[Setting A Variable Based Upon URL]

a quick revision to the code above to 3 lines

if ($url == 'www.specialpizza.com') $site = 'Special Pizza';
if ($url == 'www.mainstreetspecialpizza.com') $site = 'Main Street Pizza';
if ($url == 'www.elmstreetspecialpizza.com') $site = 'Elm Street Pizza';

$_SERVER['SERVER_NAME'] outputs www. included in the server name

[Help with GET]

since $_GET is an associative array, you can also use array_key_exists()

there are several ways to do what you want

[php pagination]

placing the code in a while loop should fix the problem of only displaying one row, however I have realized an error that i made in the previous code, I placed the closing </table> tag inside of the while loop which I should not have done....try this instead.

$result = mysql_query("SELECT * FROM `members`  WHERE username='".$_SESSION['username']."' limit $eu, $limit");
while($fetch_users_data = mysql_fetch_object($result))
{

if($bgcolor=='#f1f1f1'){$bgcolor='#ffffff';}
else{$bgcolor='#f1f1f1';}

echo "<tr >";
echo "<td align=left bgcolor=$bgcolor id='title'> <font face='Verdana' size='2'>".$fetch_users_data->statementdate."</font></td>"; 
echo "<td align=left bgcolor=$bgcolor id='title'> <font face='Verdana' size='2'>".$fetch_users_data->statementtrasct."</font></td>";
echo "<td align=left bgcolor=$bgcolor id='title'> <font face='Verdana' size='2'>".$fetch_users_data->amounttransact."</font></td>"; 
echo "<td align=left bgcolor=$bgcolor id='title'> <font face='Verdana' size='2'>".$fetch_users_data->intacctbal."</font></td>"; 
echo "<td align=left bgcolor=$bgcolor id='title'> <font face='Verdana' size='2'>".$fetch_users_data->charges."</font></td>";
echo "<td align=left bgcolor=$bgcolor id='title'> <font face='Verdana' size='2'>".$fetch_users_data->acctbal."</font></td>";
echo "</tr>"; //removed the </table> tag below
}

 

[php pagination]

since you wish to display more than one row, you will need to place you mysql_fetch_object() function inside of a while loop. Also, I recommend puting your query into a variable before using it inside of the function

$result = mysql_query("SELECT * FROM `members`  WHERE username='".$_SESSION['username']."' limit $eu, $limit");
while($fetch_users_data = mysql_fetch_object($result))
{

if($bgcolor=='#f1f1f1'){$bgcolor='#ffffff';}
else{$bgcolor='#f1f1f1';}

echo "<tr >";
echo "<td align=left bgcolor=$bgcolor id='title'> <font face='Verdana' size='2'>".$fetch_users_data->statementdate."</font></td>"; 
echo "<td align=left bgcolor=$bgcolor id='title'> <font face='Verdana' size='2'>".$fetch_users_data->statementtrasct."</font></td>";
echo "<td align=left bgcolor=$bgcolor id='title'> <font face='Verdana' size='2'>".$fetch_users_data->amounttransact."</font></td>"; 
echo "<td align=left bgcolor=$bgcolor id='title'> <font face='Verdana' size='2'>".$fetch_users_data->intacctbal."</font></td>"; 
echo "<td align=left bgcolor=$bgcolor id='title'> <font face='Verdana' size='2'>".$fetch_users_data->charges."</font></td>";
echo "<td align=left bgcolor=$bgcolor id='title'> <font face='Verdana' size='2'>".$fetch_users_data->acctbal."</font></td>";
echo "</tr>";
echo "</table>";
}

 

Edit: since "members" is not a mysql reserved word, you do not need to wrap it in backticks, however this is still acceptable

[Help creating a hyperlink with images stored in variables.....]

ah I see, that would explain it. Glad you found your error.

 

I don't recommend storing HTML code inside of a variable in the future, causes confusion when incorporating the variable into your code.

