replace text only in clause determinated by keywork - sed

I want to processes text in some files that sometime expression exist in one line and sometime in multiple line
for example in multiple line
myfunc(param1,
param2,
param3);
or in one line
myfunc(param1, param2, param3);
does exist way that sed processes text only between myfunc and ;? keywords
at first step this help to me to port all multiple line to a line. Then I can to do my manupulation one a line type
if this possible?

Sed is for simple substitutions on individual lines, that is all. For anything even slightly more interesting you should be using awk. Something like this is probably what you want:
$ cat file
myfunc(param1,
param2,
param3);
myfunc(param1, param2, param3);
$ cat tst.awk
/myfunc/ { buf=""; inBlock=1 }
inBlock {
buf = (buf==""?"":buf RS) $0
if (/;/) {
$0 = buf
sub(/param2/,"lets have a tea party")
inBlock = 0
}
}
!inBlock
$ awk -f tst.awk file
myfunc(param1,
lets have a tea party,
param3);
myfunc(param1, lets have a tea party, param3);
Just replace the sub(/param2/,"lets have a tea party") line with whatever it is you really want to do with that block of text between myfunc and ;.

You can use the following sed script:
extract.sed:
# Check for "my"
/\bmy\b/ {
# Replace everything in front of
# my (including it)
s/.*\bmy\b//
# Define a label "a"
:a
# If the line does not contain "processes"
/\bprocesses\b/!{
# Get the next line of input and append
# it to the pattern buffer
N
# Branch back to label "a"
ba
}
# Replace "processes" and everything after it
s/\bprocesses\b.*//
# Print the pattern buffer
p
}
Call it like this:
sed -nf extract.sed input.txt

Related

How to deal with sed in Tcl script with substitution

I am writing a Tcl script which inserts some text in a file behind the matched line. The following are the basic codes in the script.
set test_lists [list "test_1"\
"test_2"\
"test_3"\
"test_4"\
"test_5"
]
foreach test $test_lists {
set content "
'some_data/$test'
"
exec sed -i "/dog/a$content" /Users/l/Documents/Codes/TCL/file.txt
}
However, when I run this script, it always shows me this error:
dyn-078192:TCL l$ tclsh test.tcl
sed: -e expression #1, char 12: unknown command: `''
while executing
"exec sed -i "/dog/a$content" /Users/l/Documents/Codes/TCL/file.txt"
("foreach" body line 5)
invoked from within
"foreach test $test_lists {
set content "
'some_data/$test'
"
exec sed -i "/dog/a$content" /Users/l/Documents/Codes/TCL/file.txt
}"
(file "test.tcl" line 8)
Somehow it always tried to evaluate the first word in $contentas a command.
Any idea what should I do here to make this work?
Thanks.
You first should decide exactly what characters need to be processed by sed. (See https://unix.stackexchange.com/questions/445531/how-to-chain-sed-append-commands for why this can matter…) They might possibly be:
/dog/a\
'some_data/test_1'
which would turn a file like:
abc
dog
hij
into
abc
dog
'some_data/test_1'
hij
If that's what you want, you can then proceed to the second stage: getting those characters from Tcl into sed.
# NB: *no* newline here!
set content "'some_data/$test'"
# NB: there's a quoted backslashes and two quoted newlines here
exec sed -i "/dog/a\\\n$content\n" /Users/l/Documents/Codes/TCL/file.txt
One of the few places where you need to be careful with quoting in Tcl is when you have backslashes and newlines in close proximity.
Why not perform the text transformation directly in Tcl itself? This might reverse the order of inserted lines compared to the original code. You can fix that by lreverseing the list at a convenient time, and perhaps you will also want to do further massaging of the text to insert. That's all refinements...
set test_lists [list "'some_data/test_1'"\
"'some_data/test_2'"\
"'some_data/test_3'"\
"'some_data/test_4'"\
"'some_data/test_5'"
]
set filename /Users/l/Documents/Codes/TCL/file.txt
set REGEXP "dog"
# Read in the data; this is good even for pretty large files
set f [open $filename]
set lines [split [read $f] "\n"]
close $f
# Search for first matching line by regular expression
set idx [lsearch -regexp $lines $REGEXP]
if {$idx >= 0} {
# Found something, so do the insert in the list of lines
set lines [linsert $lines [expr {$idx + 1}] {*}$test_lists]
# Write back to the file as we've made changes
set f [open $filename "w"]
puts -nonewline $f [join $lines "\n"]
close $f
}
(an extended comment, not an answer)
Running this in the shell to clarify your desired output: is this what you want?
$ cat file.txt
foo
dog A
dog B
dog C
dog D
dog E
bar
$ for test in test_{1..5}; do content="some_data/$test"; sed -i "/dog/a$content" file.txt; done
$ cat file.txt
foo
dog A
some_data/test_5
some_data/test_4
some_data/test_3
some_data/test_2
some_data/test_1
dog B
some_data/test_5
some_data/test_4
some_data/test_3
some_data/test_2
some_data/test_1
dog C
some_data/test_5
some_data/test_4
some_data/test_3
some_data/test_2
some_data/test_1
dog D
some_data/test_5
some_data/test_4
some_data/test_3
some_data/test_2
some_data/test_1
dog E
some_data/test_5
some_data/test_4
some_data/test_3
some_data/test_2
some_data/test_1
bar

