I get a string from a command that lokos like the following:
part1=$(pip install numpy 2>&1)
part1 has stored some variable like: c:\programdata\anaconda3\lib\site-package
Now I want to append "numpy" to it. To do so I tried it with:
part1+=/numpy
and every other solution listed here:
How to concatenate strings in bash
However the output looks like this: 'c:\programdata\anaconda3\lib\site-packages'$'\r''/numpy'
What do I have to do to get rid of this effect?
I am using Windows10 and scripting .sh files.
You can use:
part1="${part1/$'\r'}/numpy"
Here "${part1/$'\r'} replaces \r by an empty string. $'\r' is special bash construct to enter escape sequences.
Related
My python script can take a series of bitwise operators as one of its arguments. They all work fine except for "=<<" which is roll left, and "=>>" which is roll right. I run my script like ./script.py -b +4,-4,=>>10,=<<1, where anything after -b can be any combination of similar operations. As soon as the terminal sees "<<" though, it just drops the cursor to a new line after the command and asks for more input instead of running the script. When it sees ">>", my script doesn't process the arguments correctly. I know it's because bash uses these characters for a specific purpose, but I'd like to get around it while still using "=>>" and "=<<" in my arguments for my script. Is there any way to do it without enclosing the argument in quotation marks?
Thank you for your help.
You should enclose the parameters that contain special symbols into single quotation marks (here, echo represents your script):
> echo '+4,-4,=>>10,=<<1'
+4,-4,=>>10,=<<1
Alternatively, save the parameters to a file (say, params.txt) and read them from the file onto the command line using the backticks:
> echo `cat params.txt`
+4,-4,=>>10,=<<1
Lastly, you can escape some offending symbols:
> echo +4,-4,=\>\>10,=\<\<1
+4,-4,=>>10,=<<1
I want to get some data from a HTTP server. What it sends me depends on what I put in a POST request.
What I put in the INPUT_TEXT field is a sequence of words. When I run the following command, I get good looking output.
$ curl http://localhost:59125/process -d INPUT_TEXT="here are some words"
I want a bash script to take some string as a command line argument, and pass it appropriately to curl. The first thing I tried was to put the following in a script:
sentence=$1
command="curl http://localhost:59125/process -d INPUT_TEXT=\"${sentence}\""
$command
I then run the script like so:
$ ./script "here are some words"
But then I get a curl Couldn't resolve host error for each of "are", "some", and "words". It would seem that "here" got correctly treated as the INPUT_TEXT, but the rest of the words were then considered to be hosts, and not part of the option.
So I tried:
command=("curl" "http://localhost:59125/process" "-d" "INPUT_TEXT='$sentence'")
${command[#]}
I got the same output as the first script. I finally got what I wanted with:
result=$(curl http://localhost:59125/process -d INPUT_TEXT="${sentence}")
echo $result
I'm still unsure as to what the distinction is. In the first two cases, when I echoed out the contents of command, I get exactly what I input from the interactive Bash prompt, which had worked fine. What caused the difference?
The following will work:
command=("curl" "http://localhost:59125/process"
"-d" "INPUT_TEXT=$sentence")
"${command[#]}"
That has two changes from yours:
I removed the incorrect quotes around $sentence since you don't want to send quotes to the server (as far as I can see).
I put double-quotes around the use of "${command[#]}". Without the double quotes, the array's elements are concatenated with spaces between them and then the result is word-split. With double quotes, the individual array elements are used as individual words.
The second point is well-explained in the bash FAQ and a bunch of SO answers dealing with quotes.
The important thing to understand is that quotes only quote when a command is parsed. A quote which is a character in a variable is just a character; it is not reinterpreted when the value of the variable expanded. Whitespace in the variable is used for word-splitting if the variable expansion is unquoted; the fact that the whitespace was quoted in the the command which defined the variable is completely irrelevant. In this sense, bash is just the same as any other programming language.
I have separated some tracks from mp3 mixes using mp3splt.
BASH: (mp3splt -c('**!!***use .cue file***!!**') [cuefile.cue] [nonstopmix.mp3] ~for anyone interested, is in the Ubu repos~)
And I ended up with these filenames: "Antares" - 01 - "Xibalba".mp3 which is not a format I prefer, now I've made it a little project to change them with a shell script but its more difficult than I anticipated.
I want to change the filename from:
"Antares" - 01 - "Xibalba".mp
to:
01-Antares_-_Xibalba.mp3
so far I've used :
for var in *.mp3; do mv $var {var/"/}; done
and I could repeat that until I'm through, delete the 0x number and add one but I'd like to do it more efficient.
Could anyone give me a pointer (!not a script!) ?
I'd still like to write it myself but there's so much options that I'm a bit lost.
so far I thought to use this program flow:
read all the filenames containing .mp3 and declare as variable $var
strip $var from quotes
select 0x number, append delimiter _ (0x_)
move 0x_ to the beginning of the string
select remaining ' - - ' and change to '-'
done
which bash programs to use? especially changing the 0x puzzles me cuz I need a loop which increments this number and test if it is present in the filename variable and then it has to be changed.
It is easy to do in python 2.x. You can use this logic in any language you want.
import string
a=raw_input('Enter the name of song')
a=a.replace('"', "")
a=a.replace('.mp', ' .mp3')
words = a.split()
print words[2]+'-'+words[0]+'_-_'+words[4]+words[5]
Logic:
I removed ", then make .mp to .mp3, then splitted the string, which created a list ( array ) and then printed the elements according to need.
Try doing this :
rename -n 's/"(\w+)"\s+-\s*(\d+)\s*-\s*"(\w+)"\.mp/$2-$1_-_$3.mp3/' *mp
from the shell prompt. It's very useful, you can put some perl tricks like I does in a substitution.
