I have the following directory structure in one of my projects:
/web
/bower_components
/bootstrap
/jquery
/typeahead.js
/views
/index.jade
I'm trying to use these components from my index.jade file in the following way:
link(src='../bower_components/bootstrap/dist/css/bootstrap.min.css', rel='stylesheet')
script(src='../bower_components/jquery/dist/jquery.min.js')
script(src='../bower_components/typeahead.js/dist/typeahead.bundle.min.js')
but it doesn't work (there's no bootstrap-related styles on this page and jQuery code doesn't work).
Why? What am I doing wrong? How can I fix it?
Should I refer to these components in a different way?
Should I place bower_components directory in another folder?
You probably need to change the file path to be relative to where you are compiling the jade file to.
For example, if you render the jade like this:
/web
/index.html
the source paths would be
link(src='bower_components/bootstrap/dist/css/bootstrap.min.css', rel='stylesheet')
script(src='bower_components/jquery/dist/jquery.min.js')
script(src='bower_components/typeahead.js/dist/typeahead.bundle.min.js')
Related
So I'm trying to make a website where I'm able to just drag and drop folders containing HTML, js, and CSS files into a "library" directory and have those files served on my flask app. There's one big problem I haven't been able to get over. The problem is that all of these HTML files link to their respective js files using relative paths. For example:
<script src="js/keyboard_input_manager.js"></script>
and these files are 2 folders deep into my (combined) templates/static folder. I could fix this by doing
<script src="library/2048/js/keyboard_input_manager.js"></script>
but that is very tedious, especially when working with so many files.
TLDR: If I were to run my flask app and load a template, it wouldn't load any js or CSS because of relative paths.
Is there any way to go about this without individually changing each path to relate to the templates folder?
Example of my current filesystem:
-FLASK PROJECT
|->library (templates/static folder)
| |->2048
| |->index.html, index.css, index.js
| |->Game2
| |->index.html, index.css, index.js
|->main.py
What I've Tried:
Using Blueprint but I can't create a whole blueprint for every file?
Messing with paths of template folder and static folder
Flask Noob - Please let me know if I'm leaving out any helpful information :) Thanks
I am using Gulp to build a deployable build for an application. I would like to cache-bust all of my .js and .css files so that when a new build is deployed, users will need to retrieve the new "cache-busted" files. For example, if an app.js file is stored in the browser's cache, I would like to have the ref to app.js in my index.html look something like this:
<script src="app/js/app.js?v=1.2"></script>
and so on for all relevant files I would like to cache bust.
Some other questions related to this problem I have:
1) How can I tell that these files are actually getting cache busted properly?
2) Is there a better way to approach this?
Here is what I am trying so far:
//compile index.html, app, vendor
gulp.task('compile-dist', function(){
var revAll = new RevAll();
gulp.src('../../backend-angular-seed/app/**')
.pipe(gulp.dest('../dist/app'));
gulp.src('../../backend-angular-seed/vendor/**')
.pipe(gulp.dest('../dist/vendor'));
gulp.src('../index.html')
.pipe(gulp.dest('../dist/'));
})
This code takes all of the code from my app/ directory (which is a result of my compiled code from my master/ directory) and builds a dist/ directory with all of my js, css, and vendor files.
After this build I have a dist/ directory that looks like this:
/dist
/css
|_app.css
/img
/js
|_ app.js
|_ base.js
/vendor
index.html
I have tried using a few different methods on modifying this dist directory to effectively have it bust the cache. I tried using gulp-cachebust as well as gulp-rev-all, but I believe both of these tools are a bit overkill for what I am trying to do.
Ideally, through Gulp, I would like to go into the index.html file made from the Gulp build, and modify all of my script tags to append the query string of "?v=1.0" on the end of all files I would like to cache bust per deploy build.
Any answers/suggestions would be greatly appreciated. Thanks so much!!!