Append to line that is preceded AND followed by empty line

I need to append an asterisk to a line, but only if said line is preceded and followed by empty lines (FYI, said empty lines will NOT have any white space in them).
Suppose I have the following file:
foo
foo
foo
foo
foo
I want the output to look like this:
foo
foo
foo
foo*
foo
I tried modifying the following awk command (found here):
awk 'NR==1 {l=$0; next}
/^$/ {gsub(/test/,"xxx", l)}
{print l; l=$0}
END {print l}' file
to suit my uses, but got all tied up in knots.
Sed or Perl solutions are, of course, welcome also!
UPDATE:
It turned out that the question I asked was not quite correct. What I really needed was code that would append text to non-empty lines that do not start with whitespace AND are followed, two lines down, by non-empty lines that also do not start with whitespace.
For this revised problem, suppose I have the following file:
foo
third line foo
fifth line foo
this line starts with a space foo
this line starts with a space foo
ninth line foo
eleventh line foo
this line starts with a space foo
last line foo
I want the output to look like this:
foobar
third line foobar
fifth line foo
this line starts with a space foo
this line starts with a space foo
ninth line foobar
eleventh line foo
this line starts with a space foo
last line foo
For that, this sed one-liner does the trick:
sed '1N;N;/^[^[:space:]]/s/^\([^[:space:]].*\o\)\(\n\n[^[:space:]].*\)$/\1bar\2/;P;D' infile
Thanks to Benjamin W.'s clear and informative answer below, I was able to cobble this one-liner together!
A sed solution:
$ sed '1N;N;s/^\(\n.*\)\(\n\)$/\1*\2/;P;D' infile
foo
foo
foo
foo*
foo
N;P;D is the idiomatic way to look at two lines at the same time by appending the next one to the pattern space, then printing and deleting the first line.
1N;N;P;D extends that to always having three lines in the pattern space, which is what we want here.
The substitution matches if the first and last line are empty (^\n and \n$) and appends one * to the line between the empty lines.
Notice that this matches and appends a * also for the second line of three empty lines, which might not be what you want. To make sure this doesn't happen, the first capture group has to have at least one non-whitespace character:
sed '1N;N;s/^\(\n[^[:space:]].*\)\(\n\)$/\1*\2/;P;D' infile
Question from comment
Can we not append the * if the line two above begins with abc?
Example input file:
foo
foo
abc
foo
foo
foo
foo
There are three foo between empty lines, but the first one should not get the * appended because the line two above starts with abc. This can be done as follows:
$ sed '1{N;N};N;/^abc/!s/^\(.*\n\n[^[:space:]].*\)\(\n\)$/\1*\2/;P;D' infile
foo
foo
abc
foo
foo*
foo*
foo
This keeps four lines at a time in the pattern space and only makes the substitution if the pattern space does not start with abc:
1 { # On the first line
N # Append next line to pattern space
N # ... again, so there are three lines in pattern space
}
N # Append fourth line
/^abc/! # If the pattern space does not start with abc...
s/^\(.*\n\n[^[:space:]].*\)\(\n\)$/\1*\2/ # Append '*' to 3rd line in pattern space
P # Print first line of pattern space
D # Delete first line of pattern space, start next cycle
Two remarks:
BSD sed requires an extra semicolon: 1{N;N;} instead of 1{N;N}.
If the first and third line of the file are empty, the second line does not get an asterisk appended because we only start checking once there are four lines in the pattern space. This could be solved by adding an extra substitution into the 1{} block:
1{N;N;s/^\(\n[^[:space:]].*\)\(\n\)$/\1*\2/}
(remember the extra ; for BSD sed), but trying to cover all edge cases makes sed even less readable, especially in one-liners:
sed '1{N;N;s/^\(\n[^[:space:]].*\)\(\n\)$/\1*\2/};N;/^abc/!s/^\(.*\n\n[^[:space:]].*\)\(\n\)$/\1*\2/;P;D' infile
One way to think about these problems is as a state machine.
start: state = 0
0: /* looking for a blank line */
if (blank line) state = 1
1: /* leading blank line(s)
if (not blank line) {
nonblank = line
state = 2
}
2: /* saw non-blank line */
if (blank line) {
output noblank*
state = 0
} else {
state = 1
}
And we can translate this pretty directly to an awk program:
BEGIN {
state = 0; # start in state 0
}
state == 0 { # looking for a (leading) blank line
print;
if (length($0) == 0) { # found one
state = 1;
next;
}
}
state == 1 { # have a leading blank line
if (length($0) > 0) { # found a non-blank line
saved = $0; # save it
state = 2;
next;
} else {
print; # multiple leading blank lines (ok)
}
}
state == 2 { # saw the non-blank line