[Help creating a hyperlink with images stored in variables.....]

can you show more code where you set your $imagetoshow variable

[Too much PHP and MySQL Query]

can you post your code please

[Help creating a hyperlink with images stored in variables.....]

you have forgotten to place you href and src attributes inside of quotations.

print "<tr><td><a href='$imagetoshow'><img src='$imagetoshow'></a></td><td>$title $subtitle</td><td>$location</td><td>$blurb</td></tr>";

[Problem Uploading Image Files to Windows Server, permission denied]

have you read my edited post?

[Help on query required]

the error isn't caused by the code that I provided, most likely your are missing a closing curly bracket somewhere. If you cant figure it out post your code and I will look at it

[Private message reply not grabbing id from_id]

<form method="post" action="#">
  <!-- form contents -->
</form>

[Help on query required]

use the complex syntax on your session inside of you query, also, add some debugging just in case that fails

<?php

	$query = "SELECT * FROM selections WHERE username = '{$_SESSION['username']}'";
	$result = mysql_query($query) or die(mysql_error());

	while($row = mysql_fetch_assoc($result))
	{
?>

[Problem Uploading Image Files to Windows Server, permission denied]

Edit: misread the title. 

 

Found this piece from php.net

 

Andrey P. 01-Mar-2011 03:23

I was trying to change permissions of a folder with chmod command with FTP connection. (I needed a writable folder to upload pictures with php)

 

I got the following respond:

"SITE CHMOD 777 uploads: command not understood"

 

The reason: Server is running under Windows system that does not allow to set file permissions via FTP. Conversely, the UNIX-running servers allow that.

 

Solutions:

 

1. If your web hosting provider has a web-based control panel that lets you set file permissions, then you need to login there and make changes.

 

2. It is possible to contact the hosting provider and ask them about this issue; maybe they can make the changes.

 

3. It is possible to change the hosting provider that has servers run on UNIX, and keep the site there.

[You have an error in your SQL syntax;]

is your movie_year field an INT?

[Private message reply not grabbing id from_id]

change

 print_r($_POST['senderID']);

to

 print_r($_POST);

to make sure all necessary data is being past except for the senderID, post your results

[Private message reply not grabbing id from_id]

why are you using javascript to validate your form instead of php? javascript can be disabled.

Is your $_POST['senderID'] empty?

Try using print_r($_POST); after your reply form has been submitted to see what is happening to that value

[Dymamic filename for image element]

No, I mean the one from which you've posted the above lines of code.

 

I'm confused

No, I mean the one from which you've posted the above lines of code.

I don't understand

i believe Pikachu means the actual file that contains

<td><img src="images/blog/<?php echo $row_Recordset1['filename'];?>"/> </td>

 

[Dymamic filename for image element]

what are the contents of your $row_Recordset1['filename']

[Private message reply not grabbing id from_id]

I notice that you are using $_SERVER['PHP_SELF'] as your action in your forms. I do not recommend doing this as there are several risks to doing this, XSS injection etc...

 

Also, can you show the code specific to handling the replies please

[PHP question]

can explain further what you mean by "response" please

[POST multiple columns of data to one row]

when you specify your $query value twice, you are actually overriding the first value of $query. I recommend doing something like this.

$section1 = $_POST['section1'];
$section2 = $_POST['section2'];
$query="INSERT INTO selections (section1, section2) VALUES ('$section1', '$section2')";
mysql_query($query) or die ("Error updating database: " . mysql_error());

[help !! Undefined variable: option in on line 19]

  if($_SESSION['user']==1){
      $option='edit';
      }
      elseif($_SESSION['user']==2){
      $option='approve/deny';
      }

im assuming that neither of theses conditions are met, therefore you $options variable is never set

[Hello! :D]

Welcome, feel free to ask us any questions along the way of your learning php

[upudating database]

http://dev.mysql.com/doc/refman/5.0/en/update.html

[looping images from mysql]

Can you post our code please