How can I match a blank line in expect? expect -r "" {send "y\n"}" causes y to be entered immediately.
|
| Type 'y' to accept:
|
<---- match this blank line!
Etan Reisner's suggestion of matching consecutive newlines works.
expect -r "\r\n\r\n" {send "y\n"}
Related
So I was looking into tr and I was playing around with the following command:
echo "test 123 new LINE" | tr -c 'A-Za-z' '[\n*]' vs echo "test 123 new LINE" | tr -c 'A-Za-z' '[\n]'. These are the different outputs:
> echo "test 123 new LINE" | tr -c 'A-Za-z' '[\n*]'
test
new
LINE
> echo "test 123 new LINE" | tr -c 'A-Za-z' '[\n]'
test]]]]]new]LINE]
Without the addition of the wildcard, it appears to be replacing each new line with a right bracket character. Taking a look at the man page (https://linuxcommand.org/lc3_man_pages/tr1.html), it says that for the argument [CHAR*], it "in SET2, copies of CHAR until length of SET1". So clearly it still replaces characters in SET1 based on SET2 but where is it getting the right bracket from?
Nothing to do with my job so no need to tell me about sed or awk, just came across this and was curious.
In your second command, the replacement set is not in one of the special formats
[CHAR*]
[CHAR*REPEAT]
[:<keyword>:]
[=CHAR=]
So it doesn't get any special treatment and the square brackets are treated literally. So the first two non-alphabetic characters are replaced with [ and \n, respectively, and all other characters are replaced with ] (because the replacement set is extended by repeating the last character).
I am pulling my hair on this one for days.. very annoying..
I know it is something to do with utf-8 and string not outputting the proper format, but cant figure out what..
This is the code:
#!/bin/bash
#test
REGURL=http://bugs.ws
CHECKURL=$(curl -m 3 -sk --head "$REGURL" | grep -i "location" | awk '{print $2}')
if [[ "${CHECKURL: -1}" != *'/'* ]] # if redirected url does not contain / at end, we need to add it
then
CHECKURL+='/'
echo "$CHECKURL"
fi
This is doing some character substitution rather than simply adding the '/' after the URL..
it works when you do it without a piped curl, grep, so i know it is something dealing with grep or curl..
Basically the outcome MUST have a forward slash at the end, ex: http://bugs.ws will end up having a location redirect of https://alphaterminte.com but I need to add a '/' to the end of "alphatermite.com", I've tried it all, I just cant get the forward slash to go after the variable result.. it keeps substituting it for the first character of the grepped result.. UGHHHH (yes this test code needs to be in bash)
The output from curl is uses carriage return + linefeed line terminators; unix tools only expect linefeed, and treat the carriage return as part of the line's content. Net result: CHECKURL has a not-normally-visible carriage return character at the end, which confuses everything.
Specifically, CHECKURL winds up containing "https://alphatermite.com<carriage return>/", which prints something like:
https://alphatermite.com
/
...except with only a carriage return (no linefeed) between, the "/" prints over top of the "h" in "https".
Solution: you could add | tr -d '\r' to the pipeline creating CHECKURL, but I'd just have awk do everything in one step:
CHECKURL=$(curl -m 3 -sk --head "$REGURL" | awk '/^[Ll]ocation:/ {sub("\r", "", $2); print $2}')
Explanation: the /^[Ll]ocation:/" part makes it only print the Location (or "location") header, and sub("\r", "", $2) deletes the carriage return from $2 before it's printed.
BTW, I'd use this to test for "/" at the end of the string:
if [[ "${CHECKURL}" != *'/' ]]
You can either extract the last character and see if it's "/", or use a wildcard pattern to check if it ends with "/"; no need to do both.
BTW2, I also recommend using lower- or mixed-case variable names, to avoid accidentally using one of the many all-caps names that have special meanings (and hence unexpected consequences).
Try bash regex
url=http://bugs.ws
re='.*/$'
[[ $url =~ $re ]] || url+='/'
EXPECTED:
Input:
###
# comment
###
var1=/opt
#var2=/app
Output:
comment
var1=/opt
var2=/app
I've tried running some codes but it doesn't print in the actual order:
grep "var1" cf.cfg
grep "#" cf.cfg | cut -d "#" -f2
Output using these codes:
var1=/opt
comment
var2=/app
Try below command.
sed '/^[ #]*$/d;s/^[ #]*//' cf.cfg
It removes..
