adding a line feed when building a string php - php

Question

i have been looking all over and i cant find the answer. i am sure this is easy
i am sending an email, and in the body are names of clients
the code looks for clients that fit the search conditions and as each one is found it goes into a loop
while($row = mysql_fetch_array($result)) {
$clients = $clients . $first_name . ???
}
the result should be
Client1
Client2
Client3
but what i keep getting is:
Client1Client2Client3
I have tried
$clients = $clients . $first_name . lf;
$clients = $clients . $first_name . cr;
$clients = $clients . $first_name . '\n';
but always the same result
TIA

Answer 1

If your output will be the command line, file, etc. Use PHP_EOL. If it will be a web browser use <br>:
while($row = mysql_fetch_array($result)) {
$clients = $clients . $first_name . PHP_EOL;
}
while($row = mysql_fetch_array($result)) {
$clients = $clients . $first_name . "<br/>";
}

Answer 2

For text e-mails, the simplest case would be:
$clients = $clients . $first_name . "\n";
Note that you need to enclose \n in double (") instead of single (`) quotes, so that it gets treated as a new line character.
If you are sending HTML e-mails, then you'll need the appropriate HTML tag, i.e. <br>

Answer 3

You can use "\n" or PHP_EOL (single quoted strings like '\n' are not interpolated)
$clients = $clients . $first_name . PHP_EOL;
If you're outputting this to the browser, newline characters are not displayed in html, so you can do something like this:
print nl2br($clients);
Or this:
header('Content-type: text/plain');
print $clients;`
Or this:
printf('<pre>%s</pre>', $clients);