if (length($0) == 0) { # followed by a blank line
print saved "*"; # BINGO!
state = 1; # to the saw a blank-line state
} else { # nope, consecutive non-blank lines
print saved; # as-is
state = 0; # to the looking for a blank line state
}
print;
next;
}
END { # cleanup, might have something saved to show
if (state == 2) print saved;
}
This is not the shortest way, nor likely the fastest, but it's probably the most straightforward and easy to understand.
EDIT
Here is a comparison of Ed's way and mine (see the comments under his answer for context). I replicated the OP's input a million-fold and then timed the runnings:
# ls -l
total 22472
-rw-r--r--. 1 root root 111 Mar 13 18:16 ed.awk
-rw-r--r--. 1 root root 23000000 Mar 13 18:14 huge.in
-rw-r--r--. 1 root root 357 Mar 13 18:16 john.awk
# time awk -f john.awk < huge.in > /dev/null
2.934u 0.001s 0:02.95 99.3% 0+0k 112+0io 1pf+0w
# time awk -f ed.awk huge.in huge.in > /dev/null
14.217u 0.426s 0:14.65 99.8% 0+0k 272+0io 2pf+0w
His version took about 5 times as long, did twice as much I/O, and (not shown in this output) took 1400 times as much memory.
EDIT from Ed Morton:
For those of us unfamiliar with the output of whatever time command John used above, here's the 3rd-invocation results from the normal UNIX time program on cygwin/bash using GNU awk 4.1.3:
$ wc -l huge.in
1000000 huge.in
$ time awk -f john.awk huge.in > /dev/null
real 0m1.264s
user 0m1.232s
sys 0m0.030s
$ time awk -f ed.awk huge.in huge.in > /dev/null
real 0m1.638s
user 0m1.575s
sys 0m0.030s
so if you'd rather write 37 lines than 3 lines to save a third of a second on processing a million line file then John's answer is the right one for you.
EDIT#3
It's the standard "time" built-in from tcsh/csh. And even if you didn't recognize it, the output should be intuitively obvious. And yes, boys and girls, my solution can also be written as a short incomprehensible mess:
s == 0 { print; if (length($0) == 0) { s = 1; next; } }
s == 1 { if (length($0) > 0) { p = $0; s = 2; next; } else { print; } }
s == 2 { if (length($0) == 0) { print p "*"; s = 1; } else { print p; s = 0; } print; next; }
END { if (s == 2) print p; }
Here's a perl filter version, for the sake of illustration — hopefully it's clear to see how it works. It would be possible to write a version that has a lower input-output delay (2 lines instead of 3) but I don't think that's important.
my #lines;
while (<>) {
# Keep three lines in the buffer, print them as they fall out
push #lines, $_;
print shift #lines if #lines > 3;
# If a non-empty line occurs between two empty lines...
if (#lines == 3 && $lines[0] =~ /^$/ && $lines[2] =~ /^$/ && $lines[1] !~ /^$/) {
# place an asterisk at the end
$lines[1] =~ s/$/*/;
}
}
# Flush the buffer at EOF
print #lines;
A perl one-liner
perl -0777 -lne's/(?<=\n\n)(.*?)(\n\n)/$1\*$2/g; print' ol.txt
The -0777 "slurps" in the whole file, assigned to $_, on which the (global) substitution is run and which is then printed.
The lookbehind (?<=text) is needed for repeating patterns, [empty][line][empty][line][empty]. It is a "zero-width assertion" that only checks that the pattern is there without consuming it. That way the pattern stays available for next matches.
Such consecutive repeating patterns trip up the /(\n\n)(.*?)(\n\n)/$1$2\*$3/, posted initially, since the trailing \n\n are not considered for the start of the very next pattern, having been just matched.
Update: My solution also fails after two consecutive matches as described above and needs the same lookback: s/(?<=\n\n)(\w+)\n\n/\1\2*\n\n/mg;
The easiest way is to use multi-line match:
local $/; ## slurp mode
$file = <DATA>;
$file =~ s/\n\n(\w+)\n\n/\n\n\1*\n\n/mg;
printf $file;
__DATA__
foo
foo
foo
foo
foo
It's simplest and clearest to do this in 2 passes:
$ cat tst.awk
NR==FNR { nf[NR]=NF; nr=NR; next }
FNR>1 && FNR<nr && NF && !nf[FNR-1] && !nf[FNR+1] { $0 = $0 "*" }
{ print }
$ awk -f tst.awk file file
foo
foo
foo
foo*
foo
The above takes one pass to record the number of fields on each line (NF is zero for an empty line) and then the second pass just checks your requirements - the current line is not the first or last in the file, it is not empty and the lines before and after are empty.
alternative awk solution (single pass)
$ awk 'NR>2 && !pp && !NF {p=p"*"}
NR>1{print p}
{pp=length(p);p=$0}
END{print p}' foo
foo
foo
foo
foo*
foo
Explanation: defer printing to next line for decision making, so need to keep previous line in p and state of the second previous line in pp (length zero assumed to be empty). Do the bookkeeping assignments and at the end print the last line.

