I'm writing a Ruby command line application in which the user has to enter a "format string" (much like Date.strptime/strftime's strings).
I tried taking them in as arguments to the command, eg
> torque "%A\n%d\n%i, %u"
but it seems that bash actually removes all backslashes from input before it can be processed (plus a lot of trouble with spaces). I also tried the highline gem, which has some more advanced input options, but that automatically escapes backslashes "\" -> "\\" and provides no alternate options.
My current solution is to do a find-and-replace: "\\n" -> "\n". This would take care of the problem, but it also seems hacky and awful.
I could have users write the string in a text file (complicated for the user) or treat some other character, like "&&", as a newline (still complicated for the user).
What is the best way for users to input escaped characters on the command line?
(UPDATE: I checked the documentation for strptime/strftime, and the format strings for those functions replace newline characters with "%n", tabs with "%t", etc. So for now I'm doing that, but any other suggestions are welcome)
What you're looking for is using single quotes instead of double quotes.
Thus:
> torque '%A\n%d\n%i, %u'
Any string quoted in single quotes 'eg.' is does not go through any expansions and is used as is.
More details can be found in the Quoting section of man bash.
From man bash:
Enclosing characters in single quotes preserves the literal value of each character within the quotes. A single quote may not occur between single quotes, even when preceded by a backslash.
p eval("\"#{gets.chomp}\"")
Example use:
\n\b # Input by the user from the keyboard
"\n\b" # Value from the script
%A\n%d\n%i, %u # Input by the user from the keyboard
"%A\n%d\n%i, %u" # Value from the script
Related
I am trying to execute script with commands:
sed -i "USER/c\export USER=${signumid}" .bashrc
sed -i "DEVENVHOME=$/c\export DEVENVHOME=${DEVENVHOME:-/home/${signumid}/CPM_WORKAREA/devenv.x}" .bashrc
I want to replace the line with string "USER" in .bashrc with export USER=${signumid} where $signumid variable is being provided through Cygwin prompt. Similarly I want to replace line with string DEVENVHOME=$ with export DEVENVHOME=${DEVENVHOME:-/home/${signumid}/CPM_WORKAREA/devenv.x} in .bashrc file, where $signumid variable is provided through Cygwin prompt.
But I am getting following errors on Cygwin termminal.:
sed: -e expression #1, char 1: unknown command: `U'
sed: -e expression #1, char 3: extra characters after command
The general syntax of a sed script is a sequence of address command arguments statements (separated by newline or semicolon). The most common command is the s substitution command, with an empty address, so we can perhaps assume that that is what you want to use here. You seem to be attempting to interpolate a shell variable $signumid which adds a bit of a complication to this exposition.
If your strings were simply static text, it would make sense to use single quotes; then, the shell does not change the text within the quotes at all. The general syntax of the s command is s/regex/replacement/ where the slash as the argument separator is just a placeholder, as we shall soon see.
sed -i 's/.*USER.*/export USER=you/
s% DEVENVHOME=\$%export DEVENVHOME=${DEVENVHOME:-/home/you/CPM_WORKAREA/devenv.x}%' .bashrc
This will find any line with USER and substitute the entire line with export USER=you; then find any line which contains DEVENVHOME=$ (with a space before, and a literal dollar character) and replace the matched expression with the long string. Because the substitution string uses slashes internally, we use a different regex separator % -- alternatively, we could backslash-escape the slashes which are not separators, but as we shall see, that quickly becomes untenable when we add the following twist. Because the dollar sign has significance as the "end of line" metacharacter in regular expressions, we backslash-escape it.
I have ignored the c\ in your attempt on the assumption that it is simply a misunderstanding of sed syntax. If it is significant, what do you hope to accomplish with it? c\export is not a valid Bash command, so you probably mean something else, but I cannot guess what.
Now, to interpolate the value of the shell variable signumid into the replacement, we cannot use single quotes, because those inhibit interpolation. You have correctly attempted to use double quotes instead (in your edited question), but that means we have to make some additional changes. Inside double quotes, backslashes are processed by the shell, so we need to double all backslashes, or find alternative constructs. Fortunately for us, the only backslash is in \$ which can equivalently be expressed as [$], so let's switch to that notation instead. Also, where a literal dollar sign is wanted in the replacement string, we backslash-escape it in order to prevent the shell from processing it.
sed -i "s/.*USER.*/export USER=$signumid/
s% DEVENVHOME=[$]%export DEVENVHOME=\${DEVENVHOME:-/home/$signumid/CPM_WORKAREA/devenv.x}%" .bashrc
Equivalenty, you could use single quotes around the parts of the script which are meant to be untouched by the shell, and then put an adjacent double-quoted string around the parts which need interpolation, like
'un$touched*by$(the!shell)'"$signumid"'more$[complex]!stuff'
This final script still rests on a number of lucky or perhaps rather unlucky guesses about what you actually want. On the first line, I have changed just USER to a regular expression which matches the entire line -- maybe that's not what you want? On the other hand, the second line makes the opposite assumption, just so you can see the variations -- it only replaces the actual text we matched. Probably one or the other needs to be changed.