Line having only # or space of any combination.
From start of line, # or space of any combination, including line as # # comment.
You can use
sed -e 's/#\s*//;/^\s*$/d' yourfile
This will remove the # character and subsequent whitespace (s/#\s*//) and print lines containing non-whitespace characters (/^\s*$/d).
In the case of a line of the form
var=1 # this is a comment
the above line will print
If you want to remove the characters before the #, you can use
sed -e 's/^[^#]*#\+\s*//;/^\s*$/d' yourfile
var=1 this is a comment
This will remove all text up to the first #.
How can I print a command output like one from rm -rv * in a single line ? I think it would need \r but I can't figure out how.
I would need to have something like this :
From:
removed /path/file1
removed /path/file2
removed /path/file3
To : Line 1 : removed /path/file1
Then : Line 1 : removed /path/file2
Then : Line 1 : removed /path/file3
EDIT : I may have been misunderstood, I want to have the whole process beeing printing in a single same line, changing as the command outputs an another line (like removed /path/file123)
EDIT2 : The output is sometimes too long to be display in on line (very long path). I would need something that considers that problem too :
/very/very/very/long/path/to/a/very/very/very/far/file/with-a-very-very-very-long-name1
/very/very/very/long/path/to/a/very/very/very/far/file/with-a-very-very-very-long-name2
/very/very/very/long/path/to/a/very/very/very/far/file/with-a-very-very-very-long-name3
Here's a helper function:
shopt -s checkwinsize # ensure that COLUMNS is available w/ window size
oneline() {
local ws
while IFS= read -r line; do
if (( ${#line} >= COLUMNS )); then
# Moving cursor back to the front of the line so user input doesn't force wrapping
printf '\r%s\r' "${line:0:$COLUMNS}"
else
ws=$(( COLUMNS - ${#line} ))
# by writing each line twice, we move the cursor back to position
# thus: LF, content, whitespace, LF, content
printf '\r%s%*s\r%s' "$line" "$ws" " " "$line"
fi
done
echo
}
Used as follows:
rm -rv -- * 2>&1 | oneline
To test this a bit more safely, one might use:
for f in 'first line' 'second line' '3rd line'; do echo "$f"; sleep 1; done | oneline
...you'll see that that test displays first line for a second, then second line for a second, then 3rd line for a second.
If you want a "status line" result that is showing the last line output by the program where the line gets over-written by the next line when it comes out you can send the output for the command through a short shell while loop like this:
YourCommand | while read line ; do echo -n "$line"$' ...[lots of spaces]... \r' ; done
The [Lots of spaces] is needed in case a shorter line comes after a longer line. The short line needs to overwrite the text from the longer line or you will see residual characters from the long line.
The echo -n $' ... \r' sends a literal carriage return without a line-feed to the screen which moves the position back to the front of the line but doesn't move down a line.
If you want the text from your command to just be output in 1 long line, then
pipe the output of any command through this sed command and it should replace the carriage returns with spaces. This will put the output all on one line. You could change the space to another delimiter if desired.
your command | sed ':rep; {N;}; s/\n/ /; {t rep};'
:rep; is a non-command that marks where to go to in the {t rep} command.
{N;} will join the current line to the next line.
It doesn't remove the carriage return but just puts the 2 lines in the buffer to be used for following commands.
s/\n/ /; Says to replace the carriage return character with a space character. They space is between the second and third/ characters.
You may need to replace \r\n depending on if the file has line feeds. UNIX files don't unless they came from a pc and haven't been converted.
{t rep}; says that if the match was found in the s/// command then go to the :rep; marker.
This will keep joining lines, removing the \n, then jumping to :rep; until there are no more likes to join.
How to I concatenate stdin to a string, like this?
echo "input" | COMMAND "string"
and get
inputstring
A bit hacky, but this might be the shortest way to do what you asked in the question (use a pipe to accept stdout from echo "input" as stdin to another process / command:
echo "input" | awk '{print $1"string"}'
Output:
inputstring
What task are you exactly trying to accomplish? More context can get you more direction on a better solution.
