Bash in Ubuntu (13.04) seems to have this quote function by default:
quote ()
{
local quoted=${1//\'/\'\\\'\'};
printf "'%s'" "$quoted"
}
This function should always return a correct shell escaped version of it's first parameter.
Is there any input string which will break this function (i.e. return a string unusable for shell input)?
Examples:
quote "A string's but a string."
'A string'\''s but a string.'
quote "A newline *doesn't*
seem to break anything..."
'A newline *doesn'\''t*
seem to break anything...'
It appears to be "bullet-proof".
A singly-quoted item in the shell is absolutely anything between single quotes, other than a single quote. The single quote can be pseudo-included by using the sequence '\'' which terminates one quoting, then backslash-quotes the quote, and then re-starts a new quoting.
This substitution is all that the function does.
The output of the function is single-quote syntax which represents the input string.
The function even defeats the "process-substitution-swallows-trailing-newlines" issue.
Ordinarily, if you have some function or command which outputs multiple lines, followed by one or more blank lines, and then capture it with process substitution like this:
FOO=$(command)
the trailing blank lines are gone. Since quote's output is wrapped in single quotes, trailing newlines in the string are protected, allowing this to work:
FOO=$("a
b
c
")
echo "$FOO"
'a
b
c
'
Clearly, the output of quote is meant to be eval-ed, and quote ensures that the result of this eval is the original argument to quote without any accidental expansion, splitting or stripping of whitespace.
Related
I want to use backticks in ruby for a programm call.
The parameter is a String variable containing one or more backticks, i.e.
"&E?##A`?". The following command yields a new label as its return value:
echo "&E?##A\`?" | nauty-labelg 2>/dev/null
From a ruby program I can call it as follows and get the correct result:
new_label = `echo "&E?##A\\\`?" | nauty-labelg 2>/dev/null`
I want to achieve the same using a variable for the label.
So I have to insert three slashes into my variable label = "&E?##A`?" in order to escape the backtick. The following seems to work, though it is not very elegant:
escaped_label = label.gsub(/`/, '\\\`').gsub(/`/, '\\\`').gsub(/`/, '\\\`')
But the new variable cannot be used in the program call:
new_label = `echo "#{escaped_label}" | nauty-labelg 2>/dev/null`
In this case I do not get an answer from nauty-labelg.
So I have to insert three slashes into my variable label = "&E?##A`?" in order to escape the backtick.
No, you only need to add one backslash for the output. To escape the ` special bash character. The other other two are only for representation proposes, otherwise it isn't valid Ruby code.
new_label = `echo "&E?##A\\\`?" | nauty-labelg 2>/dev/null`
The first backslash will escape the second one (outputting one single backslash). The third backslash escapes the ` character (outputting one single `).
You should only add backslashes before characters that have a special meaning within double quoted bash context. These special characters are: $, `, \ and \n. Those can be escaped with the following code:
def escape_bash_string(string)
string.gsub(/([$`"\\\n])/, '\\\\\1')
end
For label = "&E?##A`?" only the ` should be escaped.
escaped_string = escape_bash_string("&E?##A\`?")
puts escaped_string
# &E?##A\`?
I have a string say
string="MYSTRING"
Now I want to grep for any occurrence of "MYSTRING" (with double quotes) such that there must be parentheses at starting, basically I want to search, in which places "MYSTRING" is used as function's parameter. So,
foo( "MYSTRING" //postive
foo("MYSTRING" //positive
foo('MYSTRING' //positive
foo( 'MYSTRING' //positive
var a = "MYSTRING" //negative
I used:
string="MYSTRING"
regexstring="[(]*\"$string\""
grep -e "$regexString" <<'EOF'
foo( "MYSTRING" //postive
foo("MYSTRING" //positive
foo('MYSTRING' //positive
foo( 'MYSTRING' //positive
var a = "MYSTRING" //negative
EOF
Ideally, all the items with "positive" next to them and none of the items with "negative" will match. What needs to change to make that happen?
The easy way to do this is to use the ksh extension (adopted by bash) $'' to provide a literal string that can include backticks.
#!/usr/bin/env bash
string=MYSTRING
regexstring=$'[(][[:space:]]*[\'"]'"$string"$'[\'"]'
grep -e "$regexstring" "$#"
Breaking down this assignment:
$'[(][[:space:]]*[\'"]'
...is a string literal which evaluates to the following:
[(][[:space:]]*['"]
...thus, it matches a single (, followed by zero or more spaces, followed by either ' or ".
