从字符串中删除前导和尾随中断标记

Question

I have developed a simple PHP feed system which displays all of the logged in users posts. It works fine, however, I have come across a "bug" where if the user has made a status which contains leading and/or trailing <br> points, it'll break the status view.

Now, before anyone marks this as a duplicate; If have tried the following approaches (and two more from posts which I can no longer find) in removing the trailing and leading <br>'s:

1. Remove <br>'s from the end of a string

2. How to trim all leading/trailing
   code using php

However, they a) do nothing at all, or b) do work, but remove all the <br> points within the string, even if they are in the middle.

Attempts:

$ctx = preg_replace('/^(<br>){0,}|(<br>){0,}$/m', '', nl2br(escape($post->ctx)));

---

$ctx = preg_replace('/(<br>)+$/', '', nl2br(escape($post->ctx)));

---

$ctx = preg_replace('#(( ){0,}<br( {0,})(/{0,1})>){1,}$#i', '', nl2br(escape($post->ctx)));

---

I have been working on this for a little while now, but have come up with no proper fix. I have come close with attempt c, however, it would remove all <br> points.

All help is appreciated,
Thanks!

Answer 1

Try to replace with contents optionally wrapped by the <br />'s (note that nl2br inserts <br />):

^(?:<br />[
]+)*(.*?)(?:<br />[
]+)*$

Demo: https://regex101.com/r/7KDCuA/3

And the PHP code:

$input = "
Hello World
hello world

";
$unwrapped = preg_replace("|^(?:<br />[
]+)*(.*?)(?:<br />[
]+)*$|s", '$1', nl2br($input));
echo $unwrapped;

Demo: http://codepad.org/BT0A5583