Compress heredoc declaration to one line in bash?

问题

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.

回答1:

why not use jq? It's pretty good at managing string interpolation and it lints your structure.

$ echo '{}' >> foo.json
$ local 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"}

回答2:

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}\"}`

回答3:

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.

回答4:

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}\" }"

回答5:

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.

回答6:

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"
}

来源：https://stackoverflow.com/questions/49951601/compress-heredoc-declaration-to-one-line-in-bash

标签

json

bash

heredoc