The second part of it is a double-quoted expansion, "$string"; this should be fairly straight on its face.
The final part is $'[\'"']'; just as in the first part, the $'' string-literal syntax is used to generate a string that can contain both ' and " as contents.
By the way -- in POSIX sh, this might instead look like:
regexstring='[(][[:space:]]*['"'"'"]'"$string""['"'"]'
There, we're using single-quoted strings to hold literal double quotes, and double-quoted strings to hold literal single quotes.
I have two specific questions about the IFS. I'm aware that changing the internal field separator, IFS, changes what the bash script iterates over.
So, why is it that the length of the array doesn't change?
Here's my example:
delimiter=$1
strings_to_find=$2
OIFS=$IFS
IFS=$delimiter
echo "the internal field separator is $IFS"
echo "length of strings_to_find is ${#strings_to_find[#]}"
for string in ${strings_to_find[#]}
do
echo "string is separated correctly and is $string"
done
IFS=$OIFS
But why does the length not get affected by the new IFS?
The second thing that I don't understand is how to make the IFS affect the input arguments.
Let's say I'm expecting my input arguments to look like this:
./executable_shell_script.sh first_arg:second_arg:third_arg
And I want to parse the input arguments by setting the IFS to :. How do I do this? Setting the IFS doesn't seem to do anything. I must be doing this wrong....?
Thank you.
Bash arrays are, in fact, arrays. They are not strings which are parsed on demand. Once you create an array, the elements are whatever they are, and they won't change retroactively.
However, nothing in your example creates an array. If you wanted to create an array out of argument 2, you would need to use a different syntax:
strings_to_find=($2)
Although your strings_to_find is not an array, bash allows you to refer to it as though it were an array of one element. So ${#strings_to_find[#]} will always be one, regardless of the contents of strings_to_find. Also, your line:
for string in ${strings_to_find[#]}
is really no different from
for string in $strings_to_find
Since that expansion is not quoted, it will be word-split, using the current value of IFS.
If you use an array, most of the time you will not want to write for string in ${strings_to_find[#]}, because that just reassembles the elements of an array into a string and then word-splits them again, which loses the original array structure. Normally you will avoid the word-splitting by using double quotes:
strings_to_find=(...)
for string in "${strings_to_find[#]}"
As for your second question, the value of IFS does not alter the shell grammar. Regardless of the value of IFS, words in a command are separated by unquoted whitespace. After the line is parsed, the shell performs parameter and other expansions on each word. As mentioned above, if the expansion is not quoted, the expanded text is then word-split using the value of IFS.
If the word does not contain any expansions, no word-splitting is performed. And even if the word does contain expansions, word-splitting is only performed on the expansion itself. So, if you write:
IFS=:
my_function a:b:c
my_function will be called with a single argument; no expansion takes places, so no word-splitting occurs. However, if you use $1 unquoted inside the function, the expansion of $1 will be word-split (if it is expanded in a context in which word-splitting occurs).
On the other hand,
IFS=:
args=a:b:c
my_function $args
will cause my_function to be invoked with three arguments.
And finally,
IFS=:
args=c
my_function a:b:$args
is exactly the same as the first invocation, because there is no : in the expansion.
This is an example script based on #rici's answer :
#!/bin/bash
fun()
{
echo "Total Params : " $#
}
fun2()
{
array1=($1) # Word splitting occurs here based on the IFS ':'
echo "Total elements in array1 : "${#array1[#]}
# Here '#' before array counts the length of the array
array2=("$1") # No word splitting because we have enclosed $1 in double quotes
echo "Total elements in array2 : "${#array2[#]}
}
IFS_OLD="$IFS"
IFS=$':' #Changing the IFS
fun a:b:c #Nothing to expand here, so no use of IFS at all. See fun2 at last
fun a b c
fun abc
args="a:b:c"
fun $args # Expansion! Word splitting occurs with the current IFS ':' here
fun "$args" # preventing word spliting by enclosing ths string in double quotes
fun2 a:b:c
IFS="$IFS_OLD"
Output
Total Params : 1
Total Params : 3
Total Params : 1
Total Params : 3
Total Params : 1
Total elements in array1 : 3
Total elements in array2 : 1
Bash manpage says :
The shell treats each character of IFS as a delimiter, and splits the
results of the other expansions into words on these characters.
