Replacing code between two JavaScript comments

Question

I need to set a value of JavaScript variable from bash. Variable lives in index.html file and I'd like to use unix SED command to do that. Inside index.html I have it lake this:

<script>
/*bash_var*/ var foo = 1;  /*end_bash_var*/
</script>

I tried to do it like this:

sed -i -e 's%/*bash_var*/(.*)/*end_bash_var*//'"$ var foo = 0; /g" index.html

and few more variations of this command, but I alway get some error.

Thanks for any help.

Update

Expected output:

<script>
/*bash_var*/ var foo = 0;  /*end_bash_var*/
</script>

Why not use a JavaScript parser & code generator to do this for you? Use acorn and escodegen. — Aadit M Shah
– Aadit M Shah, Commented Nov 5, 2014 at 15:15
errors like 'unterminated substitute pattern' or 'parentheses not balanced' — Ladislav M
– Ladislav M, Commented Nov 5, 2014 at 15:22

pfnuesel · Accepted Answer · 2014-11-05 15:37:05Z

1

You need to escape the *. Moreover your ' are wrong and there's no need for a %.

sed 's#\(/\*bash_var\*/\).*\(/\*end_bash_var\*/\)#\1 var foo = 0; \2#'

Note that I use # as delimiters, rather than /. This way I can use / in my expressions without escaping them. This makes the sed command a little bit more readable. However, this is still tedious. Can you rewrite your index.htm into something like this?

<script>
//BASHVAR
var foo = 1;
</script>

Then you can use

sed '/BASHVAR/{n; s/.*/var foo = 0;/;}'

edited Nov 5, 2014 at 15:37

answered Nov 5, 2014 at 15:15

pfnuesel

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5 Comments

Ladislav M Over a year ago

Sure, I can rewrite it like that, but BASHVAR string need to be under inside inline or block comments.

Ladislav M Over a year ago

I'm getting "/BASHVAR/{n; s/.*/var f ...": bad flag in substitute command: '}'

pfnuesel Over a year ago

What version of sed are you using? It works for me on GNU sed 4.2.2. Can you try again?

Ladislav M Over a year ago

I'm not sure, how can I display the version of sed. But I'm runnig OS X 10.9.

pfnuesel Over a year ago

Okay, I'm not much help, since I don't have access to your version of sed. Did you try the last command again? I changed it slightly.

NeronLeVelu · Accepted Answer · 2014-11-05 15:44:06Z

1

sed 's#\(/\*bash_var\*/ *var foo = \).*\(; */\*end_bash_var\*/\)#\10\2#'

If it is just the value of the foo in this context

answered Nov 5, 2014 at 15:44

NeronLeVelu

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