In BASH, when we use sed command. if I have to replace a word say "sin" with "sin (x)" what should I do? Its taking the parenthesis as something else and i am getting error
example:
sed -i "3s/.*/xaxis label \" "Time (fs)" /" sigma.bfile
(I could escape quotation marks, but not the ( or ) )
Using syntax highlighting can help you:
sed -i "3s/.*/xaxis label \" "Time (fs)" /"
1 2 3 4 5
The double quote 3 ends double quote 1, so the string Time (fs) appears unquoted. Double quotes don't nest. Parentheses have special meaning in the shell. Are you sure you wanted to end the quoted string by the double quote nubmer 3?
You probably wanted just
sed -i "3s/.*/xaxis label \"Time (fs)\"/"
which could be even simpler if you replaced double quotes by single quotes
sed -i '3s/.*/xaxis label "Time (fs)"/'
Related
How can I escape double quotes inside a double string in Bash?
For example, in my shell script
#!/bin/bash
dbload="load data local infile \"'gfpoint.csv'\" into table $dbtable FIELDS TERMINATED BY ',' ENCLOSED BY '\"' LINES TERMINATED BY \"'\n'\" IGNORE 1 LINES"
I can't get the ENCLOSED BY '\"' with double quote to escape correctly. I can't use single quotes for my variable, because I want to use variable $dbtable.
Use a backslash:
echo "\"" # Prints one " character.
A simple example of escaping quotes in the shell:
$ echo 'abc'\''abc'
abc'abc
$ echo "abc"\""abc"
abc"abc
It's done by finishing an already-opened one ('), placing the escaped one (\'), and then opening another one (').
Alternatively:
$ echo 'abc'"'"'abc'
abc'abc
$ echo "abc"'"'"abc"
abc"abc
It's done by finishing already opened one ('), placing a quote in another quote ("'"), and then opening another one (').
More examples: Escaping single-quotes within single-quoted strings
Keep in mind that you can avoid escaping by using ASCII codes of the characters you need to echo.
Example:
echo -e "This is \x22\x27\x22\x27\x22text\x22\x27\x22\x27\x22"
This is "'"'"text"'"'"
\x22 is the ASCII code (in hex) for double quotes and \x27 for single quotes. Similarly you can echo any character.
I suppose if we try to echo the above string with backslashes, we will need a messy two rows backslashed echo... :)
For variable assignment this is the equivalent:
a=$'This is \x22text\x22'
echo "$a"
# Output:
This is "text"
If the variable is already set by another program, you can still apply double/single quotes with sed or similar tools.
Example:
b="Just another text here"
echo "$b"
Just another text here
sed 's/text/"'\0'"/' <<<"$b" #\0 is a special sed operator
Just another "0" here #this is not what i wanted to be
sed 's/text/\x22\x27\0\x27\x22/' <<<"$b"
Just another "'text'" here #now we are talking. You would normally need a dozen of backslashes to achieve the same result in the normal way.
Bash allows you to place strings adjacently, and they'll just end up being glued together.
So this:
echo "Hello"', world!'
produces
Hello, world!
The trick is to alternate between single and double-quoted strings as required. Unfortunately, it quickly gets very messy. For example:
echo "I like to use" '"double quotes"' "sometimes"
produces
I like to use "double quotes" sometimes
In your example, I would do it something like this:
dbtable=example
dbload='load data local infile "'"'gfpoint.csv'"'" into '"table $dbtable FIELDS TERMINATED BY ',' ENCLOSED BY '"'"'"' LINES "'TERMINATED BY "'"'\n'"'" IGNORE 1 LINES'
echo $dbload
which produces the following output:
load data local infile "'gfpoint.csv'" into table example FIELDS TERMINATED BY ',' ENCLOSED BY '"' LINES TERMINATED BY "'\n'" IGNORE 1 LINES
It's difficult to see what's going on here, but I can annotate it using Unicode quotes. The following won't work in Bash – it's just for illustration:
dbload=‘load data local infile "’“'gfpoint.csv'”‘" into ’“table $dbtable FIELDS TERMINATED BY ',' ENCLOSED BY '”‘"’“' LINES ”‘TERMINATED BY "’“'\n'”‘" IGNORE 1 LINES’
The quotes like “ ‘ ’ ” in the above will be interpreted by bash. The quotes like " ' will end up in the resulting variable.
