);
			
			
	 ?><table >
          
	 
	 				<tr><img src="Images/Image.jpg" width="380" height="300" /> </tr>
                    <tr>
						
						
						<td><a href="index.php?page=index&action=add&id=<?php echo $row['product_id']?>">Add to cart</a></td>
						
						</tr>
    <?php
			
		
     ?>
        </table>
       

This code display me first product but i would like to call 3 product how would i do that, so i have 6 products and i would like to call product N.3 to my index page. My code calls the first one, how would it call the third.

Sorry for the beginner question, here I'm trying to retrieve data from the database and display it in the table format. But only table headers are printed, and not the actual values. The count variable is echoing 2 saying that data is present and correctly retrieved. Can anyone help?

<?php
include 'connect.php';
error_reporting(E_ALL ^ E_DEPRECATED);
error_reporting(E_ERROR | E_PARSE);
$sql="SELECT * FROM `resources` as r INNER JOIN `project_resources` as pr
ON r.res_id =pr.res_id WHERE project_id='$_POST[project_id]'";
$result=mysql_query($sql);

$count=mysql_num_rows($result);

if($result === FALSE) {
die(mysql_error()); 
}
echo "$count";
echo '<table>
<tr>
<th>Resource ID</th>
<th>Resource Name</th>
<th>Email</th>
<th>Phone Number</th>
<th>Reporting Manager</th>
<th>Role</th>
<th>Designation</th>
</tr>';

while ($row = mysql_fetch_array($result)) {
echo '
<tr>
<td>'.$row['res_id'].'</td>
<td>'.$row['res_name'].'</td>
<td>'.$row['email'].'</td>
<td>'.$row['phone_number'].'</td>
<td>'.$row['reporting_manager'].'</td>
<td>'.$row['role'].'</td>
<td>'.$row['designation'].'</td>
</tr>';

}

echo '
</table>';

?>

Edited by mac_gyver, 22 September 2014 - 07:25 AM.
code tags please

Coming from the world of Excel, I can easily format numbers as $1,500.00, or 27%.  When I uploaded a large chunk of data into SQL to be read back through a table, the values all come out as exactly what they were uploaded as.  For example, I have an SQL column set as Decimal(19,4) that I want formatted like currency, but which shows up as 1500.0000 in my table, or another column with type decimal(5,2) which shows up as 5500.0000, but which I want to show up as 55%.  How do I do this?

Hi all -

I am parsing info from mysql into a php table but the date being returned is in YYYY-MM-DD format and I want it DD-MM-YYYY. How would I do this?

Code at the moment is:

Code: [Select]
$link = connect();

function get_user_posts(){

$getPosts = mysql_query('SELECT a.id as id,a.post as question, a.answer as answer, a.created as created, a.updated as updated, b.vclogin as admin
FROM user_posts as a, admin as b WHERE a.admin_id=b.adminid',connect());
return $getPosts;
}


Code: [Select]
<table class="list">
<tr>
<th><?php echo getlocal("upost.post") ?></th>
<th><?php echo getlocal("upost.answer") ?></th>
<th><?php echo getlocal("upost.admin") ?></th>
<th><?php echo getlocal("upost.created") ?></th>
<th><?php echo getlocal("upost.updated") ?></th>
<th></th>
<th></th>
</tr>
<?php 
$all_questions = get_user_posts();

while ($question = mysql_fetch_assoc($all_posts)):
?>

<tr>
<td><?php echo $post['post']?></td>
<td><?php echo $post['answer']?></td>
<td><?php echo $post['admin']?></td>
<td><?php echo $post['created']?></td>
<td><?php echo $post['updated']?></td>
<td><a href="javascript:user_post('<?php echo $post['id']?>')">edit</a></td>
<td><a href="javascript:delete_post('<?php echo $post['id']?>')">delete</a></td>
</tr>


THANKS!

