How would I compile this use Enclose.js?
I have tried and got this error
x file "was not included into executable at compilation stage. Please recompile adding it as asset or script."
I then tried to add a config file as the docks say to do but have gotten nowhere with that and now I do not know what to do so I have come hear to see if someone can help.
var fs = require('./graceful-fs');
saveToFile();
function saveToFile() {
var date = new Date();
fs.writeFileSync(__dirname + "/Output.txt", date + "\r\n");
} //save dater to file
I had to remove __dirname and use ./ in it's place.
var fs = require('./graceful-fs');
saveToFile();
function saveToFile() {
var date = new Date();
fs.writeFileSync("./Output.txt", date + "\r\n");
} //save dater to file
Related
Using gulp 3.9.1
I am attempting to return a bunch of files and perform a task that requires a var to be passed between two pipes.
I'm using node uuid to create a v3 UUID for each file path to
ultimately end up with a uuid for each page. I'm grabbing the file path with gulp-print.
I want to store that uuid value as a var. In the next pipe Im using
gulp-inject-string to write it into the page during the build.
Help: Either I need help getting the file path inside the gulp-inject-string pipe or I need to pass the var between the two different pipes. If I globally set a var with a default value outside the src it gets passed easily to the pipe(inject).
Super simplified code below:
// test code
var gulp = require('gulp');
var print = require('gulp-print');
var inject = require('gulp-inject-string');
var reload = browserSync.reload;
const uuidv3 = require('uuid/v3');
var uuid;
gulp.task('uuid', function() {
return gulp.src('**/*.html'])
// create uuid
.pipe(print(function(filepath) {
uuid = uuidv3(filepath, uuidv3.URL);
return "compiled: " + filepath + ' uuid: ' + uuid;
}))
// need to to add UUIDv3 to each page
.pipe(inject.before('</head>', '<meta name="dc.identifier" content="' + uuid + '">'))
.pipe(gulp.dest('/prod/./'))
.pipe(reload({ stream: true }));
});
It's worth noting that I need a cross platform way to get the file path starting in the root of the project and including forward slashes. The gulp(print) does this perfectly starting at the root of the project and ignoring anything upstream from that point. The format of the path is important because it's one half of the equation in creating the uuid and the uuid's must match on Mac or PC platforms.
examples:
/index.html
/dir1/file.html
/dir1/dir2/dir3/file.html
var gulp = require('gulp');
var print = require('gulp-print');
var inject = require('gulp-inject-string');
const uuidv3 = require('uuid/v3');
var tap = require('gulp-tap');
// you can declare here
var uuid;
gulp.task('pages', function() {
// or you can declare here
var uuid;
return gulp.src('**/*.html')
// bunch of stuff happens here involving templating/minifying
// create uuid
.pipe(print(function(filepath) {
// then set it here and use it further below
// it will be available
uuid = uuidv3(filepath, uuidv3.URL);
return "compiled: " + filepath + ' uuid: ' + uuid;
}))
// need to to add UUIDv3 to each page
//.pipe(inject.before('</head>', '<meta name="dc.identifier" content="' + uuid + '">\n'))
.pipe(tap(function(file, t) {
return t.through(inject.before('</head>', '<meta name="dc.identifier" content="' + uuid + '">\n');
})
.pipe(gulp.dest('/prod/./'))
.pipe(reload({stream:true}));
});
You are just creating a variable at a higher scope that you can set and refer to later. If you need a bunch of them create an array with filepath as an index. But I would try it first as just a simple value.
I solved the problem. It was an amateur mistake. I returned the statement where the var was set so the var was essentially killed. Updated code that allows the var to pass through the pipes.
var gulp = require('gulp');
var print = require('gulp-print');
var replace = require('gulp-replace');
const uuidv3 = require('uuid/v3');
var uuid;
gulp.task('build', function() {
return gulp.src('**/*.html')
// get a cross-platform filepath and create a uuid
.pipe(print(function(filepath) {
uuid = uuidv3(filepath, uuidv3.URL);
}))
// inject uuid
.pipe(replace('dc.identifier" content=""', function() {
return 'dc.identifier" content="' + uuid + '"';
}))
.pipe(gulp.dest('/prod/./'));
});
The var uuid passes through the pipes just fine now. This code creates a UUID based on a cross-platform file path and injects it into an empty dc.identifier meta tag.
I am writing a cordova plugin with a node hook to run after_prepare .
This is for Android only.
From within this hook I need to get the cordova package name, so I can copy a file to the src/com/example/myproject folder (if the package is com.example.myproject).
If I know the package name I can make this path. I have it working hardcoded now but I need this to work with two different package names.
