Re: Regular Expression Question

[Sean Hubbell]

Tim Chase wrote:

print("This is 1.\n");print("This is 2.");
print("This is 3.\n");

I would like the above to be formated like (which is just adding in the newline a print that may not have it, but only if it does not have it):

print("This is 1.\n");print("This is 2.");
print("This is 3.\n");

I presume you mean that second section to read

print("This is 1.\n");print("This is 2.\n");
print("This is 3.\n");

adding in that "\n" after "2."

On the assumption that you don't have quotation-marks escaped in your strings such as

    print ("She said \"hello\" to me");

you should be able to do something like

    %s/print\s*(\s*"[^"]*\(\\n\)\@<!\ze"/&\\n/g

which should catch most of the non-pathological cases. If quotes can span lines, such as

    print ("this is a
        really long string")

then you can change

    [^"]*

to

    \_[^"]*


HTH,

-tim

Thanks Tim.

Sean