You can remove the -n (dry-run mode switch) when your tests become valids.
There are other tools with the same name which may or may not be able to do this, so be careful.
If you run the following command (linux)
$ file $(readlink -f $(type -p rename))
and you have a result like
.../rename: Perl script, ASCII text executable
then this seems to be the right tool =)
If not, to make it the default (usually already the case) on Debian and derivative like Ubuntu :
$ sudo update-alternatives --set rename /path/to/rename
Last but not least, this tool was originally written by Larry Wall, the Perl's dad.
I'm having a nightmare from what should be the most trivial of tasks.
My final goal is issue the following command from a bash script:
sqlite3 my_db.db '.read my_file.sql'
There are two catches here:
1. The single-quotes are obligatory, and can't be replaced by, say, double-quotes
2. my_file.sql is a variable known only at run-time.
So what I need is a way to have bash build a string that on one hand contains a variable value, while on the other hand that value should be surrounded by single quotes.
I would also much prefer a solution not relying on additional tools like AWK, Perl or the like. Maybe sed if it's really necessary.
Thanks.
Thanks Jonathan and Nelson.
I tried all three suggestions, but they all failed.
For simplicity I reduced the problem to the following:
I wrote the following script (tst.sh):
#!/bin/bash
file=/tmp/1
ls "'"$file"'"
ls \'$file\'
ls "'$file'"
Then I isuues the following commands:
$ touch /tmp/1
$ ls '/tmp/1'
/tmp/1
$ ./tst.sh
'/tmp/1': No such file or directory
'/tmp/1': No such file or directory
'/tmp/1': No such file or directory
It seems the quotes were indeed added, but the resulting command was not the same as when entered manually.
Any ideas ?
Single-quotes are not obligatory. All of the following commands run sqlite3 with exactly the same arguments:
sqlite3 my_db.db '.read my_file.sql'
sqlite3 my_db.db ".read my_file.sql"
sqlite3 my_db.db .read\ my_file.sql
sqlfile="my_file.sql"
sqlite3 my_db.db ".read $sqlfile"
In all cases, the quotes (/escape) are parsed and removed before the arguments are passed to sqlite3. This is what you want. You want sqlite3 to get two arguments: my_db.db and .read my_file.sql. You do not want sqlite3 to see the quotes around the command -- that would be the equivalent of:
$ sqlite3 my_db.db
SQLite version 3.7.7 2011-06-25 16:35:41
Enter ".help" for instructions
Enter SQL statements terminated with a ";"
sqlite> '.read my_file.sql'
...>
...which, as you can see, just confuses sqlite3.
BTW, this is the same as the problem in your ls examples: you're passing single-quotes as part of the argument to ls, so it's looking for a file with single-quotes in the name and not finding it. You want the shell to remove the quotes rather than pass them to the command as part of an argument.
This will do what you say you want to do (getting single quotes to the program), but it uses double quotes:
sqlite3 my_db.db "'".read" "my_file.sql"'"
Avoiding double quotes, you can write:
sqlite3 my_db.db \'.read\ my_file.sql\'
For both of these, the second argument will be seen by sqlite3 as a string containing:
'.read my_file.sql'
If the file name is in a variable (file=my_file.sql), then:
sqlite3 my_db.db "'".read" "$file"'"
sqlite3 my_db.db \'.read\ $file\'
These notations are vulnerable to confusion if the file name contains spaces.
However, I don't think that's likely to be what you really want. The proscription on double quotes is puzzling, and the requirement for single quotes is likewise puzzling.
You can do as follows:
VAR=my_file.sql
VAR2="'.read $VAR'"
sqlite3 my_db.db $VAR2
user1860085, if you check out documentation for sqlite3 command and you will know how shell treats quotes and white spaces, you will probably come to conclusion that you want double quotes for your case.
but if you really want single quotes, here is solution:
eval sqlite3 my_db.db \'.read $VARIABLE\'
which in the fly will change to:
sqlite3 my_db.db '.read my_file.sql'
But I don't see why you could want it...
OK, problem solved !!
All that was missing is adding a little 'eval' command before the line.
So, in the simple example script I gave, changing:
ls "'$file'" to:
eval ls "'$file'"
did the job.
Thanks to all replyers :-)
I am experimenting with wildcards in bash and tried to list all the files that start with "xyz" but does not end with ".TXT" but getting incorrect results.
Here is the command that I tried:
$ ls -l xyz*[!\.TXT]
It is not listing the files with names "xyz" and "xyzTXT" that I have in my directory. However, it lists "xyz1", "xyz123".
It seems like adding [!\.TXT] after "xyz*" made the shell look for something that start with "xyz" and has at least one character after it.
Any ideas why it is happening and how to correct this command? I know it can be achieved using other commands but I am especially interested in knowing why it is failing and if it can done just using wildcards.
These commands will do what you want
shopt -s extglob
ls -l xyz!(*.TXT)
shopt -u extglob
The reason why your command doesn't work is beacause xyz*[!\.TXT] which is equivalent to xyz*[!\.TX] means xyz followed by any sequence of character (*) and finally a character in set {!,\,.,T,X} so matches 'xyzwhateveryouwant!' 'xyzwhateveryouwant\' 'xyzwhateveryouwant.' 'xyzwhateveryouwantT' 'xyzwhateveryouwantX'
EDIT: where whateveryouwant does not contain any of !\.TX
I don't think this is doable with only wildcards.
Your command isn't working because it means:
Match everything that has xyz followed by whatever you want and it must not end with sequent character: \, .,T and X. The second T doesn't count as far as what you have inside [] is read as a family of character and not as a string as you thought.
You don't either need to 'escape' . as long as it has no special meaning inside a wildcard.
At least, this is my knowledge of wildcards.