All jobs are executed by a shell, so start that shell snippet by a command to change the directory.

    cd /path/to/directory &amp;&amp; ./bin/myapp

Concerning the use of `&amp;&amp;` instead of `;`: normally it doesn&#39;t make a difference, but if the `cd` command fails (e.g. because the directory doesn&#39;t exist) with `&amp;&amp;` the application isn&#39;t executed, whereas with `;` it&#39;s executed (but **not in the intended directory**).

Reading `man 5 crontab` should tell you that there&#39;s a `HOME` variable set which can be redefined in the file. It becomes your working directory. You can set `PATH` for the command(s) too. Of course this affects all the cron schedule lines.

E.G. 
&gt; Several environment variables are set up automatically by the cron(8)
  daemon.  `SHELL` is set to `/bin/sh`, and `LOGNAME` and `HOME` are set from
  the `/etc/passwd` line of the crontab&#180;s owner.  `HOME` and `SHELL` can be
  overridden by settings in the crontab; `LOGNAME` can not.
&gt;
&gt; (Note: the `LOGNAME` variable is sometimes called `USER` on `BSD` systems
  and is also automatically set).

Depending on your cron of course, but mine also has `MAILTO`, `MAILFROM` `CONTENT_TYPE`, `CRON_TZ`, `RANDOM_DELAY`, and `MLS_LEVEL`.
 
So for your hypothetical app I&#39;d recommend a file name `/etc/cron.d/hypothetical` containing:
```sh
# Runs hypothetical app @ 00:01Z in its local path for reading its config or something.
SHELL=/bin/sh
HOME=/where/the/app/is
PATH=/where/the/app/is:/bin:/usr/local/bin:/usr/bin:/bin:/usr/sbin:/sbin
CRON_TZ=UTC
1 0 * * * theappuser hypothetical --with arguments 
```

For example with `docker-compose` relying on the cwd `docker-compose.yml`:
```bash
SHELL=/bin/sh
HOME=/path/to/composed-app
5 5 * * * root docker-compose restart -t 10 service-name
```

I am new to hibernate and need to use one-to-many and many-to-one relations. It is a bi-directional relationship in my objects, so that I can traverse from either direction. `mappedBy` is the recommended way to go about it, however, I couldn&#39;t understand it. Can someone explain:

- what is the recommended way to use it?
- what purpose does it solve?

For the sake of my example, here are my classes with annotations:

- `Airline` **OWNS many** `AirlineFlights`
- **Many** `AirlineFlights` belong to **ONE** `Airline`

**Airline**:

    @Entity 
    @Table(name=&quot;Airline&quot;)
    public class Airline {
    	private Integer idAirline;
    	private String name;

    	private String code;
    	
    	private String aliasName;
    	private Set&lt;AirlineFlight&gt; airlineFlights = new HashSet&lt;AirlineFlight&gt;(0);
    	
    	public Airline(){}
    	
    	public Airline(String name, String code, String aliasName, Set&lt;AirlineFlight&gt; flights) {
    		setName(name);
    		setCode(code);
    		setAliasName(aliasName);
    		setAirlineFlights(flights);
    	}
    
    	@Id
    	@GeneratedValue(strategy=GenerationType.IDENTITY)
    	@Column(name=&quot;IDAIRLINE&quot;, nullable=false)
    	public Integer getIdAirline() {
    		return idAirline;
    	}
    	
    	private void setIdAirline(Integer idAirline) {
    		this.idAirline = idAirline;
    	}
    	
    	@Column(name=&quot;NAME&quot;, nullable=false)
    	public String getName() {
    		return name;
    	}
    	public void setName(String name) {
    		this.name = DAOUtil.convertToDBString(name);
    	}
    	
    	@Column(name=&quot;CODE&quot;, nullable=false, length=3)
    	public String getCode() {
    		return code;
    	}
    	public void setCode(String code) {
    		this.code = DAOUtil.convertToDBString(code);
    	}
    	
    	@Column(name=&quot;ALIAS&quot;, nullable=true)
    	public String getAliasName() {
    		return aliasName;
    	}
    	public void setAliasName(String aliasName) {
    		if(aliasName != null)
    			this.aliasName = DAOUtil.convertToDBString(aliasName);
    	}
    
    	@OneToMany(fetch=FetchType.LAZY, cascade = {CascadeType.ALL})
    	@JoinColumn(name=&quot;IDAIRLINE&quot;)
    	public Set&lt;AirlineFlight&gt; getAirlineFlights() {
    		return airlineFlights;
    	}
    
    	public void setAirlineFlights(Set&lt;AirlineFlight&gt; flights) {
    		this.airlineFlights = flights;
    	}
    }

**AirlineFlights:**

    @Entity
    @Table(name=&quot;AirlineFlight&quot;)
    public class AirlineFlight {
    	private Integer idAirlineFlight;
    	private Airline airline;
    	private String flightNumber;
    	
    	public AirlineFlight(){}
    	
    	public AirlineFlight(Airline airline, String flightNumber) {
    		setAirline(airline);
    		setFlightNumber(flightNumber);
    	}
    