Finally, notice how the two separate sed commands have been conflated into a single script. Many newcomers do not realize that sed is a scripting language which accepts an arbitrary number of commands in a script, and simply treat it as a "replace" program with a funny syntax.
Another common source of confusion is the evaluation order. The shell processes the double-quoted string even before sed starts to execute, so if you have mistakes in the quoting, you can easily produce syntax errors in the sed script which lead to rather uninformative error messages (because what sed tells you in the error message is based on what the script looks like after the shell's substutions). For example, if signumid contains slashes, it will produce syntax errors, because sed will see those as terminating separators for the s/// command. An easy workaround is to switch to a separator which does not occur in the value of signumid.
I have program whose textual output I want to directly execute in a shell. How shall I format the output of this program such that the paths with spaces are accepted by the shell ?
$(echo ls /folderA/folder\ with\ spaces/)
Some more info: the program that generates the output is coded in Haskell (source). It's a simple program that keeps a list of my favorite commands. It prints the commands with 'cmdl -l'. I can then choose one command to execute with 'cmdl -g12' for command number 12. Thanks for pointing out that instead of $( ) use 'cmdl -g12 | bash', I wasn't aware of that...
How shall I format the output of this program such that the paths with
spaces are accepted by the shell ?
The shell cannot distinguish between spaces that are part of a path and spaces that are separator between arguments, unless those are properly quoted. Moreover, you actually need proper quoting using single quotes ('...') in order to "shield" all those characters combinations that might otherwise have special meaning for the shell (\, &, |, ||, ...).
Depending the language used for your external tool, their might be a library available for that purpose. As as example, Python has pipes.quote (shlex.quote on Python 3) and Perl has String::ShellQuote::shell_quote.
I'm not quite sure I understand, but don't you just want to pipe through the shell?
For a program called foo
$ foo | sh
To format output from your program so Bash won't try to space-separate them into arguments either update, probably easiest just to double-quote them with any normal quoting method around each argument, e.g.
mkdir "/tmp/Joey \"The Lips\" Fagan"
As you saw, you can backslash the spaces alternatively, but I find that less readable ususally.
EDIT:
If you may have special shell characters (&|``()[]$ etc), you'll have to do it the hard/proper way (with a specific escaper for your language and target - as others have mentioned.
It's not just spaces you need to worry about, but other characters such as [ and ] (glob a.k.a pathname-expansion characters) and metacharacters such as ;, &, (, ...
You can use the following approach:
Enclose the string in single quotes.
Replace existing single quotes in the string with '\'' (which effectively breaks the string into multiple parts with spliced in \-escaped single quotes; the shell then reassembles the parts into a single string).
Example:
I'm good (& well[1];) would encode to 'I'\''m good (& well[1]);'
Note how single-quoting allows literal use of the glob characters and metacharacters.
Since single quotes themselves can never be used within single-quoted strings (there's not even an escape), the splicing-in approach described above is needed.
As described by #mklement0, a safe algorithm is to wrap every argument in a pair of single quotes, and quote single quotes inside arguments as '\''. Here is a shell function that does it:
function quote {
typeset cmd="" escaped
for arg; do
escaped=${arg//\'/\'\\\'\'}
cmd="$cmd '$escaped'"
done
printf %s "$cmd"
}
$ quote foo "bar baz" "don't do it"
'foo' 'bar baz' 'don'\''t do it'
In C#, there is a verbatim string so that,
string c = "hello \t world"; // hello world
string d = #"hello \t world"; // hello \t world
I am new to shell script, is there a similar method in shell?
Because I have many folders with the name like "Apparel & Accessories > Clothing > Activewear", I want to know if there is a easy way to process the escape characters without write so many .
test.sh
director="Apparel & Accessories > Clothing > Activewear"
# any action to escape spaces, &, > ???
hadoop fs -ls $director
For definining the specific string in your example, Apparel & Accessories > Clothing > Activewear, either double quotes or single quotes will work; referring to it later is a different story, however:
In the shell (any POSIX-compatible shell), how you refer to a variable is just as important as how you define it.
