How to extract link from curl response in bat file?

Question

cmd/bat

I make a request via curl from a bat file. From the request response body

{
   "href":"https://uploader9m.disk.yandex.net:443/upload-target/20160623T092605.135.utd.eb5dbfed7bwwb6imitb6xku41-k9m.5744252",
   "method":"PUT",
   "templated":false
}

I want to get the value of "href", i.e. the string

"https://uploader9m.disk.yandex.net:443/upload-target/20160623T092605.135.utd.eb5dbfed7bwwb6imitb6xku41-k9m.5744252"

How can this be done simply and correctly?
Is there an option to have curl immediately return the "href" value?

Answer 1

I used the jq console utility , as a result, getting href from the response for my case looks like this:

for /f "delims=" %%i in ('curl -H %H% %link_to_get_upload%') do set response=%%i
echo %response% | C:\jq.exe ".href" > %file%
set /p upload_url= < %file%