Remove characters between markers in a bash variable - bash

Question

I'm trying to remove unknown characters between 2 known markers from a variable using bash.
eg
string="This text d #! more text jsdlj end and mo{re ;re end text.text"
I want to remove all the characters between the last word "text " (before the end word) and the first occurance thereafter called "end" . ie between the last occurance of the word "text " after that the first occurance of the word "end", but keeping both these markers)
result="This text d #! more text end and mo{re ;re end text.text"
I'll be using it as part of a find -print0 | xargs -0 bash -c 'command; command...etc.' script.
I've tried
echo $string | sed 's/[de][ex][ft][^\-]*//' ;
but that does it from the first "ext" and "-" (not the last "ext" before the end marker) and also does not retain the markers.
Any suggestions?
EDIT: So far the outcomes are as follows:
string="text text text lk;sdf;-end end 233-end.txt"
start="text "
end="-end"
Method 1
[[ $string =~ (.*'"${start}"').*('"${end}"'.*) ]] || :
nstring="${BASH_REMATCH[1]}${BASH_REMATCH[2]}" ;
echo "$nstring" ;
>"text text text -end.txt"
Required output = "text text text -end end 233-end.txt"
Method 2
temp=${cname%'"$end"'*}
nend=${cname#"$temp"}
nstart=${temp%'"$start"'*}
echo "$nstart$nend"
>"text text -end.txt"
Required output = "text text text -end end 233-end.txt"
Method 3
nstring=$(sed -E "s/(.*'"$start"').*('"$end"')/\1\2/" <<< "$string")
echo "$nstring";
>"text text text -end.txt"
Required output = "text text text -end end 233-end.txt"
Method 4
nstring=$(sed -En "s/(^.*'"$start"').*('"$end"'.*$)/\1\2/p" <<< "$string")
echo "$nstring" ;
>"text text text -end.txt"
Required output = "text text text -end end 233-end.txt"

Answer 1

Using Bash's Regex match:
#!/usr/bin/env bash
string='This text and more text jsdlj-end.text'
[[ $string =~ (.*text\ ).*(-end.*) ]] || :
printf %s\\n "${BASH_REMATCH[1]}${BASH_REMATCH[2]}"

Answer 2

UPDATE: question has been updated with more details for dealing with a string that contains multiple start and end markers.
The new input string:
This text d #! more text jsdlj end and mo{re ;re end text.text
Test case:
start marker = 'text'
end marker = 'end'
objective = remove all text between last start marker and before the first end marker (actually replace all said text with a single space)
Input with all markers in bold:
This text d #! more text jsdlj end and mo{re ;re end text.text
Input with the two markers of interest in bold:
This text d #! more text jsdlj end and mo{re ;re end text.text
Desired result:
This text d #! more text end and mo{re ;re end text.text
While we can use sed to remove the desired text (replace <space>jsdlj<space> with <space>), we have to deal with the fact that sed does greedy matching (fine for finding the 'last' start marker) but does not do non-greedy matching (needed to find the 'first' end marker). We can get around this limitation by switching out our end marker with a single-character replacement, simulate a non-greedy match, then switch back to the original end marker.
m1='text' # start marker
m2='end' # end marker
string="This text d #! more text jsdlj end and mo{re ;re end text.text"
sed -E "s/${m2}/#/g;s/(^.*${m1})[^#]*(#.*$)/\1 \2/;s/#/${m2}/g" <<< "${string}"
Where:
-E - enable Extended regex support (includes capture groups)
s/${m2}/#/g - replace our end marker with the single character # (OP needs to determine what character cannot show up in expected input strings)
(^.*${m1}) - 1st capture group; greedy match from start of string up to last start marker before ...
[^#]* - match everything that's not the # character
(#.*$) - 2nd capture group; everything from # character until end of string
\1 \2 - replace entire string with 1st capture group + <space> + 2nd capture group
s/#/${m2}/g - replace single character # with our end marker
This generates:
This text d #! more text end and mo{re ;re end text.text
Personally, I'd probably opt for a more straight forward parameter expansion approach (similiar to Jetchisel's answer) but that could be a bit problematic for inline xargs processing ... ???
Original answer
One sed idea using capture groups:
$ string="This text and more text jsdlj-end.text"
$ sed -En 's/(^.*text ).*(-end.*$)/\1\2/p' <<< "${string}"
This text and more text -end.text
Where:
-En - enable Extended regex support (and capture groups) and (-n) disable default printing of pattern space
(^.*text ) - first capture group = start of line up to last text
.* - everything between the 2 capture groups
(-end.*$) - second capture group = from -end to end of string
\1\2/p - print the contents of the 2 capture groups.
Though this runs into issues if there are multiple -end strings on the 'end' of the string, eg:
$ string="This text and more text jsdlj-end -end.text"
$ sed -En 's/(^.*text ).*(-end.*$)/\1\2/p' <<< "${string}"
This text and more text -end.text
Whether this is correct or not depends on the desired output (and assuming this type of 'double' ending string is possible).

Answer 3

With Parameter Expansion.
string="This text and more text jsdlj-end.text"
temp=${string%-*}
end=${string#"$temp"}
start=${temp% *}
echo "$start$end"

Answer 4

This is a bit tricky using only a posix extended regex (ERE), but easy with a perl compatible regex (PCRE). Therefore, we switch from sed to perl:
To get the last text (that still has a end afterwards), put a .* in front. The closest end to that text can then be matched using a non-greedy .*?.
Here we also put \b around text and end to avoid matching parts of other words (for example, the word send should not be matched even though it contains end too).
perl -pe 's/(.*\btext\b).*?(\bend\b)/\1 \2/' <<< "$string"