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'
The problem
I have multiple property lines in a single string separated by \n like this:
LINES2="Abc1.def=$SOME_VAR\nAbc2.def=SOMETHING_ELSE\n"$LINES
The LINES variable
might contain an undefined set of characters
may be empty. If it is empty, I want to avoid the trailing \n.
I am open for any command line utility (sed, tr, awk, ... you name it).
Tryings
I tried this to no avail
sed -z 's/\\n$//g' <<< $LINES2
I also had no luck with tr, since it does not accept regex.
Idea
There might be an approach to convert the \n to something else. But since $LINES can contain arbitrary characters, this might be dangerous.
Sources
I skim read through the following questions
How can I replace a newline (\n) using sed?
sed with literal string--not input file
Here's one solution:
LINES2="Abc1.def=$SOME_VAR"$'\n'"Abc2.def=SOMETHING_ELSE${LINES:+$'\n'$LINES}"
The syntax ${name:+value} means "insert value if the variable name exists and is not empty." So in this case, it inserts a newline followed by $LINES if $LINES is not empty, which seems to be precisely what you want.
I use $'\n' because "\n" is not a newline character. A more readable solution would be to define a shell variable whose value is a single newline.
It is not necessary to quote strings in shell assignment statements, since the right-hand side of an assignment does not undergo word-splitting nor glob expansion. Not quoting would make it easier to interpolate a $'\n'.
It is not usually advisable to use UPPER-CASE for shell variables because the shell and the OS use upper-case names for their own purposes. Your local variables should normally be lower case names.
So if I were not basing the answer on the command in the question, I would have written:
lines2=Abc1.def=$someVar$'\n'Abc2.def=SOMETHING_ELSE${lines:+$'\n'$lines}
I'm partially successfully using sed to replace variables in a text file. I'm stuck on an exception.
A script reads input from a list - say the $roll_symbol is C20.
sed replaces C20, GC20, and KC20 (because C20 matches part of the string).
I searched the web and tried the variations I found - no success.
I tried these variations without success:
escape the reserved character $
escape braces
escape both
use double quotes instead of single quotes.
*the best version so far (but only partially):
sed -i 's/'${roll_symbol}'/'${roll_symbol}\,${contract_month}'/g' $OUTPUT_DIRECTORY/$OUTPUT_FILE;
You need to tell sed what characters are legal before the start of your match to limit where it can match. To only match at start-of-word boundaries try \<.
sed -i "s/\<${roll_symbol}/${roll_symbol},${contract_month}/g" "$OUTPUT_DIRECTORY/$OUTPUT_FILE";
I have read a BASH script, and found the following line:
lines="$lines"$'\n'
After testing, I know the meaning of this line is adding a "\n" after the string "$lines".
But after checking the bash manual, I can't find "$" can be used as a concatenated symbol. Could anyone give explainations on this usage of "$"? Thanks very much in advance!
A slightly closer read of the Bash Manual under Quoting would reveal where this gem is hidden. Specifically:
Words of the form $'string' are treated specially. The word expands to
string, with backslash-escaped characters replaced as specified by the
ANSI C standard.
Used specifically in the context of \n it provdes a new line. You most often see this form of quoting used in regard to the Bash IFS (internal field separator) whose default is space tab newline written:
IFS=$' \t\n'
This question already has answers here:
Using different delimiters in sed commands and range addresses
(3 answers)
Closed 1 year ago.
I saw a bash command sed 's%^.*/%%'
Usually the common syntax for sed is sed 's/pattern/str/g', but in this one it used s%^.* for the s part in the 's/pattern/str/g'.
My questions:
What does s%^.* mean?
What's meaning of %% in the second part of sed 's%^.*/%%'?
The % is an alternative delimiter so that you don't need to escape the forward slash contained in the matching portion.
So if you were to write the same expression with / as a delimiter, it would look like:
sed 's/^.*\///'
which is also kind of difficult to read.
Either way, the expression will look for a forward slash in a line; if there is a forward slash, remove everything up to and including that slash.
the usually used delimiter is / and the usage is sed 's/pattern/str/'.
At times you find that the delimiter is present in the pattern. In such a case you have a choice to either escape the delimiter found in the pattern or to use a different delimiter. In your case a different delimiter % has been used.
The later way is recommended as it keeps your command short and clean.