You don&#39;t need to use `fs.close` after `fs.readFile`, `fs.writeFile`, or `fs.appendFile` as they don&#39;t return a `fd` (file descriptor). Those open the file, operate on it, and then close it for you. 

The streams returned by `fs.createReadStream` and `fs.createWriteStream` close after the stream ends but may be closed early. If you have paused a stream, you must call close on the stream to close the `fd` or resume the stream and let it end after emitting all its data.

But if you call `fs.open` or any of the others that give a `fd`, you must eventually `fs.close` the `fd` that you are given.

I&#39;ve got a PHP application on a free Heroku account. Today for no apparent reason I got Heroku&#39;s Application Error screen for about an hour. No downtimes on Heroku&#39;s status report page.

I&#39;d like to check the logs to figure out why the application was down.. But have no idea how to do this! Can somebody please help me?

Heroku: How can you check Heroku Error Logs?

I am trying to use this code for NET.reflector using Reflexil. I am trying to replace code with this:

    if(Input.GetKeyDown(KeyCode.Keypad5)) { 
    int i = 0; 
    Character localPlayer = PlayerClient.GetLocalPlayer().controllable.GetComponent&lt;Character&gt;(); 
    foreach (UnityEngine.Object obj2 in UnityEngine.Object.FindObjectsOfType(typeof(LootableObject))) 
    { 
        if (obj2 != null) 
        { 
            i++; 
            LootableObject loot = (LootableObject) obj2; 
            Debug.Log(&quot;Loot &quot;+i+&quot;: &quot;+loot.transform.position.ToString()); 
            CCMotor ccmotor = localPlayer.ccmotor; 
            if(ccmotor != null &amp;&amp; tpPos1 != Vector3.zero) { 
                ccmotor.Teleport(loot.transform.position); 
                Notice.Popup(&quot;&quot;, &quot;Teleported to &quot;+loot.name, 1.5f); 
            } 
            break; 
        } 
    } 
}


But it gives me an error when I try to compile:

    Line: 1 Column: 1 Error Number: CS0116	Error Message: &quot;A namespace does not directly contain members such as fields or methods&quot;


This is Unity code I think. I am not that experienced. Could anyone fix this for me? Or tell me what to do? Thanks!


&quot;A namespace cannot directly contain members such as fields or methods&quot;

I can&#39;t find more about fs.close explain in nodejs API.
I want to know what the scenario call fs.close is necessary.
for example:
&lt;pre&gt;
var fs =  require(&#39;fs&#39;);
fs.writeFile(&quot;/home/a.tex&quot;,&quot;abc&quot;); or like fs.appendFile(&quot;/home/a.tex&quot;,&quot;close&quot;)
fs.close(); //is it necessary?
&lt;/pre&gt;

Are there any effects if i don&#39;t call fs.close?

Any help is appreciated. 

What the scenario call fs.close is necessary

I can't find more about fs.close explain in nodejs API.
I want to know what the scenario call fs.close is necessary.
for example:
<pre>var fs = require('fs');
fs.writeFile("/home/a.tex","abc"); or like fs.appendFile("/home/a.tex","close")
fs.close(); //is it necessary?
</pre>
Are there any effects if i don't call fs.close?
Any help is appreciated.

What is the PHP [`die()`]/(http://www.php.net/manual/de/function.die.php) equivalent in Node.js?

https://www.php.net/manual/de/function.exit.php

What is PHP exit()/die() equivalent in Node.js

Is the following the right way to create a directory if it doesn&#39;t exist?

It should have full permission for the script and readable by others.

    var dir = __dirname + &#39;/upload&#39;;
    if (!path.existsSync(dir)) {
        fs.mkdirSync(dir, 0744);
    }




How to create a directory if it doesn&#39;t exist using Node.js

We have a fairly simple node.js app, but due to AWS Elastic Beanstalk deployment mechanism, it takes about 5 minutes to roll-out a new version (via `git aws.push`) even after a single file commit.

I.e. the commit itself (and upload) is fast (only 1 file to push), but then Elastic Beanstalk fetches whole package from S3, unzips it and runs `npm install`, which causes node-gyp to compile some modules. Upon installation/building completion, Elastic Beanstalk wipes `/var/app/current` and replaces it with the new app version.

Needless to say, constant node_modules rebuilding is not necessary, and rebuilding that takes 30 seconds on my old Macbook Air, takes &gt;5 mins on a ec2.micro instance, not fun.

I see two approaches here:

 1. tweak `/opt/containerfiles/ebnode.py` and play with node_modules location to avoid its removal and rebuilding upon deployment. 
 2. set up a git repo on Elastic Beanstalk EC2 instance and basically re-write deployment procedure ourselves, so /var/app/current receives pushes and runs `npm install` only when necessary (which makes Elastic Beanstalk to look like OpsWorks..)

Both options lack grace and are prone to issues when Amazon updates their Elastic Beanstalk hooks and architecture.

Maybe somebody has a better idea how to avoid constant rebuilding of node_modules that are already present in the app dir? Thank you.

avoid rebuilding node_modules in elastic beanstalk

I&#39;ve mentioned that my application uses different version of NodeJS when running from `sudo`.

    $ node -v
    v0.10.23
    $ sudo node -v
    v0.11.8-pre

This **v0.11.8-pre** caused me some problems, so I definitely don&#39;t want to use it, but I can&#39;t change it for root.

    $ sudo nvm use v0.10.23
    sudo: nvm: command not found

I&#39;ve tried to install nvm from root user, but got error &quot;NVM already installed&quot;, but still nvm not found when running from `sudo`. What is my problem?