If I give the same treatment to the earlier example, it looks like this:
echo “I like to use” ‘"double quotes"’ “sometimes”
Store the double quote character in a variable:
dqt='"'
echo "Double quotes ${dqt}X${dqt} inside a double quoted string"
Output:
Double quotes "X" inside a double quoted string
Check out printf...
#!/bin/bash
mystr="say \"hi\""
Without using printf
echo -e $mystr
Output: say "hi"
Using printf
echo -e $(printf '%q' $mystr)
Output: say \"hi\"
Make use of $"string".
In this example, it would be,
dbload=$"load data local infile \"'gfpoint.csv'\" into table $dbtable FIELDS TERMINATED BY ',' ENCLOSED BY '\"' LINES TERMINATED BY \"'\n'\" IGNORE 1 LINES"
Note (from the man page):
A double-quoted string preceded by a dollar sign ($"string") will cause the string to be translated according to the current locale. If the current locale is C or POSIX, the dollar sign is ignored. If the string is translated and replaced, the replacement is double-quoted.
For use with variables that might contain spaces in you Bash script, use triple quotes inside the main quote, e.g.:
[ "$(date -r """$touchfile""" +%Y%m%d)" -eq "$(date +%Y%m%d)" ]
Add "\" before double quote to escape it, instead of \
#! /bin/csh -f
set dbtable = balabala
set dbload = "load data local infile "\""'gfpoint.csv'"\"" into table $dbtable FIELDS TERMINATED BY ',' ENCLOSED BY '"\""' LINES TERMINATED BY "\""'\n'"\"" IGNORE 1 LINES"
echo $dbload
# load data local infile "'gfpoint.csv'" into table balabala FIELDS TERMINATED BY ',' ENCLOSED BY '"' LINES TERMINATED BY "''" IGNORE 1 LINES
I have this file:
http://www.beta.inegi.org.mx/contenidos/proyectos/enchogares/especiales/endireh/2016/tabulados/I_Cuestionario_general_estimaciones_endireh2016.xlsx
http://www.beta.inegi.org.mx/contenidos/proyectos/enchogares/especiales/endireh/2016/tabulados/IV_Ingresos_y_recursos_estimaciones_endireh2016.xlsx
http://www.beta.inegi.org.mx/contenidos/proyectos/enchogares/especiales/endireh/2016/tabulados/VI_ambito_escolar_estimaciones_endireh2016.xlsx
http://www.beta.inegi.org.mx/contenidos/proyectos/enchogares/especiales/endireh/2016/tabulados/VII_ambito_laboral_estimaciones_endireh2016.xlsx
http://www.beta.inegi.org.mx/contenidos/proyectos/enchogares/especiales/endireh/2016/tabulados/VIII_ambito_comunitario_estimaciones_endireh2016.xlsx
http://www.beta.inegi.org.mx/contenidos/proyectos/enchogares/especiales/endireh/2016/tabulados/IX_Atencion_Obstetrica_estimaciones_endireh2016.xlsx
http://www.beta.inegi.org.mx/contenidos/proyectos/enchogares/especiales/endireh/2016/tabulados/X_ambito_familiar_estimaciones_endireh2016.xlsx
And this bash script:
while read p; do
echo "\"$p\""
done < file.txt
I would expect the same file but with double quotes around each line, but this is what bash is outputting:
"http://www.beta.inegi.org.mx/contenidos/proyectos/enchogares/especiales/endireh/2016/tabulados/I_Cuestionario_general_estimaciones_endireh2016.xlsx
"http://www.beta.inegi.org.mx/contenidos/proyectos/enchogares/especiales/endireh/2016/tabulados/IV_Ingresos_y_recursos_estimaciones_endireh2016.xlsx
"http://www.beta.inegi.org.mx/contenidos/proyectos/enchogares/especiales/endireh/2016/tabulados/VI_ambito_escolar_estimaciones_endireh2016.xlsx
"http://www.beta.inegi.org.mx/contenidos/proyectos/enchogares/especiales/endireh/2016/tabulados/VII_ambito_laboral_estimaciones_endireh2016.xlsx
"http://www.beta.inegi.org.mx/contenidos/proyectos/enchogares/especiales/endireh/2016/tabulados/VIII_ambito_comunitario_estimaciones_endireh2016.xlsx
"http://www.beta.inegi.org.mx/contenidos/proyectos/enchogares/especiales/endireh/2016/tabulados/IX_Atencion_Obstetrica_estimaciones_endireh2016.xlsx
"http://www.beta.inegi.org.mx/contenidos/proyectos/enchogares/especiales/endireh/2016/tabulados/X_ambito_familiar_estimaciones_endireh2016.xlsx
Anyone know why bash is behaving this way? And how to output both " double quotes? (beginning and end)
I'm near certain that the line endings on your input file are CR/LF rather than just LF. This would output:
";
the web address;
a CR returning the cursor to the beginning of the line;
"; and, finally,
moving to a new line.
