I'm trying to pull variables from an API in json format and then put them back together with one variable changed and fire them back as a put.
Only issue is that every value has quote marks in it and must go back to the API separated by commas only.
example of what it should see with redacted information, variables inside the **'s:
curl -skv -u redacted:redacted -H Content-Type: application/json -X PUT -d'{properties:{basic:{request_rules:[**"/(req) testrule","/test-body","/(req) test - Admin","test-Caching"**]}}}' https://x.x.x.x:9070/api/tm/1.0/config/active/vservers/xxx-xx
Obviously if I fire them as a plain array I get spaces instead of commas. However I tried outputting it as a plain string
longstr=$(echo ${valuez[#]})
output=$(echo $longstr |sed -e 's/" /",/g')
And due to the way bash is interpreted it seems to either interpret the quotes wrong or something else. I guess it might well be the single ticks encapsulating after the PUT -d as well but I'm not sure how I can throw a variable into something that has single ticks.
If I put the raw data in manually it works so it's either the way the variable is being sent or the single ticks. I don't get an error and when I echo the line out it looks perfect.
Any ideas?
valuez=( "/(req) testrule" "/test-body" "/(req) test - Admin" "test-Caching" )
# Temporarily set IFS to some character which is known not to appear in the array.
oifs=$IFS
IFS=$'\014'
# Flatten the array with the * expansion giving a string containing the array's elements separated by the first character of $IFS.
d_arg="${valuez[*]}"
IFS=$oifs
# If necessary, quote or escape embedded quotation marks. (Implementation-specific, using doubled double quotes as an example.)
d_arg="${d_arg//\"/\"\"}"
# Substitute the known-to-be-absent character for the desired quote+separator+quote.
d_arg="${d_arg//$'\014'/\",\"}"
# Prepend and append quotes.
d_arg="\"$d_arg\""
# insert the prepared arg into the final string.
d_arg="{properties:{basic:{request_rules:[${d_arg}]}}}"
curl ... -d"$d_arg" ...
if you have gnu awk with version 4 and above, which support FPAT
output=$(echo $longstr |awk '$1=$1' FPAT="(\"[^\"]+\")" OFS=",")
Explanation
FPAT #
This is a regular expression (as a string) that tells gawk to create the fields based on text that matches the regular expression. Assigning a value to FPAT overrides the use of FS and FIELDWIDTHS for field splitting. See Splitting By Content, for more information.
If gawk is in compatibility mode (see Options), then FPAT has no special meaning, and field-splitting operations occur based exclusively on the value of FS.
valuez=( "/(req) testrule" "/test-body" "/(req) test - Admin" "test-Caching" )
csv="" sep=""
for v in "${valuez[#]}"; do csv+="$sep\"$v\""; sep=,; done
echo "$csv"
"/(req) testrule","/test-body","/(req) test - Admin","test-Caching"
If it's something you need to do repeatedly, but it into a function:
toCSV () {
local csv sep val
for val; do
csv+="$sep\"$val\""
sep=,
done
echo "$csv"
}
csv=$(toCSV "${valuez[#]}")
Related
Should or should I not wrap quotes around variables in a shell script?
For example, is the following correct:
xdg-open $URL
[ $? -eq 2 ]
or
xdg-open "$URL"
[ "$?" -eq "2" ]
And if so, why?
General rule: quote it if it can either be empty or contain spaces (or any whitespace really) or special characters (wildcards). Not quoting strings with spaces often leads to the shell breaking apart a single argument into many.
$? doesn't need quotes since it's a numeric value. Whether $URL needs it depends on what you allow in there and whether you still want an argument if it's empty.
I tend to always quote strings just out of habit since it's safer that way.
In short, quote everything where you do not require the shell to perform word splitting and wildcard expansion.
Single quotes protect the text between them verbatim. It is the proper tool when you need to ensure that the shell does not touch the string at all. Typically, it is the quoting mechanism of choice when you do not require variable interpolation.
$ echo 'Nothing \t in here $will change'
Nothing \t in here $will change
$ grep -F '#&$*!!' file /dev/null
file:I can't get this #&$*!! quoting right.
Double quotes are suitable when variable interpolation is required. With suitable adaptations, it is also a good workaround when you need single quotes in the string. (There is no straightforward way to escape a single quote between single quotes, because there is no escape mechanism inside single quotes -- if there was, they would not quote completely verbatim.)
$ echo "There is no place like '$HOME'"
There is no place like '/home/me'
No quotes are suitable when you specifically require the shell to perform word splitting and/or wildcard expansion.