Update - responding to comment:
#NoamRoss
The more idiomatic way of doing what you want is then:
echo 'http://dx.doi.org/'"$(pbpaste)"
The $(...) syntax is called command substitution. In short, it executes the commands enclosed in a new subshell, and substitutes the its stdout output to where the $(...) was invoked in the parent shell. So you would get, in effect:
echo 'http://dx.doi.org/'"rsif.2012.0125"
use cat - to read from stdin, and put it in $() to throw away the trailing newline
echo input | COMMAND "$(cat -)string"
However why don't you drop the pipe and grab the output of the left side in a command substitution:
COMMAND "$(echo input)string"
I'm often using pipes, so this tends to be an easy way to prefix and suffix stdin:
echo -n "my standard in" | cat <(echo -n "prefix... ") - <(echo " ...suffix")
prefix... my standard in ...suffix
There are some ways of accomplish this, i personally think the best is:
echo input | while read line; do echo $line string; done
Another can be by substituting "$" (end of line character) with "string" in a sed command:
echo input | sed "s/$/ string/g"
Why i prefer the former? Because it concatenates a string to stdin instantly, for example with the following command:
(echo input_one ;sleep 5; echo input_two ) | while read line; do echo $line string; done
you get immediatly the first output:
input_one string
and then after 5 seconds you get the other echo:
input_two string
On the other hand using "sed" first it performs all the content of the parenthesis and then it gives it to "sed", so the command
(echo input_one ;sleep 5; echo input_two ) | sed "s/$/ string/g"
will output both the lines
input_one string
input_two string
after 5 seconds.
This can be very useful in cases you are performing calls to functions which takes a long time to complete and want to be continuously updated about the output of the function.
You can do it with sed:
seq 5 | sed '$a\6'
seq 5 | sed '$ s/.*/\0 6/'
In your example:
echo input | sed 's/.*/\0string/'
I know this is a few years late, but you can accomplish this with the xargs -J option:
echo "input" | xargs -J "%" echo "%" "string"
And since it is xargs, you can do this on multiple lines of a file at once. If the file 'names' has three lines, like:
Adam
Bob
Charlie
You could do:
cat names | xargs -n 1 -J "%" echo "I like" "%" "because he is nice"
Also works:
seq -w 0 100 | xargs -I {} echo "string "{}
Will generate strings like:
string 000
string 001
string 002
string 003
string 004
...
The command you posted would take the string "input" use it as COMMAND's stdin stream, which would not produce the results you are looking for unless COMMAND first printed out the contents of its stdin and then printed out its command line arguments.
It seems like what you want to do is more close to command substitution.
http://www.gnu.org/software/bash/manual/html_node/Command-Substitution.html#Command-Substitution
With command substitution you can have a commandline like this:
echo input `COMMAND "string"`
This will first evaluate COMMAND with "string" as input, and then expand the results of that commands execution onto a line, replacing what's between the ‘`’ characters.
cat will be my choice: ls | cat - <(echo new line)
With perl
echo "input" | perl -ne 'print "prefix $_"'
Output:
prefix input
A solution using sd (basically a modern sed; much easier to use IMO):
# replace '$' (end of string marker) with 'Ipsum'
# the `e` flag disables multi-line matching (treats all lines as one)
$ echo "Lorem" | sd --flags e '$' 'Ipsum'
Lorem
Ipsum#no new line here
You might observe that Ipsum appears on a new line, and the output is missing a \n. The reason is echo's output ends in a \n, and you didn't tell sd to add a new \n. sd is technically correct because it's doing exactly what you are asking it to do and nothing else.
However this may not be what you want, so instead you can do this:
# replace '\n$' (new line, immediately followed by end of string) by 'Ipsum\n'
# don't forget to re-add the `\n` that you removed (if you want it)
$ echo "Lorem" | sd --flags e '\n$' 'Ipsum\n'
LoremIpsum
If you have a multi-line string, but you want to append to the end of each individual line:
$ ls
foo bar baz
$ ls | sd '\n' '/file\n'
bar/file
baz/file
foo/file
I want to prepend my sql script with "set" statement before running it.
So I echo the "set" instruction, then pipe it to cat. Command cat takes two parameters : STDIN marked as "-" and my sql file, cat joins both of them to one output. Next I pass the result to mysql command to run it as a script.
echo "set #ZERO_PRODUCTS_DISPLAY='$ZERO_PRODUCTS_DISPLAY';" | cat - sql/test_parameter.sql | mysql
p.s. mysql login and password stored in .my.cnf file