Someone please help to explain how this work? About the single quote it should not interpret anything but it is not working as what i expected. I expect to get echo $testvar value exactly '"123b"'.
a="testvar"
b="'"123b"'"
eval $a='$b'
echo $testvar
'123b'
a="testvar"
b='"123b"'
eval $a='$b'
echo $testvar
"123b"
a="testvar"
b='"123b"'
eval $a=$b
echo $testvar
123b
Guessing that you want testvar to be <single-quote><double-quote>123b<double-quote><single-quote>:
testvar=\'\"123b\"\'
Consider this in C or Java:
char* str = "123b";
printf("%s\n", str);
String str = "123b";
System.out.println(str);
Why does this write 123b when we clearly used double quotes? Why doesn't it write "123b", with quotes?
The answer is that the quotes are not part of the data. The quotes are used by the programming language to determine where strings start and stop, but they're not in any way part of the string. This is just as true for Bash as for C and Java.
Just like there's no way in Java to differentiate Strings created with "123" + "b" and "123b", there's no way in Bash to tell that b='"123b"' used single quotes in its definition, as opposed to e.g. b=\"123b\".
If given a variable you want to assign its value surrounded by single quotes, you can use e.g.
printf -v testvar "'%s'" "$b"
But this just adds new literal single quotes around a string. It doesn't and cannot care how b was originally quoted, because that information is stored.
To instead add a layer of escaping to a variable, so that when evaluated once it turns into a literal string identical to your input, you can use:
printf -v testvar "%q" "$b"
This will produce a value which is quoted equivalently but not necessarily identically to your original definition. For "value" (a literal with double quotes in it), it may produce \"value\" or '"value"' or '"'value'"' which all evaluate exactly to "value".
I know there is a better way to do this.
What is the better way?
How do you do a string replace on a string variable in bash?
For Example: (using php because that's what I know)
$path = "path/to/directory/foo bar";
$path = str_replace(" ", "\ ", "$path");
echo $path;
returns:
path/to/directory/foo\ bar
To perform the specific replacement in bash:
path='path/to/directory/foo bar'
echo "${path// /\\ }"
Don't use prefix $ when assigning to variables in bash.
No spaces are allowed around the =.
Note that path is assigned with single quotes, whereas the string replacement occurs in double quotes - this distinction is important: bash does NOT interpret single-quoted strings, whereas you can refer to variables (and do other things) in double-quoted strings; (also, not quoting a variable reference at all has other ramifications, often undesired - in general, double-quote your variable references)
Explanation of string replacement "${path// /\\ }":
In order to perform value substitution on a variable, you start with enclosing the variable name in {...}
// specifies that ALL occurrences of the following search pattern are to be replaced (use / to replace the first occurrence only).
/ separates the search pattern, (a single space), from the replacement string, \\ .
The replacement string, \ , must be represented as \\ , because \ has special meaning as an escape char. and must therefore itself be escaped for literal use.
The above is an instance of what bash (somewhat cryptically) calls shell parameter expansion and also parameter expansion and [parameter and] variable expansion. There are many more flavors, such as for extracting a substring, providing a default value, stripping a prefix or suffix, ... - see the BashGuide page on the topic or the manual.
As for what types of expressions are supported in the search and replacement strings:
The search expression is a globbing pattern of the same type used in filename expansion (e.g, *.txt); for instance, v='dear me'; echo "${v/m*/you}" yields 'dear you'. Note that the longest match will be used.
Additionally, the first character of the pattern has special meaning in this context:
/, as we've seen above, causes all matching occurrences of the pattern to be replaced - by default, only the first one is replaced.
# causes the rest of the pattern to only match at the beginning of the input variable
% only matches at the end
The replacement expression is a string that is subject to shell expansions; while there is no support for backreferences, the fact that the string is expanded allows you to have the replacement string reference other variables, contain commands, with $(...), ...; e.g.:
v='sweet home'; echo "${v/home/$HOME}" yields, for instance, 'sweet /home/jdoe'.
v='It is now %T'; echo "${v/\%T/$(date +%T)}" yields, for instance, It is now 10:05:17.
o1=1 o2=3 v="$o1 + $o2 equals result"; echo "${v/result/$(( $o1 + $o2 ))}" yields '1 + 3 equals 4' (I think)
There are many more features and subtleties - refer to the link above.
How about sed? Is that what you're looking for?
#!/bin/bash
path="path/to/directory/foo bar"
new_path=$(echo "$path" | sed 's/ /\\ /g')
echo "New Path: '$new_path"
But as #n0rd pointed out in his comment, is probably better just quoting the path when you want to use it; something like...
path="path/to/directory/foo bar"
echo "test" > "$path"