Word splitting (aka token splitting);
$ words="foo bar baz"
$ for word in $words; do
> echo "$word"
> done
foo
bar
baz
By contrast:
$ for word in "$words"; do echo "$word"; done
foo bar baz
(The loop only runs once, over the single, quoted string.)
$ for word in '$words'; do echo "$word"; done
$words
(The loop only runs once, over the literal single-quoted string.)
Wildcard expansion:
$ pattern='file*.txt'
$ ls $pattern
file1.txt file_other.txt
By contrast:
$ ls "$pattern"
ls: cannot access file*.txt: No such file or directory
(There is no file named literally file*.txt.)
$ ls '$pattern'
ls: cannot access $pattern: No such file or directory
(There is no file named $pattern, either!)
In more concrete terms, anything containing a filename should usually be quoted (because filenames can contain whitespace and other shell metacharacters). Anything containing a URL should usually be quoted (because many URLs contain shell metacharacters like ? and &). Anything containing a regex should usually be quoted (ditto ditto). Anything containing significant whitespace other than single spaces between non-whitespace characters needs to be quoted (because otherwise, the shell will munge the whitespace into, effectively, single spaces, and trim any leading or trailing whitespace).
When you know that a variable can only contain a value which contains no shell metacharacters, quoting is optional. Thus, an unquoted $? is basically fine, because this variable can only ever contain a single number. However, "$?" is also correct, and recommended for general consistency and correctness (though this is my personal recommendation, not a widely recognized policy).
Values which are not variables basically follow the same rules, though you could then also escape any metacharacters instead of quoting them. For a common example, a URL with a & in it will be parsed by the shell as a background command unless the metacharacter is escaped or quoted:
$ wget http://example.com/q&uack
[1] wget http://example.com/q
-bash: uack: command not found
(Of course, this also happens if the URL is in an unquoted variable.) For a static string, single quotes make the most sense, although any form of quoting or escaping works here.
wget 'http://example.com/q&uack' # Single quotes preferred for a static string
wget "http://example.com/q&uack" # Double quotes work here, too (no $ or ` in the value)
wget http://example.com/q\&uack # Backslash escape
wget http://example.com/q'&'uack # Only the metacharacter really needs quoting
The last example also suggests another useful concept, which I like to call "seesaw quoting". If you need to mix single and double quotes, you can use them adjacent to each other. For example, the following quoted strings
'$HOME '
"isn't"
' where `<3'
"' is."
can be pasted together back to back, forming a single long string after tokenization and quote removal.
$ echo '$HOME '"isn't"' where `<3'"' is."
$HOME isn't where `<3' is.
This isn't awfully legible, but it's a common technique and thus good to know.
As an aside, scripts should usually not use ls for anything. To expand a wildcard, just ... use it.
$ printf '%s\n' $pattern # not ``ls -1 $pattern''
file1.txt
file_other.txt
$ for file in $pattern; do # definitely, definitely not ``for file in $(ls $pattern)''
> printf 'Found file: %s\n' "$file"
> done
Found file: file1.txt
Found file: file_other.txt
(The loop is completely superfluous in the latter example; printf specifically works fine with multiple arguments. stat too. But looping over a wildcard match is a common problem, and frequently done incorrectly.)
A variable containing a list of tokens to loop over or a wildcard to expand is less frequently seen, so we sometimes abbreviate to "quote everything unless you know precisely what you are doing".
Here is a three-point formula for quotes in general:
Double quotes
In contexts where we want to suppress word splitting and globbing. Also in contexts where we want the literal to be treated as a string, not a regex.
Single quotes
In string literals where we want to suppress interpolation and special treatment of backslashes. In other words, situations where using double quotes would be inappropriate.
No quotes
In contexts where we are absolutely sure that there are no word splitting or globbing issues or we do want word splitting and globbing.
