When I try to do build with steal/buildjs of Javascript MVC, envJS silently fail to compress and check JavaScript files for errors.
I read documentation for envJS and found there that if in some case you try to build empty .js files envJS will fail.
Obviously it is in my case, and later I founded some of empty .js files.
I need a way to add checking for empty .js and .ejs files before build.
What would be the most efficient way to do check for empty .js and .ejs when running steal/buildjs?
I would use find:
find . -type f \( -name *.js -o -name *.ejs \) -size 0 -print
Change -print to -delete when you're sure.
Related
I need to include all *.js files and exclude only *.spec.js files under a directory.
I tried the following ways, it's either including all js or excluding all js files
'built/code/js/libs/preact/**/!(*.spec).js') but it fails with the message syntax error near unexpected token `('.
If I do this way - 'built/code/js/libs/preact/**/!*.spec.js') it excludes all js files
If I add in 2 different lines like below it excludes all js files
('built/code/js/libs/preact/**/*.js' +
'built/code/js/libs/preact/**/!*.spec.js')
Please let me know if there's a way to exclude spec files and include all other js files.
I have multiple minified files in multiple folder in a single folder for minified js.
I wish to create single js file from all of minified js.
Now I am using type command to concatenate all files like
type .\\v2\\dist\\js\\*.js >> .\\build\\a.min.js && type .\\v2\\dist\\js\\config\\*.js >> .\\build\\a.min.js && ...
Like wise I need to append all recursive folders.
Is there any clean way to do it.
Please don't suggest using Gulp or Grunt as we are already in process of removing them & using Webpack. Any code using webpack or using npm or simple command line is welcome.
If on Windows, you can use command line like below (refer to #dbenham's answer):
for /r "c:\javascripts" %F in (*.js) do #type "%F" >>concatenated.js
If on Linux or Mac, you can use find:
find ./v2/dist/js/ -name '*.js' -exec cat {} \; > ./build/a.min.js
Hello evryone,
I'm trying to find a way to make the r.js not to copy all the files form my /lib/ directory except for (for example) jquery.js and require.js.
I'm using fileExclusionRegExp option to exclude all *.js files except for the above mentioned.
fileExclusionRegExp: '\/lib\/(?!jquery|require).*\.js'
But after the optimization, I can still see that other files have been copied over too.
Is there anything I'm doing wrong ? Or is the regex incorrect?
Thanks in advance
The problem you've run into is that you are trying to make fileExclusionRegExp match the entire path of a file, but r.js uses it only to test against the base name of files. You can infer this from the description of the option and its default value. The description says:
//When the optimizer copies files from the source location to the
//destination directory, it will skip directories and files that start
//with a ".".
The default value is:
/^\./
If this were to be tested against a full path, it would not be able to exclude files that start with a period if they are in a subdirectory. We also find confirmation of this behavior in this issue report.
If you have only one file named require.js and one file named jquery.js, then you can get by with this regexp:
fileExclusionRegExp: /^(?!jquery|require).*\.js$/
Otherwise, fileExclusionRegExp is not going to do it, and you should use a build step to clean your directory, as suggested by the author of RequireJS in the issue report I mention above.
we have a problem at work, we are using require js but our folder structure is a bit different,
we have the following:
--js folder
--Folder
---some base js files
-Folder
---main
--src
---require.js
--- require JS modules
--plugin js files
--more js files
We would like to minify all these JS files to a SINGLe js file for production as such
---js folder
--min-all.js
Is this possible?
if so how? ..
Any help would be appreciated!
Thanks!
I just thought I would clarify that the other Folders contain standard non modular javascript files, they can be a mix of plugins or simple javascript helpers.
The short answer is: yes, RequireJS can do this.
Basically, you will need to create one JS file that requires all of the resources that you want minified. Then you will point the optimizer at that file and it will mash them all together.
require(["one", "../another/two", "folder/three", "folder/inner/four" ... ]);
If that file was called myfile.js, you would run the optimizer with similar parameters to this:
node r.js -o name=myfile out=optimized.js
If you have libraries or other files that you do not want included into the final optimized file, you would use the excludeShallow flag. e.g.
node r.js -o name=myfile out=optimized.js excludeShallow=jquery.min
There are more options so you should check out their optimization documentation if you haven't yet.
This is probably a quick question. I'd like to run the YUI compressor so that, instead of overwriting all of the javascript files in the input directory, it dumps the output files into a subdirectory called min. I would like to do something like
java -jar yuicompressor-2.4.7.jar -o *.js .\min\*.js c:\MyJavascriptDirectory\*.*
but instead it just overwrites the existing files with the minified files. Does anyone know the correct syntax to accomplish this?
Thanks!
I'm not sure of the correct syntax to do that, but someone has made a modification so that you can set up your output directory as follows
java -jar yuicompressor.jar --output-dir /some/folder/for/compressed/js *js
This allows you to have a source folder and a compressed folder. As opposed to having source and a bunch of -min.js files in the same folder.
You can download the mod at the bottom of the page here http://yuilibrary.com/projects/yuicompressor/ticket/2528131
java -jar yuicompressor.jar -o '.js$:-min.js' *.js will minify all .js files and save them as -min.js then you could just move all those files: mkdir min; mv *-min.js min/.
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