Capture the output to a file and pass it through a dump utility like od -xcb, that should show you the raw bytes being output.
As a test, creating a file consisting of the two lines 123<CR> and 456, I see:
pax> while read p; do echo "\"$p\""; done <testfile
"123
"456"
which seems to indicate the problem is as described.
If you're having trouble escaping the leading and trailing double quotes, you can just use single quotes around your echo statement. Any double quotes inside of single quotes have no significance in terms of defining a string literal, and vice versa:
while read p; do
echo '"$p"'
done < file.txt
I have strings which contain double quotes like this one:
"[{"clientid":"*", "identityzone":"*"}]"
I would like to use set or grep to delete the double quotes at the beginning and at the end of it, the output should look like :
[{"clientid":"*", "identityzone":"*"}]
I have used : sed -e 's/\"//g' but this deletes all the " in a string
You need to use line anchors
$ echo '"[{"clientid":"*", "identityzone":"*"}]"' | sed 's/^"//; s/"$//'
[{"clientid":"*", "identityzone":"*"}]
^" match " only at start of line
"$ match " only at end of line
You can also combine them using | as sed 's/^"\|"$//g'
See Overview of basic regular expression syntax
easy:
sed 's/^\"\(.*\)\"$/\1/g' <<<'"[{"clientid":"*", "identityzone":"*"}]"'
I am trying to replace a string in a line with another string with quotation marks in a file, say, $FILE. I'm trying to use sed.
I want to replace m = uniform(0, 0, 1) by m.LoadFile("run2_initial_$NF.ovf")
I am using this:
sed -i 's#m = uniform(0, 0, 1)#m.LoadFile("run2_initial_$NF.ovf")#g' $FILE
What I am getting is m = uniform(0, 0, 1) replaced by m.LoadFile(run2_initial_$NF.ovf)
That is, sed is just ignoring the quotation marks in the replacement string.
Am I doing something stupid?
Please suggest.
Edit: The quotation mark is now working fine, when I try now. Though the $NF is not being replaced by a number :(
What I got in the new file is:
m.LoadFile("run2_initial_$NF.ovf")
whereas I wanted: m.LoadFile("run2_initial_3.ovf")
If you use double quotes, the variables inside the string will be replaced by their values, if you use single quotes, they won't.
What you need to do here is to replace the single quotes with double quotes, and escape the double quotes you already had:
$ echo "m = uniform(0, 0, 1)" | \
sed "s#m = uniform(0, 0, 1)#m.LoadFile(\"run2_initial_$NF.ovf\")#g"
m.LoadFile("run2_initial_3.ovf")
How can I escape double quotes inside a double string in Bash?
For example, in my shell script
#!/bin/bash
dbload="load data local infile \"'gfpoint.csv'\" into table $dbtable FIELDS TERMINATED BY ',' ENCLOSED BY '\"' LINES TERMINATED BY \"'\n'\" IGNORE 1 LINES"
I can't get the ENCLOSED BY '\"' with double quote to escape correctly. I can't use single quotes for my variable, because I want to use variable $dbtable.
Use a backslash:
echo "\"" # Prints one " character.
A simple example of escaping quotes in the shell:
$ echo 'abc'\''abc'
abc'abc
$ echo "abc"\""abc"
abc"abc
It's done by finishing an already-opened one ('), placing the escaped one (\'), and then opening another one (').
Alternatively:
$ echo 'abc'"'"'abc'
abc'abc
$ echo "abc"'"'"abc"
abc"abc
It's done by finishing already opened one ('), placing a quote in another quote ("'"), and then opening another one (').