Examples
Double quotes
literal strings with whitespace ("StackOverflow rocks!", "Steve's Apple")
variable expansions ("$var", "${arr[#]}")
command substitutions ("$(ls)", "`ls`")
globs where directory path or file name part includes spaces ("/my dir/"*)
to protect single quotes ("single'quote'delimited'string")
Bash parameter expansion ("${filename##*/}")
Single quotes
command names and arguments that have whitespace in them
literal strings that need interpolation to be suppressed ( 'Really costs $$!', 'just a backslash followed by a t: \t')
to protect double quotes ('The "crux"')
regex literals that need interpolation to be suppressed
use shell quoting for literals involving special characters ($'\n\t')
use shell quoting where we need to protect several single and double quotes ($'{"table": "users", "where": "first_name"=\'Steve\'}')
No quotes
around standard numeric variables ($$, $?, $# etc.)
in arithmetic contexts like ((count++)), "${arr[idx]}", "${string:start:length}"
inside [[ ]] expression which is free from word splitting and globbing issues (this is a matter of style and opinions can vary widely)
where we want word splitting (for word in $words)
where we want globbing (for txtfile in *.txt; do ...)
where we want ~ to be interpreted as $HOME (~/"some dir" but not "~/some dir")
See also:
Difference between single and double quotes in Bash
What are the special dollar sign shell variables?
Quotes and escaping - Bash Hackers' Wiki
When is double quoting necessary?
I generally use quoted like "$var" for safe, unless I am sure that $var does not contain space.
I do use $var as a simple way to join lines:
lines="`cat multi-lines-text-file.txt`"
echo "$lines" ## multiple lines
echo $lines ## all spaces (including newlines) are zapped
Whenever the https://www.shellcheck.net/ plugin for your editor tells you to.
I have this which works to declare a JSON string in a bash script:
local my_var="foobar"
local json=`cat <<EOF
{"quicklock":"${my_var}"}
EOF`
The above heredoc works, but I can't seem to format it any other way, it literally has to look exactly like that lol.
Is there any way to get the command to be on one line, something like this:
local json=`cat <<EOF{"quicklock":"${my_var}"}EOF`
that would be so much nicer, but doesn't seem to take, obviously simply because that's not how EOF works I guess lol.
I am looking for a shorthand way to declare JSON in a file that:
Does not require a ton of escape chars.
That allows for dynamic interpolation of variables.
Note: The actual JSON I want to use has multiple dynamic variables with many key/value pairs. Please extrapolate.
I'm not a JSON guy, don't really understand the "well-formed" arguments in the discussion above, but, you can use a 'here-string' rather than a 'here-document', like this:
my_var="foobar"
json=`cat <<<{\"quicklock\":\"${my_var}\"}`
why not use jq? It's pretty good at managing string interpolation and it lints your structure.
$ echo '{}' >> foo.json
$ declare myvar="assigned-var"
$ jq --arg ql "$myvar" '.quicklock=$ql' foo.json
the text that comes out on the other end of that call to jq can then be cat into a file or whatever you wanna do. text would look something like this:
{"quicklock": "assigned-var"}
You can do this with printf:
local json="$(printf '{"quicklock":"%s"}' "$my_var")"
(Never mind that SO's syntax highlighting looks odd here. Posix shell command substitution allows nesting one level of quotes.)
A note (thanks to Charles Duffy's comment on the question): I'm assuming $my_var is not controlled by user input. If it is, you'll need to be careful to ensure it is legal for a JSON string. I highly recommend barring non-ASCII characters, double quotes, and backslashes. If you have jq available, you can use it as Charles noted in the comments to ensure you have well-formed output.
You can define your own helper function to address the situation with missing bash syntax:
function begin() { eval echo $(sed "${BASH_LINENO[0]}"'!d;s/.*begin \(.*\) end.*/\1/;s/"/\\\"/g' "${BASH_SOURCE[0]}"); }
Then you can use it as follows.
my_var="foobar"
json=$(begin { "quicklock" : "${my_var}" } end)
echo "$json"
This fragment displays the desired output:
{ "quicklock" : "foobar" }
This is just a proof of concept. You can define your syntax in any way you want (such as end of the input by the custom EOF string, correctly escape invalid characters). For example, since Bash allows function identifiers using characters other than alphanumeric characters, it is possible to define such a syntax:
json=$(/ { "quicklock" : "${my_var}" } /)
Moreover, if you relax the first criterion (escape characters), ordinary assignment will nicely solve this problem:
json="{ \"quicklock\" : \"${my_var}\" }"
How about just using the shell's natural concatenation of strings? If you concatenate ${mybar} rather than interpolate it, you can avoid escapes and get everything on one line:
my_var1="foobar"
my_var2="quux"
json='{"quicklock":"'${my_var1}'","slowlock":"'$my_var2'"}'
That said, this is a pretty crude scheme, and as others have pointed out you'll have problems if the variables, say, contain quote characters.
Since no escape chars is strong requirement here is a here-doc based solution:
#!/bin/bash
my_var='foobar'
read -r -d '' json << EOF
{
"quicklock": "$my_var"
}
EOF
echo "$json"
It will give you the same output as the first solution I mentioned.