Can&#39;t use NVM from root (or sudo)

For each module I have some files that need to be copied over to the build directory, and am looking for a way to minimize the repeated code from this: 
  
    gulp.src(&#39;./client/src/modules/signup/index.js&#39;)
      .pipe(gulp.dest(&#39;./build/public/js/signup&#39;));

    gulp.src(&#39;./client/src/modules/admin/index.js&#39;)
      .pipe(gulp.dest(&#39;./build/public/js/admin&#39;));

to something like this:

    gulp.src(&#39;./client/src/modules/(.*)/index.js&#39;)
      .pipe(gulp.dest(&#39;./build/public/js/$1&#39;));

Obviously the above doesn&#39;t work, so is there a way to do this, or an npm that already does this?

Thanks


Looking for way to copy files in gulp and rename based on parent directory

My code looks like this:

    def storescores():

       hs = open(&quot;hst.txt&quot;,&quot;a&quot;)
       hs.write(name)
       hs.close() 

so if I run it and enter &quot;Ryan&quot;
then run it again and enter &quot;Bob&quot;
the file hst.txt looks like 

    RyanBob 

instead of

    Ryan
    Bob

How do I fix this?

How to append new data onto a new line

I&#39;m trying to style a file upload button to my personal preferences, but I couldn&#39;t find any really solid ways to do this without JS. I did find [two](https://stackoverflow.com/q/3226167/1256925) [other](https://stackoverflow.com/q/3606876/1256925) questions about this subject, but the answers there either involved JavaScript, or suggested [Quirksmode&#39;s approach](http://www.quirksmode.org/dom/inputfile.html).

My major issue with this Quirksmode&#39;s approach is that the file button will still have the browser-defined dimensions, so it won&#39;t automatically adjust to whatever&#39;s used as button that&#39;s placed below it. I&#39;ve made some code, based on it, but it will just take up the space the file button would normally take up, so it won&#39;t at all fill the parent div like I want it to.

**HTML:**

    &lt;div class=&quot;myLabel&quot;&gt;
        &lt;input type=&quot;file&quot;/&gt;
        &lt;span&gt;My Label&lt;/span&gt;
    &lt;/div&gt;

**CSS:**

    .myLabel {
        position: relative;
    }
    .myLabel input {
        position: absolute;
        z-index: 2;
        opacity: 0;
        width: 100%;
        height: 100%;
    }

[**This fiddle**](http://jsfiddle.net/KY42f/1/) demonstrates how this approach is quite flawed. In Chrome, clicking the `!!` below the second demo button will open the file dialog anyway, but also in all other browsers, the file button doesn&#39;t take up the correct areas of the button.

Is there any more solid way to style the file upload button, without any JavaScript, and preferably using as little &#39;hacky&#39; coding as possible (since hacking usually brings other problems along with it, such as the ones in the fiddle)?

Cross-browser custom styling for file upload button

I&#39;ve searched all over stackoverflow / google for this, but can&#39;t seem to figure it out. 

I&#39;m scraping social media links of a given URL page, and the function returns an object with a list of URLs. 

When I try to write this data into a different file, it outputs to the file as `[object Object]` instead of the expected: 
    [ &#39;https://twitter.com/#!/101Cookbooks&#39;,
    &#39;http://www.facebook.com/101cookbooks&#39;] 
as it does when I `console.log()` the results.

This is my sad attempt to read and write a file in Node, trying to read each line(the url) and input through a function call `request(line, gotHTML)`: 

    fs.readFileSync(&#39;./urls.txt&#39;).toString().split(&#39;\n&#39;).forEach(function (line){
    	console.log(line); 
    	var obj = request(line, gotHTML); 
    	console.log(obj); 
    	fs.writeFileSync(&#39;./data.json&#39;, obj , &#39;utf-8&#39;); 
    });   

for reference -- the `gotHTML` function: 

    function gotHTML(err, resp, html){ 
    	var social_ids = []; 
    
        if(err){
    		return console.log(err); 
    	} else if (resp.statusCode === 200){ 
    		var parsedHTML = $.load(html); 
    
    		parsedHTML(&#39;a&#39;).map(function(i, link){
    			var href = $(link).attr(&#39;href&#39;);
    			for(var i=0; i&lt;socialurls.length; i++){
    				if(socialurls[i].test(href) &amp;&amp; social_ids.indexOf(href) &lt; 0 ) {
    					social_ids.push(href); 
    				}; 
    			}; 
    		})
    	};
    
    	return social_ids;
    };






Write objects into file with Node.js

    with open(&quot;games.txt&quot;, &quot;w&quot;) as text_file:
        print(driver.current_url)
        text_file.write(driver.current_url + &quot;\n&quot;)

I&#39;m using this code right now, but when it writes to the file it overwrites the old content. How can I simply add to it without erasing the content that&#39;s already there.

How to write to a file without overwriting current contents?

Several times here on SO I&#39;ve seen people using `rt` and `wt` modes for reading and writing files.

For example:

    with open(&#39;input.txt&#39;, &#39;rt&#39;) as input_file:
         with open(&#39;output.txt&#39;, &#39;wt&#39;) as output_file: 
             ...

I don&#39;t see the modes [documented][1], but since `open()` doesn&#39;t throw an error - looks like it&#39;s pretty much legal to use.

What is it for and is there any difference between using `wt` vs `w` and  `rt` vs `r`?



  [1]: https://docs.python.org/2/library/functions.html#open

Content Type	Original Author	Original Content on Stackoverflow
Question	L.T	View Question on Stackoverflow
Solution 1 - node.js	Dan D.	View Answer on Stackoverflow

What the scenario call fs.close is necessary

node.js Problem Overview

node.js Solutions

Solution 1 - node.js

"A namespace cannot directly contain members such as fields or methods"

Heroku: How can you check Heroku Error Logs?

Attributions