Is there a way to get the package name from within this code in the plugin hook?
module.exports = function(ctx){
var fs = ctx.requireCordovaModule('fs');
var path = ctx.requireCordovaModule('path');
var deferral = ctx.requireCordovaModule('q').defer();
//get package name here
//do other stuff
}
I have done a lot of research but have not been able to find this.
Thanks.
It doesn't look like it is available off of the context object, but you could try to parse the config.xml.
module.exports = function(context) {
var fs = require('fs');
var path = require('path');
var config_xml = path.join(context.opts.projectRoot, 'config.xml');
var et = context.requireCordovaModule('elementtree');
var data = fs.readFileSync(config_xml).toString();
var etree = et.parse(data);
console.log(etree.getroot().attrib.id);
};
The local-webserver plugin uses a similar strategy for reading config properties.
Here my compilation from different answers that works in 2021.
I use it to update some parameters in Xcode project for plugins compilation.
You can see that I am getting here app id and name from config.xml
And you can add it to after_prepare hook:
<hook src="scripts/addBuildSettingsToXcode.js" type="after_prepare" />
#!/usr/bin/env node
let fs = require('fs');
let xcode = require('xcode');
let path = require('path');
let et = require('elementtree');
module.exports = function (context) {
//console.log(context);
function addBuildPropertyToDebugAndRelease(prop, value) {
console.log('Xcode Adding ' + prop + '=' + value);
myProj.addBuildProperty(prop, value, 'Debug');
myProj.addBuildProperty(prop, value, 'Release');
}
function updateBuildPropertyToDebugAndRelease(prop, value) {
console.log('Xcode Updating ' + prop + '=' + value );
myProj.updateBuildProperty(prop, value, 'Debug');
myProj.updateBuildProperty(prop, value, 'Release');
}
// Getting app id and name from config.xml
let config_xml = path.join(context.opts.projectRoot, 'config.xml');
let data = fs.readFileSync(config_xml).toString();
let etree = et.parse(data);
let appId = etree.getroot().attrib.id ;
let appName = etree.getroot().find('name')['text'];
// Building project path
let projectPath = 'platforms/ios/' + appName + '.xcodeproj/project.pbxproj';
// Opening Xcode project and parsing it
myProj = xcode.project(projectPath);
myProj = myProj.parseSync();
// Common properties
addBuildPropertyToDebugAndRelease('DEVELOPMENT_TEAM', 'CGXXXXXXX');
addBuildPropertyToDebugAndRelease('CODE_SIGN_IDENTITY', '"Apple Development"');
// Compilation properties
addBuildPropertyToDebugAndRelease('ALWAYS_EMBED_SWIFT_STANDARD_LIBRARIES', 'YES');
// Save project file
fs.writeFileSync(projectPath, myProj.writeSync());
};
I would like to know how exactly can I manipulate the output of my Gulp plugin so, for example, no matter how many files are passed to the plugin, it will wrap the output with a string. Currently I cannot know when does the last file is done.
The super simplified example below will iterate on 3 files and will create a new file named output.js and in it there will be three times the string xxx (xxxxxxxxx).
I would like the plugin itself to wrap the contents so the output will
be: +xxxxxxxxx+.
How can I do this?
Thanks!
Gulpfile
var gulp = require('gulp');
var concat = require('gulp-concat');
var foo = require('./index');
gulp.task('default', function() {
gulp.src([a.html, b.html, c.html])
.pipe(foo())
.pipe(concat('output.js'))
.pipe(gulp.dest('./test/output'))
});
The most basic gulp plugin (index.js):
var through2 = require('through2'),
gutil = require('gulp-util');
var PLUGIN_NAME = 'foo';
module.exports = function( options ){
// through2.obj(fn) is a convenience wrapper around
// through2({ objectMode: true }, fn)
return through2.obj(function( file, enc, callback ){
file.contents = new Buffer( 'xxx' );
this.push(file);
callback();
});
}
I understand the files are currently simply returned modified, but what I don't understand is how to append text and return the concatenated result that I want, while keeping it OK with Gulp working standards.
The "real" plugin should actually wrap the files results with:
var foo = { FILES_CONTENT }
where FILES_CONTENT will actually be a a concatenated string of all the files:
"file_name" : "file_content",
"file_name" : "file_content",
...
I would make the following changes to your gulpfile.js:
var gulp = require('gulp');
var foo = require('./index.js');
gulp.task('default', function() {
return gulp.src(['a.html', 'b.html', 'c.html'])
.pipe(foo({fileName:'output.js', varName:'bar'}))
.pipe(gulp.dest('./test/output'))
});
Since your foo() plugin itself will concatenate all the files, there's no need to use gulp-concat at all. Instead your plugin should accept an option fileName that provides the name of the generated file. I've also added another option varName that will provide the name of the var in the output file.