Just be careful, if you would put first EOF inside double quotes:
read -r -d '' json << "EOF"
$my_var would not be considered as a variable but as a plain text, so you would get this output:
{
"quicklock": "$my_var"
}
I have a specific problem. I make email variable in bash script like:
tDestinationEmail=email#email.com
and if I enter an email after the sign = my json should look like this:
{"ToAddresses": ["email#email.com"]}
I try to do it with
tDestinationEmailJSON='{"ToAddresses": ["$tDestinationEmail"]}'
but on the output i getting
{"ToAddresses": ["$tDestinationEmail"]}
Maybe whom know solution how i can do it ? Please suggest!
jq is the right tool for JSON data manipulation:
tDestinationEmail="email#email.com"
jq -nc --arg email "$tDestinationEmail" '{ToAddresses: [$email]}'
The output:
{"ToAddresses":["email#email.com"]}
From man page of bash:
"Enclosing characters in single quotes preserves the literal value of each character within the quotes."
So when you call $tDestinationEmail will be treated as string and will not call the value of the variable.
Using double quotes:
tDestinationEmailJSON='{"ToAddresses": ['"\"$tDestinationEmail\""']}'
The code below doesn't do what I need it to. I want to pass $STR to Underscore and pluck out the "name" attribute from the JSON data.
#!/bin/bash
STR='['$#']';
RESULT=`underscore pluck --data '+$STR+' name`;
echo $RESULT;
JSON Data:
{"maxResults":1,"resultList":[{"#class":"com.sohnar.trafficlite.transfer.crm.refactor.ClientCRMEntryTO","id":331458,"version":2,"dateCreated":"2017-05-31T13:20:22.960+0000","dateModified":"2017-06-05T14:23:59.961+0000","lastUpdatedUserId":71954,"name":"ACME_CLIENT","website":null,"description":null,"billingLocation":null,"primaryLocation":null,"crmEntryType":"CLIENT","industryType":null,"accountManagerId":103049,"crmClientClassificationListItemId":{"id":12405},"companyProfile":{"id":486024,"version":1,"dateCreated":"2017-05-31T13:20:22.960+0000","dateModified":"2017-06-05T14:23:59.962+0000","sourceOfBusinessListItemId":null,"creditTermsListItemId":{"id":4215},"relationshipSince":"2017-05-30T23:00:00.000+0000","turnover":0,"employees":0,"taxNumber":null,"companyNumber":null,"nominalCode":null,"accountPackageId":null,"optOutMarketing":false,"optOutEmail":false,"optOutTelephone":false,"notes":null},"colorCode":0,"externalCode":"SAP-01","clientState":"CLIENT","defaultCustomRateSetId":null,"preferredCurrencyId":{"id":48},"freeTags":[]}],"windowSize":5,"currentPage":1}
Several problems here. First you should use double quotes, not single. Double quotes allows the expansion of variable. Second, I am assuming that you don't want the + as part of the data - they are not a string concatenation operator in bash (there is no need for one).
str="[$#]"
result=$(underscore pluck --data "$str" name)
echo $result
I used the $( ) notation rather than back-ticks ( `` ). Back-ticks are considered deprecated and difficult to read.
I have replaced uppercase variable names with lower case. This is because there are many uppercase variable names used by the shell so using uppercase risks a name collision. Best to use lower or mixed-case variable names.
I have also removed the superfluous semi-colons ;. They are not doing any harm, but they are of no use either. In bash a semi-colon is a statement separator, whereas in languages like C and Java they are a statement terminator.
Edit: double-quoted $#
I figured it out...
#!/bin/bash
name=$(underscore select .name --outfmt text < get.crm.client.acme.json)
echo $name
I am navigating a Java-based CLI menu on a remote machine with expect inside a bash script and I am trying to extract something from the output without leaving the expect session.
Expect command in my script is:
expect -c "
spawn ssh user#host
expect \"#\"
send \"java cli menu command here\r\"
expect \"java cli prompt\"
send \"java menu command\"
"
###I want to extract a specific string from the above output###
Expect output is:
Id Name
-------------------
abcd 12 John Smith
I want to extract abcd 12 from the above output into another expect variable for further use within the expect script. So that's the 3rd line, first field by using a double-space delimiter. The awk equivalent would be: awk -F ' ' 'NR==3 {$1}'
The big issue is that the environment through which I am navigating with Expect is, as I stated above, a Java CLI based menu so I can't just use awk or anything else that would be available from a bash shell.