More examples: Escaping single-quotes within single-quoted strings
Keep in mind that you can avoid escaping by using ASCII codes of the characters you need to echo.
Example:
echo -e "This is \x22\x27\x22\x27\x22text\x22\x27\x22\x27\x22"
This is "'"'"text"'"'"
\x22 is the ASCII code (in hex) for double quotes and \x27 for single quotes. Similarly you can echo any character.
I suppose if we try to echo the above string with backslashes, we will need a messy two rows backslashed echo... :)
For variable assignment this is the equivalent:
a=$'This is \x22text\x22'
echo "$a"
# Output:
This is "text"
If the variable is already set by another program, you can still apply double/single quotes with sed or similar tools.
Example:
b="Just another text here"
echo "$b"
Just another text here
sed 's/text/"'\0'"/' <<<"$b" #\0 is a special sed operator
Just another "0" here #this is not what i wanted to be
sed 's/text/\x22\x27\0\x27\x22/' <<<"$b"
Just another "'text'" here #now we are talking. You would normally need a dozen of backslashes to achieve the same result in the normal way.
Bash allows you to place strings adjacently, and they'll just end up being glued together.
So this:
echo "Hello"', world!'
produces
Hello, world!
The trick is to alternate between single and double-quoted strings as required. Unfortunately, it quickly gets very messy. For example:
echo "I like to use" '"double quotes"' "sometimes"
produces
I like to use "double quotes" sometimes
In your example, I would do it something like this:
dbtable=example
dbload='load data local infile "'"'gfpoint.csv'"'" into '"table $dbtable FIELDS TERMINATED BY ',' ENCLOSED BY '"'"'"' LINES "'TERMINATED BY "'"'\n'"'" IGNORE 1 LINES'
echo $dbload
which produces the following output:
load data local infile "'gfpoint.csv'" into table example FIELDS TERMINATED BY ',' ENCLOSED BY '"' LINES TERMINATED BY "'\n'" IGNORE 1 LINES
It's difficult to see what's going on here, but I can annotate it using Unicode quotes. The following won't work in Bash – it's just for illustration:
dbload=‘load data local infile "’“'gfpoint.csv'”‘" into ’“table $dbtable FIELDS TERMINATED BY ',' ENCLOSED BY '”‘"’“' LINES ”‘TERMINATED BY "’“'\n'”‘" IGNORE 1 LINES’
The quotes like “ ‘ ’ ” in the above will be interpreted by bash. The quotes like " ' will end up in the resulting variable.
If I give the same treatment to the earlier example, it looks like this:
echo “I like to use” ‘"double quotes"’ “sometimes”
Store the double quote character in a variable:
dqt='"'
echo "Double quotes ${dqt}X${dqt} inside a double quoted string"
Output:
Double quotes "X" inside a double quoted string
Check out printf...
#!/bin/bash
mystr="say \"hi\""
Without using printf
echo -e $mystr
Output: say "hi"
Using printf
echo -e $(printf '%q' $mystr)
Output: say \"hi\"
Make use of $"string".
In this example, it would be,
dbload=$"load data local infile \"'gfpoint.csv'\" into table $dbtable FIELDS TERMINATED BY ',' ENCLOSED BY '\"' LINES TERMINATED BY \"'\n'\" IGNORE 1 LINES"
Note (from the man page):
A double-quoted string preceded by a dollar sign ($"string") will cause the string to be translated according to the current locale. If the current locale is C or POSIX, the dollar sign is ignored. If the string is translated and replaced, the replacement is double-quoted.
For use with variables that might contain spaces in you Bash script, use triple quotes inside the main quote, e.g.:
[ "$(date -r """$touchfile""" +%Y%m%d)" -eq "$(date +%Y%m%d)" ]
Add "\" before double quote to escape it, instead of \
#! /bin/csh -f
set dbtable = balabala
set dbload = "load data local infile "\""'gfpoint.csv'"\"" into table $dbtable FIELDS TERMINATED BY ',' ENCLOSED BY '"\""' LINES TERMINATED BY "\""'\n'"\"" IGNORE 1 LINES"
echo $dbload
# load data local infile "'gfpoint.csv'" into table balabala FIELDS TERMINATED BY ',' ENCLOSED BY '"' LINES TERMINATED BY "''" IGNORE 1 LINES