sed / awk替换包含任何可能的可打印字符的复杂字符串

Question

I need to replace a complex string that can contain any printable characters. This is a real example pattern that I've been unable to replace using sed .

<!-- %cmd: for F in $(find ../[09]* -maxdepth 1 -type d -printf "%P\n" ) | grep -v "^$"; do echo "<li><a href=\"$F\">$F</a></li>"; done -->

I'm even using a non-printable sed delimiter in order to avoid conflicts since the original string should be composed only of printable characters:

DELIM=$(echo -en "\001");

But it doesn't work. I've tried many things, and I can't figure out what I'm missing. Eg:

echo "BEFORE $PATTERN AFTER" | sed -e "s${DELIM}${PATTERN}${DELIM}NEWTEXT${DELIM}"

---

UPDATE-1:

The provided solution should print all the lines of the container text, while replacing the matching string with a new one.

pattern='<!-- %cmd: for F in $(find ../[0-9]* -maxdepth 0 -type d | sed "s/^\.\.\///"); do echo "<li><a href=\"$F\">$F</a></li>"; done -->'
container='
<h2>Title</h2>
<ul>
    <!-- %cmd: for F in $(find ../[0-9]* -maxdepth 0 -type d | sed "s/^\.\.\///"); do echo "<li><a href=\"$F\">$F</a></li>"; done -->
</ul>
'

---

UPDATE-2:

After a couple of iterations, this is the working, final and accepted answer built by @anubhava:

awk -v repl="newtext" 'FNR==NR {
    a = a $0; next
} n = index($0, a) {
    $0 = substr($0, 1, n-1) repl substr($0, n+length(a))
} 1' < (printf '%s\n' "$pattern") <(printf '%s' "$container")

Code demo

Answer 1

You can use conrol characters in sed's delimiter like this:

pattern='foobar'
delim=$'\01'
echo "before $pattern after" | sed "s${delim}${pattern}${delim}newtext${delim}"

before newtext after

---

Update:

As your pattern contain all sorts of special meta characters, it is better to ditch regex (sed) and use non-regex replacement using awk :

pattern='<!-- %cmd: for F in $(find ../[09]* -maxdepth 1 -type d -printf "%P\n" ) | grep -v "^$"; do echo "<li><a href=\"$F\">$F</a></li>"; done -->'

awk -v repl="newtext" 'FNR==NR {
    a = a $0; next
}
n = index($0, a) {
    $0 = substr($0, 1, n-1) repl substr($0, n+length(a))
} 1' <(printf "%s\n" "$pattern") <(echo "before $pattern after")

before newtext after

Code Demo