Getting out from the Java menu, processing the output and then getting in again is not an option as the login process lasts for 15 seconds so I need to remain inside and extract what I need from the output using expect internal commands only.
You can use regexp in expect itself directly with the use of -re flag. Thanks to Donal on pointing out the single quote and double quote issues. I have given solution using both ways.
I have created a file with the content as follows,
Id Name
-------------------
abcd 12 John Smith
This is nothing but your java program's console output. I have tested this in my system with this. i.e. I just simulated your program's output with cat. You just replace the cat code with your program commands. Simple. :)
Double Quotes :
#!/bin/bash
expect -c "
spawn ssh user#domain
expect \"password\"
send \"mypassword\r\"
expect {\\\$} { puts matched_literal_dollar_sign}
send \"cat input_file\r\"; # Replace this code with your java program commands
expect -re {-\r\n(.*?)\s\s}
set output \$expect_out(1,string)
#puts \$expect_out(1,string)
puts \"Result : \$output\"
"
Single Quotes :
#!/bin/bash
expect -c '
spawn ssh user#domain
expect "password"
send "mypasswordhere\r"
expect "\\\$" { puts matched_literal_dollar_sign}
send "cat input_file\r"; # Replace this code with your java program commands
expect -re {-\r\n(.*?)\s\s}
set output $expect_out(1,string)
#puts $expect_out(1,string)
puts "Result : $output"
'
As you can see, I have used {-\r\n(.*?)\s\s}. Here the braces prevent any variable substitutions. In your output, we have a 2nd line with full of hyphens. Then a newline. Then your 3rd line content. Let's decode the regex used.
-\r\n is to match one literal hyphen and a new line together. This will match the last hyphen in the 2nd line and the newline which in turn make it to 3rd line now. So, .*? will match the required output (i.e. abcd 12) till it encounters double space which is matched by \s\s.
You might be wondering why I need parenthesis which is used to get the sub-match patterns.
In general, expect will save the expect's whole match string in expect_out(0,string) and buffer all the matched/unmatched input to expect_out(buffer). Each sub match will be saved in subsequent numbering of string such as expect_out(1,string), expect_out(2,string) and so on.
As Donal pointed out, it is better to use single quote's approach since it looks less messy. :)
It is not required to escape the \r with the backslash in case of double quotes.
Update :
I have changed the regexp from -\r\n(\w+\s+\w+)\s\s to -\r\n(.*?)\s\s.
With this way - your requirement - such as match any number of letters and single spaces until you encounter first occurrence of double spaces in the output
Now, let's come to your question. You have mentioned that you have tried -\r\n(\w+)\s\s. But, there is a problem here with \w+. Remember \w+ will not match space character. Your output has some spaces in it till double spaces.
The use of regexp will matter based on your requirements on the input string which is going to get matched. You can customize the regular expressions based on your needs.
Update version 2 :
What is the significance of .*?. If you ask separately, I am going to repeat what you commented. In regular expressions, * is a greedy operator and ? is our life saver. Let us consider the string as
Stackoverflow is already overflowing with number of users.
Now, see the effect of the regular expression .*flow as below.
* matches any number of characters. More precisely, it matches the longest string possible while still allowing the pattern itself to match. So, due to this, .* in the pattern matched the characters Stackoverflow is already over and flow in pattern matched the text flow in the string.
Now, in order to prevent the .* to match only up to the first occurrence of the string flow, we are adding the ? to it. It will help the pattern to behave as non-greedy manner.
Now, again coming back to your question. If we have used .*\s\s, then it will match the whole line since it is trying to match as much as possible. This is common behavior of regular expressions.
Update version 3:
Have your code in the following way.
x=$(expect -c "
spawn ssh user#host
expect \"password\"
send \"password\r\"
expect {\\\$} { puts matched_literal_dollar_sign}
send \"cat input\r\"
expect -re {-\r\n(.*?)\s\s}
if {![info exists expect_out(1,string)]} {
puts \"Match did not happen :(\"
exit 1
}
set output \$expect_out(1,string)
#puts \$expect_out(1,string)
puts \"Result : \$output\"
")
y=$?
# $x now contains the output from the 'expect' command, and $y contains the
# exit status
echo $x
echo $y;
If the flow happened properly, then exit code will have value as 0. Else, it will have 1. With this way, you can check the return value in bash script.
Have a look at here to know about the info exists command.