Not getting desire output - php

Question

What I want to do is,
I am initializing $message variable for mail body.
And I has to iterate through loop if multiple files are there.
So how can i iterate loop and store values in a variable.
Like,
$message = "Dear " . $fname . " <br/>
<b>Manuscript and Other Documents :</b> <br/>
Source File : " .$file1 . " <br/>
Source PDF File : ".$file2 . " <br/>
Cover Letter : " . $file3 . "<br/>
Supplementary Files : " . while($row=mysql_fetch_array($supplementary)){ echo $row[0] } . ";
So how can i do this if i have multiple supplementary files?

Answer 1

To do this, simply append the data to the message variable within the while loop using the string concatenation operator.
For example:
$message = "Dear " . $fname . " <br/><b>Manuscript and Other Documents :</b><br/>
Source File : " .$file1 . "<br/>Source PDF File : ".$file2 . " <br/>
Cover Letter : " . $file3 . "<br/>Supplementary Files : ";
while($row = mysql_fetch_array($supplementary)) {
$message .= echo $row[0] . '<br />';
}

Answer 2

First, do this:
$suplArray = array();
while($row=mysql_fetch_array($supplementary)){
$suplArray[] = $row[0];
}
Now, you have the strings in an array. I don't know what format you want, but you can do something like
$suplStr = implode(',', $suplArray);
Then in your string, you can have:
... "Supplementary Files : " . $suplStr;