sed editing multiple lines

Sed editing is always a new challenge to me when it comes to multiple line editing. In this case I have the following pattern:
RECORD 4,4 ,5,48 ,7,310 ,10,214608 ,12,199.2 ,13,-19.2 ,15,-83 ,17,35 \
,18,0.8 ,21,35 ,22,31.7 ,23,150 ,24,0.8 ,25,150 ,26,0.8 ,28,25 ,29,6 \
,30,1200 ,31,1 ,32,0.2 ,33,15 ,36,0.4 ,37,1 ,39,1.1 ,41,4 ,80,2 \
,82,1000 ,84,1 ,85,1
which I want to convert into:
#RECORD 4,4 ,5,48 ,7,310 ,10,214608 ,12,199.2 ,13,-19.2 ,15,-83 ,17,35 \
# ,18,0.8 ,21,35 ,22,31.7 ,23,150 ,24,0.8 ,25,150 ,26,0.8 ,28,25 ,29,6\
# ,30,1200 ,31,1 ,32,0.2 ,33,15 ,36,0.4 ,37,1 ,39,1.1 ,41,4 ,80,2 \
# ,82,1000 ,84,1 ,85,1
Besides this I would like to preserve the entirety of these 4 lines (which may be more or less than 4 (unpredictable as the appear in the input) into one (long) line without the backslashes or line wraps.
Two tasks in one so to say.
sed is mandatory.
It's not terribly clear how you recognize the blocks you want to comment out, so I'll use blocks from a line that starts with RECORD and process as long as there are backslashes at the end (if your requirements differ, the patterns used will need to be amended accordingly).
For that, you could use
sed '/^RECORD/ { :a /\\$/ { N; ba }; s/[[:space:]]*\\\n[[:space:]]*/ /g; s/^/#/ }' filename
This works as follows:
/^RECORD/ { # if you find a line that starts with
# RECORD:
:a # jump label for looping
/\\$/ { # while there's a backslash at the end
# of the pattern space
N # fetch the next line
ba # loop.
}
# After you got the whole block:
s/[[:space:]]*\\\n[[:space:]]*/ /g # remove backslashes, newlines, spaces
# at the end, beginning of lines
s/^/#/ # and put a comment sign at the
# beginning.
}
Addendum: To keep the line structure intact, instead use
sed '/^RECORD/ { :a /\\$/ { N; ba }; s/\(^\|\n\)/&#/g }' filename
This works pretty much the same way, except the newline-removal is removed, and the comment signs are inserted after every line break (and once at the beginning).
Addendum 2: To just put RECORD blocks onto a single line:
sed '/^RECORD/ { :a /\\$/ { N; ba }; s/[[:space:]]*\\\n[[:space:]]*/ /g }' filename
This is just the first script with the s/^/#/ bit removed.
Addendum 3: To isolate RECORD blocks while putting them onto a single line at the same time,
sed -n '/^RECORD/ { :a /\\$/ { N; ba }; s/[[:space:]]*\\\n[[:space:]]*/ /g; p }' filename
The -n flag suppresses the normal default printing action, and the p command replaces it for those lines that we want printed.
To write those records out to a file while commenting them out in the normal output at the same time,
sed -e '/^RECORD/ { :a /\\$/ { N; ba }; h; s/[[:space:]]*\\\n[[:space:]]*/ /g; w saved_records.txt' -e 'x; s/\(^\|\n\)/&#/g }' foo.txt
There's actually new stuff in this. Shortly annotated:
#!/bin/sed -f
/^RECORD/ {
:a
/\\$/ {
N
ba
}
# after assembling the lines
h # copy them to the hold buffer
s/[[:space:]]*\\\n[[:space:]]*/ /g # put everything on a line
w saved_records.txt # write that to saved_records.txt
x # swap the original lines back
s/\(^\|\n\)/&#/g # and insert comment signs
}
When specifying this code directly on the command line, it is necessary to split it into several -e options because the w command is not terminated by ;.
This problem does not arise when putting the code into a file of its own (say foo.sed) and running sed -f foo.sed filename instead. Or, for the advanced, putting a #!/bin/sed -f shebang on top of the file, chmod +xing it and just calling ./foo.sed filename.
Lastly, to edit the input file in-place and print the records to stdout, this could be amended as follows:
sed -i -e '/^RECORD/ { :a /\\$/ { N; ba }; h; s/[[:space:]]*\\\n[[:space:]]*/ /g; w /dev/stdout' -e 'x; s/\(^\|\n\)/&#/g }' filename
The new things here are the -i flag for inplace editing of the file, and to have /dev/stdout as target for the w command.
sed '/^RECORD.*\\$/,/[^\\]$/ s/^/#/
s/^RECORD.*/#&/' YourFile
After several remark of #Wintermute and more information from OP
Assuming:
line with RECORD at start are a trigger to modify the next lines
structure is the same (no line with \ with a RECORD line following directly or empty lines)
Explain:
take block of line starting with RECORD and ending with \
add # in front of each line
take line (so after ana eventual modification from earlier block that leave only RECORD line without \ at the end or line without record) and add a # at the start if starting with RECORD

