How can I escape a double quote inside double quotes? - bash

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

Related

How to concatenate string variables that contain space? [duplicate]

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

tr command: strange behavior with | and \

Let's say I have a file test.txt with contents:
+-foo.bar:2.4
| bar.foo:1.1:test
\| hello.goobye:3.3.3
\|+- baz.yeah:4
I want to use the tr command to delete all instances of the following set of characters:
{' ', '+', '-', '|', '\'}
Done some pretty extensive research on this but found no clear/concise answers.
This is the command that works:
input:
cat test.txt | tr -d "[:blank:]|\\\+-"
output:
foo.bar:2.4
bar.foo:1.1:test
hello.goobye:3.3.3
baz.yeah:4
I experimented with many combinations of that set and I found out that the '-' was being treated as a range indicator (like... [a-z]) and therefore must be put at the end. But I have two main questions:
1) Why must the backslash be double escaped in order to be included in the set?
2) Why does putting the '|' at the end of the set string cause the tr program to delete everything in the file except for trailing new line characters?
Like this:
tr -d '\-|\\+[:blank:] ' < file
You have to escape the - because it is used for denoting ranges of characters like:
tr -d '1-5'
and must therefore being escaped if you mean a literal hyphen. You can also put it at the end. (learned that, thanks! :) )
Furthermore the \ must be escaped when you mean a literal \ because it has a special meaning needed for escape sequences.
The remaining characters must not being escaped.
Why must the \ being doubly escaped in your example?
It's because you are using a "" (double quoted) string to quote the char set. A double quoted string will be interpreted by the shell, a \\ in a double quoted string means a literal \. Try:
echo "\+"
echo "\\+"
echo "\\\+"
To avoid to doubly escape the \ you can just use single quotes as in my example above.
Why does putting the '|' at the end of the set string cause the tr program to delete everything in the file except for trailing new line characters?
Following CharlesDuffy's comment having the | at the end means also that you had the unescaped - not at the end, which means it was describing a range of characters where the actual range depends on the position you had it in the set.
another approach is to define the allowed chars
$ tr -cd '[:alnum:]:.\n' <file
foo.bar:2.4
bar.foo:1.1:test
hello.goobye:3.3.3
baz.yeah:4
or, perhaps delete all the prefix non-word chars
$ sed -E 's/\W+//' file

Why is bash ignoring the ending double quote (")

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

Bash - how to add newlines to string in loop with command substitution?

I have a for loop thats iterating over a list of items, gathering statistics. The constructed string is then being sent as the body of an email using the mail command. My question is: what is the proper way to output newlines, as opposed to escaped \n characters as my script is currently doing?
Here's the relavent bash code:
for x in $([..]); do
BODY="${BODY}\nReport for ${x}: \nSuccess: `get_logs $x 1`\nAttempts: `get_logs $x`"
done;
As I said in the comments, you can use:
echo -e
The -e flag enables the interpretation of backslash escapes
You can use newlines inside quoted strings, the only downside of this is that you cannot indent the new lines, because the indent will be part of the string.
for x in $([..]); do
BODY="${BODY}
Report for ${x}:
Success: $(get_logs $x 1)
Attempts: $(get_logs $x)"
done;
Alternatively, append each line separately to make your code more readable.
for x in $([..]); do
BODY+="Report for ${x}:"$'\n'
BODY+="Success: $(get_logs $x 1)"$'\n'
BODY+="Attempts: $(get_logs $x)"$'\n'
done;
$'\n' is an ANSI C string and will be replaced by a line feed.

How do I store up to 2 names into my variables if needed and also how do I escape a possible single quote in a name?

PhoneBook.txt has names delimited by :
ex= first:last:number - simple
ex= first first:last last:number - 2 firstnames and/or 2 last names
ex= f'irst:l'ast:number - first name or last name with a single quote in it.
My current script
#!/bin/bash
firstName="$1"
lastName="$2"
if [[ "$firstName" == "" ]]; then
read -p "Please enter a first name: " firstName
fi
if [[ "$lastName" == "" ]]; then
read -p "Please enter a last name: " lastName
fi
grep "$firstName:$lastName" PhoneBook.txt
The -F argument tells grep to search for a fixed string instead of a regex.
Use double quotes to delimit your arguments on the command line, e.g.:
./phonebook "first1 first2" "last o'last"
You're free to use single quotes in a double-quoted string.
Just be aware that double-quoted strings are subject to parameter (variable) and command substitution, so instances of literal $ chars. must be escaped as \$.
Similarly, embedded double quotes must be escaped as \".
Note that, by contrast, there is no direct way to embed single quotes in a single-quoted string - you have to break the string apart and insert an escaped single quote, \'; e.g., to single-quote isn't, use 'isn'\''t'; however, you can freely embed double quotes.

Resources