Image name will come from database.
It can be displayed in divs or table cells.
Which is better using divs or table cells?
Can anyone post sample code?
http://img28.imageshack.us/i/44233003.png/

Hello,   I am trying to display the data from two tables with proper format. But Its not happening   Here is my 1st table - orders   2.PNG   10.12KB   0 downloads   and my 2nd table - order_line_items    1.PNG   14.92KB   0 downloads   I want to display like this     3.PNG   4.03KB   0 downloads   Here is my code

$query = $mysqli->query("SELECT orders.order_id, orders.company_id, orders.order_for, order_line_items.order_id, order_line_items.item, order_line_items.unit,SUM(order_line_items.unit_cost * order_line_items.quantity) AS 'Total', order_line_items.tax from orders INNER JOIN order_line_items ON orders.order_id = order_line_items.order_id where orders.order_quote = 'Order' GROUP BY order_line_items.id"); ?>

<table id="dt_hScroll" class="table table-striped">
						<thead><tr>
						<th>Order ID</th>
						<th>Company</th>
						<th>Contact Person</th>
						<th>Products</th>
                        <th>Total</th>
                        
                        
						
</tr>

</thead>
<tbody> <?php

while($row = $query->fetch_array())
{
?>
<tr>
<td><?php echo $row['order_id']; ?></td>
<td><?php echo $row['company_id']; ?></td>
<td><?php echo $row['contact_person'] ?></td>
<td><?php echo $row['item']; ?></td>
<td><?php echo $row['Total']; ?> %</td>

</tr>
<?php

}	

But here order ID, Company ID, Contact Person are also repeating thrice with item in order_line_items table   Please suggest me how to do this

Hi

I have the following for an input form,  I was under the impression that ucwords would format the text into Initial Characters, but this does not do that?  What have I done wrong?

$ven_name = trim(ucwords(mysql_prep($_POST['ven_name'])));

can anione help me to display imageid along wid each image in table format...here is my code which takes image from user....n d id can b anithing i mean its ur choice u can start wid first image by giving it imageid 1 n can continue til d image ends


<?php
mysql_connect("localhost","root","");
@mysql_select_db(proj) or die( "Unable to select database");
$query="SELECT * FROM form";
$result=mysql_query($query);

//define a maxim size for the uploaded images in Kb
 define ("MAX_SIZE","100");

//This function reads the extension of the file. It is used to determine if the file  is an image by checking the extension.
 function getExtension($str) {
         $i = strrpos($str,".");
         if (!$i) { return ""; }
         $l = strlen($str) - $i;
         $ext = substr($str,$i+1,$l);
         return $ext;
 }

//This variable is used as a flag. The value is initialized with 0 (meaning no error  found) 
//and it will be changed to 1 if an errro occures. 
//If the error occures the file will not be uploaded.
 $errors=0;
//checks if the form has been submitted
 if(isset($_POST['Submit']))
 {
    //reads the name of the file the user submitted for uploading
    $image=$_FILES['image']['name'];
    //if it is not empty
    if ($image)
    {
    //get the original name of the file from the clients machine
       $filename = stripslashes($_FILES['image']['name']);
    //get the extension of the file in a lower case format
        $extension = getExtension($filename);
       $extension = strtolower($extension);
    //if it is not a known extension, we will suppose it is an error and will not  upload the file, 
   //otherwise we will do more tests
 if (($extension != "jpg") && ($extension != "jpeg") && ($extension != "png") && ($extension != "gif"))
       {
      //print error message
          echo '<h1>Unknown extension!</h1>';
          $errors=1;
       }
       else
       {
//get the size of the image in bytes
 //$_FILES['image']['tmp_name'] is the temporary filename of the file
 //in which the uploaded file was stored on the server
 $size=filesize($_FILES['image']['tmp_name']);

//compare the size with the maxim size we defined and print error if bigger
if ($size > MAX_SIZE*1024)
{
   echo '<h1>You have exceeded the size limit!</h1>';
   $errors=1;
}

//we will give an unique name, for example the time in unix time format
$image_name=time().'.'.$extension;
//the new name will be containing the full path where will be stored (images folder)
$newname="upload/".$image_name;
//we verify if the image has been uploaded, and print error instead
$copied = copy($_FILES['image']['tmp_name'], $newname);
if (!$copied)
{
   echo '<h1>Copy unsuccessfull!</h1>';
   $errors=1;
}}}}

//If no errors registred, print the success message
 if(isset($_POST['Submit']) && !$errors)
 {
    $sql=mysql_query("INSERT INTO image (img) VALUES('$newname')");
   if($sql)
   {
    echo "<h1>File Uploaded Successfully! Try again!</h1>";
   }
   
 }

 ?>

 <!--next comes the form, you must set the enctype to "multipart/frm-data" and use an input type "file" -->
 <form name="newad" method="post" enctype="multipart/form-data"  action="">
 <table>
 
    <tr><td><input type="file" name="image"></td></tr>
    <tr><td><input name="Submit" type="submit" value="Upload image"></td></tr>
 </table>   
 <p><a href="resizeimg.php">resize the images </a></p>
 </form>
 <p>&nbsp;</p>
 


here is my code for displaying d images