To safely refer to a previously defined variable without side-effects, enclose it in double quotes, e.g., "$directory".
To define [a variable as] a literal (verbatim) string:
(By contrast, to define a variable with embedded variable references or embedded command substitutions or embedded arithmetic expressions, use double quotes (").)
If your string contains NO single quotes:
Use a single-quoted string, e.g.:
directory='Apparel & Accessories > Clothing > Activewear'
A single-quoted string is not subject to any interpretation by the shell, so it's generally the safest option for defining a literal. Note that the string may span multiple lines; e.g.:
multiline='line 1
line 2'
If your string DOES contain single quotes (e.g., I'm here.) and you want a solution that works in all POSIX-compatible shells:
Break the string into multiple (single-quoted) parts and splice in single-quote characters:
Note: Sadly, single-quoted strings cannot contain single quotes, not even with escaping.
directory='I'\''m here.'
The string is broken into into single-quoted I, followed by literal ' (escaped as an unquoted string as \'), followed by single-quoted m here.. By virtue of having NO spaces between the parts, the result is a single string containing a literal single quote after I.
Alternative: if you don't mind using a multiline statement, you can use a quoted here document, as described at the bottom.
If your string DOES contain single quotes (e.g., I'm here.) and you want a solution that works in bash, ksh, and zsh:
Use ANSI-C quoting:
directory=$'I\'m here.'
Note: As you can see, ANSI-C quoting allows for escaping single quotes as \', but note the additional implications: other \<char> sequences are subject to interpretation, too; e.g., \n is interpreted as a newline character - see http://www.gnu.org/software/bash/manual/bash.html#ANSI_002dC-Quoting
Tip of the hat to #chepner, who points out that the POSIX-compatible way of directly including a single quote in a string to be used verbatim is to use read -r with a here document using a quoted opening delimiter (the -r option ensures that \ characters in the string are treated as literals).
# *Any* form of quoting, not just single quotes, on the opening EOF will work.
# Note that $HOME will by design NOT be expanded.
# (If you didn't quote the opening EOF, it would.)
read -r directory <<'EOF'
I'm here at $HOME
EOF
Note that here documents create stdin input (which read reads in this case). Therefore, you cannot use this technique to directly pass the resulting string as an argument.
use strong quotes i.e. 'string', allowing escape char or special char for string.
e.g. declare director='Apparel & Accessories > Clothing > Activewear'
also using declare is a good practice while declaring variable.
I want to run a command from a bash script which has single quotes and some other commands inside the single quotes and a variable.
e.g. repo forall -c '....$variable'
In this format, $ is escaped and the variable is not expanded.
I tried the following variations but they were rejected:
repo forall -c '...."$variable" '
repo forall -c " '....$variable' "
" repo forall -c '....$variable' "
repo forall -c "'" ....$variable "'"
If I substitute the value in place of the variable the command is executed just fine.
Please tell me where am I going wrong.
Inside single quotes everything is preserved literally, without exception.
That means you have to close the quotes, insert something, and then re-enter again.
'before'"$variable"'after'
'before'"'"'after'
'before'\''after'
Word concatenation is simply done by juxtaposition. As you can verify, each of the above lines is a single word to the shell. Quotes (single or double quotes, depending on the situation) don't isolate words. They are only used to disable interpretation of various special characters, like whitespace, $, ;... For a good tutorial on quoting see Mark Reed's answer. Also relevant: Which characters need to be escaped in bash?
Do not concatenate strings interpreted by a shell
You should absolutely avoid building shell commands by concatenating variables. This is a bad idea similar to concatenation of SQL fragments (SQL injection!).
Usually it is possible to have placeholders in the command, and to supply the command together with variables so that the callee can receive them from the invocation arguments list.
For example, the following is very unsafe. DON'T DO THIS
script="echo \"Argument 1 is: $myvar\""
/bin/sh -c "$script"
If the contents of $myvar is untrusted, here is an exploit:
myvar='foo"; echo "you were hacked'
Instead of the above invocation, use positional arguments. The following invocation is better -- it's not exploitable:
script='echo "arg 1 is: $1"'
/bin/sh -c "$script" -- "$myvar"
Note the use of single ticks in the assignment to script, which means that it's taken literally, without variable expansion or any other form of interpretation.
The repo command can't care what kind of quotes it gets. If you need parameter expansion, use double quotes. If that means you wind up having to backslash a lot of stuff, use single quotes for most of it, and then break out of them and go into doubles for the part where you need the expansion to happen.
repo forall -c 'literal stuff goes here; '"stuff with $parameters here"' more literal stuff'
Explanation follows, if you're interested.