Use sed to replace word in 2-line pattern

I try to use sed to replace a word in a 2-line pattern with another word. When in one line the pattern 'MACRO "something"' is found then in the next line replace 'BLOCK' with 'CORE'. The "something" is to be put into a reference and printed out as well.
My input data:
MACRO ABCD
CLASS BLOCK ;
SYMMETRY X Y ;
Desired outcome:
MACRO ABCD
CLASS CORE ;
SYMMETRY X Y ;
My attempt in sed so far:
sed 's/MACRO \([A-Za-z0-9]*\)/,/ CLASS BLOCK ;/MACRO \1\n CLASS CORE ;/g' input.txt
The above did not work giving message:
sed: -e expression #1, char 30: unknown option to `s'
What am I missing?
I'm open to one-liner solutions in perl as well.
Thanks,
Gert
Using a perl one-liner in slurp mode:
perl -0777 -pe 's/MACRO \w+\n CLASS \KBLOCK ;/CORE ;/g' input.txt
Or using a streaming example:
perl -pe '
s/^\s*\bCLASS \KBLOCK ;/CORE ;/ if $prev;
$prev = $_ =~ /^MACRO \w+$/
' input.txt
Explanation:
Switches:
-0777: Slurp files whole
-p: Creates a while(<>){...; print} loop for each line in your input file.
-e: Tells perl to execute the code on command line.
When in one line the pattern 'MACRO "something"' is found then in the
next line replace 'BLOCK' with 'CORE'.
sed works on lines of input. If you want to perform substitution on the next line of a specified pattern, then you need to add that to the pattern space before being able to do so.
The following might work for you:
sed '/MACRO/{N;s/\(CLASS \)BLOCK/\1CORE/;}' filename
Quoting from the documentation:
`N'
Add a newline to the pattern space, then append the next line of
input to the pattern space. If there is no more input then sed
exits without processing any more commands.
If you want to make use of address range as in your attempt, then you need:
sed '/MACRO/,/CLASS BLOCK/{s/\(CLASS\) BLOCK/\1 CORE/}' filename
I'm not sure why do you need a backreference for substituting the macro name.
You could try this awk command also,
awk '{print}/MACRO/ {getline; sub (/BLOCK/,"CORE");{print}}' file
It prints all the lines as it is and do the replacing action on seeing a word MACRO on a line.
Since getline has so many pitfall I try not to use it, so:
awk '/MACRO/ {a++} a==1 {sub(/BLOCK/,"CORE")}1' file
MACRO ABCD
CLASS CORE ;
SYMMETRY X Y ;
This could do it
#!awk -f
BEGIN {
RS = ";"
}
/MACRO/ {
sub("BLOCK", "CORE")
}
{
printf s++ ? ";" $0 : $0
}
"line" ends with ;
sub BLOCK for CORE in "lines" with MACRO
print ; followed by "line" unless first line

