echo ! question

[Michael297]

I'm trying to achieve the output where the remote ip is 1.2.3.4(for example):
1.2.3.4 !--IPADDR=1.2.3.4-->

I've tried just about everything...here is one variation:
<?php
echo "<";
echo "!";
echo "--IPADDR=";
echo $_SERVER['REMOTE_ADDR'];
echo
"-->";
?>

No matter what I try the ! screws it up.
If I do:
<?php
echo "!"';
?>

It seems to work, but as soon as I try
<?php
echo "<";
echo "!"';
?>

Nothing displays...any help is appreciated.

Thaks.

[Derokorian]

<!-- [stuff] -->

This is the syntax for an HTML comment. Perhaps you simply need to right click on the page and select View Source to see the comment.

[bradgrafelman]

Also note there's no need to use separate echo statements...

echo "<!--IPADDR=$_SERVER[REMOTE_ADDR]-->";

// or...

echo "<!--IPADDR=" . $_SERVER['REMOTE_ADDR'] . "-->";

// or...

printf( "<!--IPADDR=%s-->", $_SERVER['REMOTE_ADDR'] );

[cretaceous]

Michael297;11025407 wrote:

<?php
echo "<";
echo "!"';
?>

may not be the problem everywhere but this one has an extra single quote:
echo "!"';