When you run a command from the shell, what that command receives as arguments is an array of null-terminated strings. Those strings may contain absolutely any non-null character.
But when the shell is building that array of strings from a command line, it interprets some characters specially; this is designed to make commands easier (indeed, possible) to type. For instance, spaces normally indicate the boundary between strings in the array; for that reason, the individual arguments are sometimes called "words". But an argument may nonetheless have spaces in it; you just need some way to tell the shell that's what you want.
You can use a backslash in front of any character (including space, or another backslash) to tell the shell to treat that character literally. But while you can do something like this:
reply=\”That\'ll\ be\ \$4.96,\ please,\"\ said\ the\ cashier
...it can get tiresome. So the shell offers an alternative: quotation marks. These come in two main varieties.
Double-quotation marks are called "grouping quotes". They prevent wildcards and aliases from being expanded, but mostly they're for including spaces in a word. Other things like parameter and command expansion (the sorts of thing signaled by a $) still happen. And of course if you want a literal double-quote inside double-quotes, you have to backslash it:
reply="\"That'll be \$4.96, please,\" said the cashier"
Single-quotation marks are more draconian. Everything between them is taken completely literally, including backslashes. There is absolutely no way to get a literal single quote inside single quotes.
Fortunately, quotation marks in the shell are not word delimiters; by themselves, they don't terminate a word. You can go in and out of quotes, including between different types of quotes, within the same word to get the desired result:
reply='"That'\''ll be $4.96, please," said the cashier'
So that's easier - a lot fewer backslashes, although the close-single-quote, backslashed-literal-single-quote, open-single-quote sequence takes some getting used to.
Modern shells have added another quoting style not specified by the POSIX standard, in which the leading single quotation mark is prefixed with a dollar sign. Strings so quoted follow similar conventions to string literals in the ANSI standard version of the C programming language, and are therefore sometimes called "ANSI strings" and the $'...' pair "ANSI quotes". Within such strings, the above advice about backslashes being taken literally no longer applies. Instead, they become special again - not only can you include a literal single quotation mark or backslash by prepending a backslash to it, but the shell also expands the ANSI C character escapes (like \n for a newline, \t for tab, and \xHH for the character with hexadecimal code HH). Otherwise, however, they behave as single-quoted strings: no parameter or command substitution takes place:
reply=$'"That\'ll be $4.96, please," said the cashier'
The important thing to note is that the single string that gets stored in the reply variable is exactly the same in all of these examples. Similarly, after the shell is done parsing a command line, there is no way for the command being run to tell exactly how each argument string was actually typed – or even if it was typed, rather than being created programmatically somehow.
Below is what worked for me -
QUOTE="'"
hive -e "alter table TBL_NAME set location $QUOTE$TBL_HDFS_DIR_PATH$QUOTE"
EDIT: (As per the comments in question:)
I've been looking into this since then. I was lucky enough that I had repo laying around. Still it's not clear to me whether you need to enclose your commands between single quotes by force. I looked into the repo syntax and I don't think you need to. You could used double quotes around your command, and then use whatever single and double quotes you need inside provided you escape double ones.
just use printf
instead of
repo forall -c '....$variable'
use printf to replace the variable token with the expanded variable.
For example:
template='.... %s'
repo forall -c $(printf "${template}" "${variable}")
Variables can contain single quotes.
myvar=\'....$variable\'
repo forall -c $myvar
I was wondering why I could never get my awk statement to print from an ssh session so I found this forum. Nothing here helped me directly but if anyone is having an issue similar to below, then give me an up vote. It seems any sort of single or double quotes were just not helping, but then I didn't try everything.
check_var="df -h / | awk 'FNR==2{print $3}'"
getckvar=$(ssh user#host "$check_var")
echo $getckvar
What do you get? A load of nothing.
Fix: escape \$3 in your print function.
Does this work for you?
eval repo forall -c '....$variable'
I have a bash script,
echo 'abcd'
in shell, I want to show ab'c'd and I have tried following approach but without success
echo 'ab\'c\'d'
I am asking is it possible to show single quote in single quoted text?
From the bash manual section on Single Quotes:
A single quote may not occur between single quotes, even when preceded by a backslash.
You'll need to use double quotes instead. It's not pretty, but the following gives the output you are looking for:
echo 'ab'"'"'c'"'"'d'
A bash-specific feature, not part of POSIX, is a $'...'-quoted string:
echo $'ab\'c\'d'
Such a string behaves identically to a single-quoted string, but does allow for a selection of \-escaped characters (such as \n, \t, and yes, \').