disable histexpand and variable substitution - bash

I need to create many thousands of files and I have no control over the text strings that need to be used for the paths and names. Trouble is, these strings might contain ",',“,‘,! or $, all of which can cause problems.
I can replace all quotes with the normal apostrophe, but I want to keep any $ or ! characters and so far not been able to figure how to do this.
Set +H will turn off history expansion, but I cannot make it work within the same command. So:-
set +H; echo "Hi, this is $10 Tom!"
still generates the error due to the !. If I run them as 2 separate commands, histexpand stays off and it works, but running them both just separated by the ; does NOT change histexpand.
The problem is that the commands will be run in their own shell and I can't seem to run multiple lines (doShellScript in Apple's JXA) so need them on the same line just separated by the ;.
So how can I make the set +H stick for the immediately following command?
Also, is it possible to stop the shell from interpreting $ as preceding a variable and just deal with it as a regular character? I can't just quote the whole string with single quotes since it may contain ' within itself.
Suggestions?

Related

How to read arguments with ampersand in bash [duplicate]

I'm building a shell script (trying to be POSIX compliant) and I'm stuck in an issue.
The script is supposed to receive an URL and do some things with it's content.
myscript www.pudim.com.br/?&args=ok
The thing is, the ampersand symbol is interpreted as a command additive, and giving to my script only the www.pudim.com.br/? part as an argument.
I know that the right workaround would be to surround the URL with quotes but, because I need to use this script several times in a row, I wanted to paste the URL's without having to embrace it with quotes every time.
Is there some way to get the full URL argument, somehow bypassing the ampersand?
Quotes for full URL
Wrapping the URL in quotes will be your only chance. See popular shell utility curl, as it states for its core argument URL:
When using [] or {} sequences when invoked from a command line prompt,
you probably have to put the full URL within double quotes to avoid
the shell from interfering with it. This also goes for other
characters treated special, like for example '&', '?' and '*'.
See also this question and that.
Extra argument(s) for specifying query parameters
You can also pass query parameter (key-value pair) as separate argument. So you can bypass & as separator. See curl's -F option:
-F, --form <name=content>
Read URL from STDIN
If your script allows user interaction you could read the unescaped URL (including metachars as &) from an uninterpreted input-source. See this tutorial.
You can escape just the ampersand; quotes effectively escape every character between them.
myscript www.pudim.com.br/\?\&args=ok # The ? should be escaped as well
There is no solution that lets you avoid all quoting, as & is a shell metacharacter whose unquoted meaning cannot be disabled. The & terminates the preceding command, causing it to be run in a background process; adding some redundant whitespace, you attempt is the same as
myscript www.pudim.com.br/? &
args=ok
Unescaped, the ? will cause the URL to be treated as a pattern to expand. However, it's unlikely the pattern will match any existing file, and bash's default behavior is to treat an unmatched pattern literally. (The failglob option will treat it as an error, and the nullglob option will make the URL disappear completely from the command line, but neither option is enabled by default.)

Passing variables to vim edit in a bash script [duplicate]

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'

Expand variables in shell command [duplicate]

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'

How to pass a file name containing spaces as an argument to a command-line program?

Whenever I try to add a file with whitespace in its name to git, it shows an Error as
"fatal: pathspec 'Tic' did not match any files"
Since I was New to git and Linux based terminal I have no idea how to do it.
Screen-shot or my error:
You should be able to quote the filename (eg. "file name"), or use an escape sequence (eg. file\< space >name).
To expand on #ergonaut's correct answer, just for the sake of clarity, this is actually neither a git nor a Linux issue. This is just a general requirement for command lines across the board.
On any command line, each word (or in this case, string of words) is evaluated separately as either a command or a parameter to a command. So, for example, git's add command is expecting a single parameter to come immediately after it (a filename). In this case, the next word it sees is just "Tic". Since it's only looking for a single parameter, it stops evaluating anything else at that point and that's why it's complaining about not being able to find the "Tic" file. When the words are enclosed in quotes, the entire string is evaluated as a single parameter, therefore fixing the issue.
Always wrap filenames that contain spaces with quotes when using them on the command line. Or even better, avoid using spaces in filenames. :-)
Try:
git add Tic\ tac\ toe.c
\ is used to escape special characters, though this is more bash related rather than git specific.
Alternatively, you could put the name of the file in quotes.
git add "Tic tac toe.c"
Start writing the first word and then press "Tab" button to use autocomplete
and you will be happy ))

How to deal with colons ":" in filename in shell script

I have a large group of files with ":" in their file names. When I'm in the interactive shell, typing "\" and then hitting tab does the trick to get them recognized as valid inputs to commands, but not so in the shell script.
I've tried
less file:name.txt
less file\:name.txt
less 'file\:name.txt
less 'file:name.txt'
and it's not recognized as a valid file.
However on the interactive command line I type less, followed by first file, then I type \, and then hit the TAB key, everything then works...
How do I do this in the shell script?
Use double quotes:
less "file:name.txt"
Bash recognizes the value within the double quotes as a full string.
As seen in Using quotes to include spaces and characters in filenames:
If you want to work with files with spaces or special characters in
the filename, you may have to use quotes.
This is working even when you don't put anything.
you can try:
less file and it recognizes it for you...
But you can use less "file:name.txt" as string in your bash.

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