I'll assume that a.html, b.html and c.html are simple HTML files, something like this:
<h1 class="header">a</h1>
As you've already realized you need to concat all the files in the plugin itself. That's not really difficult however and doesn't require a lot of code. Here's a index.js which does exactly that:
var through2 = require('through2'),
gutil = require('gulp-util'),
path = require('path'),
File = require('vinyl');
var PLUGIN_NAME = 'foo';
module.exports = function(options) {
var files = { };
var outputFile = null;
return through2.obj(function(file, enc, callback){
outputFile = outputFile || file;
var filePath = path.relative(file.base, file.path);
files[filePath] = file.contents.toString();
callback();
}, function(callback) {
outputFile = outputFile ? outputFile.clone() : new File();
outputFile.path = path.resolve(outputFile.base, options.fileName);
outputFile.contents = new Buffer(
'var ' + options.varName + ' = ' +
JSON.stringify(files, null, 2) + ';'
);
this.push(outputFile);
callback();
});
}
Since you want to output a key/value mapping from file names to file contents our transformFunction just stores both of those things in a regular JavaScript object files. None of the input files themselves are emitted. Their names and contents are just stored until we have all of them.
The only tricky part is making sure that we respect the .base property of each file as is customary for gulp plugins. This allows the user to provide a custom base folder using the base option in gulp.src().
Once all files have been processed through2 calls the flushFunction. In there we create our output file with the provided fileName (once again making sure we respect the .base property).
Creating the output file contents is then just a matter of serializing our files object using JSON.stringify() (which automatically takes care of any escaping that has to be done).
The resulting ./test/output/output.js will then look like this:
var bar = {
"a.html": "<h1 class=\"header\">a</h1>\n",
"b.html": "<h1 class=\"header\">b</h1>\n",
"c.html": "<h1 class=\"header\">c</h1>\n"
};
You should use the gulp pipeline technique (standard).
This means that you can use the gulp-insert package in order
to add the string xxx.
var insert = require('gulp-insert');
.pipe(insert.append('xxx')); // Appends 'xxx' to the contents of every file
You can also prepend, append and wrap with this package and it support of course the gulp standards.
So the full example will be:
var gulp = require('gulp');
var concat = require('gulp-concat');
var foo = require('./index');
var insert = require('gulp-insert');
gulp.task('default', function() {
gulp.src([a.html, b.html, c.html])
.pipe(foo()
.pipe(insert.append('xxx'))
.pipe(concat('output.js'))
.pipe(gulp.dest('./test/output'))
});
I'd like to create an archive with archiver and put some files in it. Client's side, an user will click on a button and it'll generate the archive (server's side).
I'm using Express.js, this is my server side code where the archive will be generated. I did something like this :
app.get('/export/siteoccupancy', function(req,res){
if(_.isEmpty(req.query)){
res.status(404).send('requires a start and end date');
}else{
//getting paramas
var sDate = req.query.startDate;
var eDate = req.query.endDate;
}
var fs = require('fs');
var archiver = require('archiver');
var archive = archiver('zip');
archive.on('err',function(err){
res.status(500).send({error : err.message});
});
res.on('close',function(){
console.log('Archive size : %d b',archive.pointer());
return res.status(200).send('OK').end();
});
res.attachment('data-export.zip');
archive.pipe(res);
var stream = fs.createWriteStream("data-report.txt')");
stream.once('open',function(fd) {
stream.write('test1');
stream.write('\n test2');
stream.write('\n test3');
stream.end();
});
archive.append(stream);
archive.finalize();
});
This is totally new for me and I'd like to understand why the console tells me the stream file is empty ?
Error: append: entry name must be a non-empty string value
at Archiver.append
Regards
you can only append a readStream, because an archive can only take data from readstreams. You can use the method named archive.append, and you should pass as a second argument with the name property to name the file. Like so :
archive.append(myReadStream,{ name : 'myTest.txt'});
i have a small problem, when i try to copy one file from my tmp dir to my ftp dir the writen file is empty. I have no error, i don't understand what i'm doing wrong
var ftpPath = "/var/www/ftp/",
zipPath = "/var/www/tmp/",
file = "test";
fs.createReadStream(zipPath + file).pipe(fs.createWriteStream(ftpPath + file));
My test file contain loremipsum sample.
If you have any solution, i take it, this is the only line that bug in my app :(
First, make sure that the file /var/www/tmp/test exists, is a file, and has the right permissions for the user you start the script with.
Second, make sure that /var/www/ftp/ has writing permissions.
Then the following code should work :
var readerStream = fs.createReadStream('/var/www/tmp/test');
var writerStream = fs.createWriteStream('/var/www/ftp/test');
readerStream.pipe(writerStream);
Edit :
try debugging using this snippet :
var data;
var readerStream = fs.createReadStream('/var/www/tmp/test');
readerStream.on('data', function(data) {
data += data;
});
readerStream.on('end', function() {
console.log(data);
});