If appending query string is all you want then i recommend using gulp-cache-bust.
var cachebust = require('gulp-cache-bust');
gulp.src('./dist/index.html')
.pipe(cachebust({
type: 'timestamp'
}))
.pipe(gulp.dest('./dist'));
Here is the turorial for it: https://www.npmjs.com/package/gulp-cache-bust
I have created Dynamic web project and it has .html, .css and .js files. I group these file in respective folders like .js file in javascripts folder and .html file in views folder but i don't able to access these file in project. I used eclipse IDE for this. Is there need to configure path for these folder?
You need to put the JSP file in /index.jsp instead of in /WEB-INF/jsp/index.jsp. This way the whole servlet is superflous by the way.
WebContent
|-- META-INF
|-- WEB-INF
| -- web.xml
-- index.jsp
If you're absolutely positive that you need to invoke a servlet this strange way, then you should map it on an URL pattern of /index.jsp instead of /index. You only need to change it to get the request dispatcher from request instead of from config and get rid of the whole init() method.
These are not Java source files, so it makes no sense to configure them as such. By default in a Dynamic Web Project you only see the src folder under Java Resources. Other folders will be listed at the bottom of the tree. This is by design.
Or if you meant, that you do not see them when you move into the folder by an external file manager: press F5 on the project.
Its based on from which file you are trying to access those files.
If it is in the same folder where your working project file is, then you can use just the file name. no need of path.
If it is in the another folder which is under the same parent folder of your working project file then you can use location like in the following /javascript/sample.js
In your example if you are trying to access your js file from your html file you can use the following location
../javascript/sample.js
the prefix../ will go to the parent folder of the file(Folder upward journey)
I got answer to my question...
Now my directory structure is
WebContent
--javascripts
--stylesheets
--viwes
--META-INF
--WEB-INF
Note: view contain html files
To change path of my welcome html file i made bit change in web.xml present in WEB-INFfolder.
<welcome-file-list>
<welcome-file>/views/welcome.html</welcome-file>
</welcome-file-list>
I searched over the internet but I could not find an answer to my question. I am just trying to figure out a clean way to structure my CSS and JS files inside my project. Let us say for example I have a CSS folder and I have a custom my.css file and also I have a Scripts folder and I have inside it a myscript.js. I understand that my.css will go in the CSS folder and myscript.js will go under the Scripts folder in this folder setup :
root
css
my.css
Scripts
myscript.js
My question is, if I want to use a jQuery plugin like jstree for example. This library require me to add one js and one css file. Should I keep these files under the Scripts and CSS folder ?
Plan A
root
css
my.css
jstree.css
Scripts
myscript.js
jstree.js
Or should I separate them into a different folder for cleaner structure like this
PLan B
root
css
jstreeFolder ---> jstree.css
my.css
Scripts
jstreFolder ---> jstree.js
myscript.js
Is this structure acceptable? Any standard ways for achieving this ?
Any help is appreciated.
Root
-- styles
|-- your_file.css
-- scripts
|-- your_file.js
-- libs (or plugins)
|-- jquery
|-- jquery-ui
...
I think this structure will be nicer and all the third party libraries (no matter css or javascript libs) that you've chosen should be located under libs directory. It will make maintenance be easier and clear.
For mainentance reasons i prefer another approach. The library jstreee organise the javascript files, css-files and images in a certain way.
/libs/
/jstree/ // <-- folder
/themes/ <-- folder
/default/ <-- folder
style.css
32px.png
jstree.js
jstree.search.js
/other-plugin/
I put everything under libs and in a folder with the name of the library. This way the external dependency of a library is clear and the internal path structure (css files may point to images) of the external library is untouched.
I'm wondering if it's possible to configure the RequireJS optimizer to fit with our current project structure.
The site directory is structured as below...
root
project1
scripts
main.js
main.min.js
project2
scripts
main.js
main.min.js
project3
scripts
main.js
main.min.js
I was wondering if it's possible to have a "main" file sitting at the root level that will optimize all the child project main.js files and place them within their respective directories. I noticed the multi-page optimizer example on the Requirejs homepage but i'm unsure how to configure that to work for my use case.
Is it just one main.js file per project? I think when I used require js modules, it optimized with this behavior, but in a separate build/distribution directory
see
http://www.bennadel.com/blog/2404-Compiling-Optimizing-A-Subset-Of-A-RequireJS-Application.htm