find the line number where a specific word appears with “sed” on tcl shell

I need to search for a specific word in a file starting from specific line and return the line numbers only for the matched lines.
Let's say I want to search a file called myfile for the word my_word and then store the returned line numbers.
By using shell script the command :
sed -n '10,$ { /$my_word /= }' $myfile
works fine but how to write that command on tcl shell?
% exec sed -n '10,$ { /$my_word/= }' $file
extra characters after close-brace.
I want to add that the following command works fine on tcl shell but it starts from the beginning of the file
% exec sed -n "/$my_word/=" $file
447431
447445
448434
448696
448711
448759
450979
451006
451119
451209
451245
452936
454408
I have solved the problem as follows
set lineno 10
if { ! [catch {exec sed -n "/$new_token/=" $file} lineFound] && [string length $lineFound] > 0 } {
set lineNumbers [split $lineFound "\n"]
foreach num $lineNumbers {
if {[expr {$num >= $lineno}] } {
lappend col $num
}
}
}
Still can't find a single line that solve the problem
Any suggestions ??
I don't understand a thing: is the text you are looking for stored inside the variable called my_word or is the literal value my_word?
In your line
% exec sed -n '10,$ { /$my_word/= }' $file
I'd say it's the first case. So you have before it something like
% set my_word wordtosearch
% set file filetosearchin
Your mistake is to use the single quote character ' to enclose the sed expression. That character is an enclosing operator in sh, but has no meaning in Tcl.
You use it in sh to group many words in a single argument that is passed to sed, so you have to do the same, but using Tcl syntax:
% set my_word wordtosearch
% set file filetosearchin
% exec sed -n "10,$ { /$my_word/= }" $file
Here, you use the "..." to group.
You don't escape the $ in $my_word because you want $my_word to be substitued with the string wordtosearch.
I hope this helps.
After a few trial-and-error I came up with:
set output [exec sed -n "10,\$ \{ /$myword/= \}" $myfile]
# Do something with the output
puts $output
The key is to escape characters that are special to TCL, such as the dollar sign, curly braces.
Update
Per Donal Fellows, we do not need to escape the dollar sign:
set output [exec sed -n "10,$ \{ /$myword/= \}" $myfile]
I have tried the new revision and found it works. Thank you, Donal.
Update 2
I finally gained access to a Windows 7 machine, installed Cygwin (which includes sed and tclsh). I tried out the above script and it works just fine. I don't know what your problem is. Interestingly, the same script failed on my Mac OS X system with the following error:
sed: 1: "10,$ { /ipsum/= }": extra characters at the end of = command
while executing
"exec sed -n "10,$ \{ /$myword/= \}" $myfile"
invoked from within
"set output [exec sed -n "10,$ \{ /$myword/= \}" $myfile]"
(file "sed.tcl" line 6)
I guess there is a difference between Linux and BSD systems.
Update 3
I have tried the same script under Linux/Tcl 8.4 and it works. That might mean Tcl 8.4 has nothing to do with it. Here is something else that might help: Tcl comes with a package called fileutil, which is part of the tcllib. The fileutil package contains a useful tool for this case: fileutil::grep. Here is a sample on how to use it in your case:
package require fileutil
proc grep_demo {myword myfile} {
foreach line [fileutil::grep $myword $myfile] {
# Each line is in the format:
# filename:linenumber:text
set lineNumber [lindex [split $line :] 1]
if {$lineNumber >= 10} { puts $lineNumber}
}
}
puts [grep_demo $myword $myfile]
Here is how to do it with awk
awk 'NR>10 && $0~f {print NR}' f="$my_word" "$myfile"
This search for all line larger than line number 10 that contains word in variable $my_word in file name stored in variable myfile