Multiline curl command

Question

I am trying to modify a curl request that was captured with Google Chrome Dev Tools.

Here is what the command looks like

curl "http://WEBSITE" -H "Host: WEBSITE" -H "Accept: text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8" -H "Accept-Language: en-US,en;q=0.5" --compressed -H "Content-Type: multipart/form-data; boundary=---------------------------1184875127259" --data-binary "-----------------------------1184875127259"^  "Content-Disposition: form-data; name=""FORM1"""^  "FORM1DATA"^ "-----------------------------1184875127259"^  "Content-Disposition: form-data; name=""FORM2"""^  "FORM2DATA"^ "-----------------------------1184875127259"^  "Content-Disposition: form-data; name=""FORM3"""^  "FORM3DATA"^ "-----------------------------1184875127259"^  "Content-Disposition: form-data; name=""embed"""^  "true"^ "---------------------------1184875127259--"^ "" 

Form# is the name of the form and Form#Data is the data I submitted in the forms.

How would I make this be a single line curl request I can just copy into my command line and have it do the same thing that my browser did?

Answer 1

For Linux and MacOS: Use the \ escape character:

curl "http://WEBSITE" -H "Host: WEBSITE" \ -H "Accept: text/html,application/xhtml+xml \ ,application/xml;q=0.9,*/*;q=0.8" 

For Windows: Use the ^ escape character:

curl "http://WEBSITE" -H "Host: WEBSITE" ^ -H "Accept: text/html,application/xhtml+xml ^ ,application/xml;q=0.9,*/*;q=0.8" 

Answer 2

NOTE: watch out for the tendency to indent on multiple line commands, as it will embed spaces and screw up the curl command. the sed command replaces embedded spaces within the variables with the %20 string so that spaces can be used embedded in the strings you pass as variables

messageout="The rain in Spain stays mainly in the plains" summaryout="This is a test record" alertnameout="Test Alert"   curl -v -silent request POST "URL.com?\ summary=`echo $summaryout | sed -e 's/ /%20/g'`&\ alertname=`echo $alertnameout | sed -e 's/ /%20/g'`&\ message=`echo $messageout | sed -e 's/ /%20/g'`"