    	@Id
    	@GeneratedValue(generator=&quot;identity&quot;)
    	@GenericGenerator(name=&quot;identity&quot;, strategy=&quot;identity&quot;)
    	@Column(name=&quot;IDAIRLINEFLIGHT&quot;, nullable=false)
    	public Integer getIdAirlineFlight() {
    		return idAirlineFlight;
    	}
    	private void setIdAirlineFlight(Integer idAirlineFlight) {
    		this.idAirlineFlight = idAirlineFlight;
    	}
    	
    	@ManyToOne(fetch=FetchType.LAZY)
    	@JoinColumn(name=&quot;IDAIRLINE&quot;, nullable=false)
    	public Airline getAirline() {
    		return airline;
    	}
    	public void setAirline(Airline airline) {
    		this.airline = airline;
    	}
    	
    	@Column(name=&quot;FLIGHTNUMBER&quot;, nullable=false)
    	public String getFlightNumber() {
    		return flightNumber;
    	}
    	public void setFlightNumber(String flightNumber) {
    		this.flightNumber = DAOUtil.convertToDBString(flightNumber);
    	}
    }

**EDIT:**

Database schema:

AirlineFlights has the idAirline as ForeignKey and Airline has no idAirlineFlights. This makes, AirlineFlights as the owner/identifying entity ?

Theoretically, I would like airline to be the owner of airlineFlights.

![enter image description here][1]


  [1]: https://i.stack.imgur.com/gQvDs.png

Can someone explain mappedBy in JPA and Hibernate?

How can I retrieve the remote git address of a repo?

I tried git remote but that just lists the branches. 

How can I retrieve the remote git address of a repo?

I would like to set a job to run daily in the root crontab. But I would like it to execute it from a particular directory so it can find all the files it needs, since the application has a bunch of relative paths.

Anyway, can I tell crontab to run from a particular directory?

Crontab - Run in directory

<p>I would like to set a job to run daily in the root crontab. But I would like it to execute it from a particular directory so it can find all the files it needs, since the application has a bunch of relative paths.</p>
<p>Anyway, can I tell crontab to run from a particular directory?</p>


I need to check the duration of a group of audio files.
Is there a simple way to do this on the unix command-line?

    &gt; duration *

I have the amazing SoX app which has an option called stats that generates a set of audio info including duration. I am looking for a way to get only duration.
I am flexible on output format, could be any of sample length, hh:mm:ss, or seconds.
The latter would be my preference. 

check duration of audio files on the command-line

Let&#39;s say I wanted to benchmark two programs: foo.py and bar.py.

Are a couple thousand runs and the respective averages of `time python foo.py` and `time python bar.py` adequate enough for profiling and comparing their speed?
&lt;br/&gt;&lt;br /&gt;&lt;br /&gt;
&lt;b&gt;Edit:&lt;/b&gt; Additionally, if the execution of each program was sub-second (assume it wasn&#39;t for the above), would `time` still be okay to use?

Is the UNIX `time` command accurate enough for benchmarks?

I am trying to rsync directory A of server1 with directory B of server2.

Sitting in the directory A of server1, I ran the following commands.

    rsync -av * server2::sharename/B

but the interesting thing is, it synchronizes all files and directories except .htaccess or any hidden file in the directory A. Any hidden files within subdirectories get synchronized.

I also tried the following command:

    rsync -av --include=&quot;.htaccess&quot; * server2::sharename/B

but the results are the same.

Any ideas why hidden files of A directory are not getting synchronized and how to fix it. I am running as root user.

thanks



rsync not synchronizing .htaccess file

I&#39;d like to know if there is any tip to make **`grep`** as fast as possible. I have a rather large base of text files to search in the quickest possible way. I&#39;ve made them all lowercase, so that I could get rid of `-i` option. This makes the search much faster.

Also, I&#39;ve found out that `-F` and `-P` modes are quicker than the default one. I use the former when the search string is not a regular expression (just plain text), the latter if regex is involved.

Does anyone have any experience in speeding up `grep`? Maybe compile it from scratch with some particular flag (I&#39;m on Linux CentOS), organize the files in a certain fashion or maybe make the search parallel in some way?

Fastest possible grep

I&#39;m trying to construct an array in bash of the filenames from my camera:

    FILES=(2011-09-04 21.43.02.jpg
    2011-09-05 10.23.14.jpg
    2011-09-09 12.31.16.jpg
    2011-09-11 08.43.12.jpg)

As you can see, there is a space in the middle of each filename.

I&#39;ve tried wrapping each name in quotes, and escaping the space with a backslash, neither of which works.

When I try to access the array elements, it continues to treat the space as the elementdelimiter.

How can I properly capture the filenames with a space inside the name?

Bash array with spaces in elements

How do I check if directory exists in Python?

I use a `FileUploader` control in my application. I want to save a file to a specified folder. If this folder does not exist, I want to first create it, and then save my file to this folder.  If the folder already exists, then just save the file in it.

How can I do this?


If a folder does not exist, create it

How can I find all files in directory with the extension .csv in python?

Find all CSV files in a directory using Python

In my app I want to copy a file to the other hard disk so this is my code:

     #include &lt;windows.h&gt;
    
    using namespace std;
    
    int main(int argc, char* argv[] )
    {
        string Input = &quot;C:\\Emploi NAm.docx&quot;;
        string CopiedFile = &quot;Emploi NAm.docx&quot;;
        string OutputFolder = &quot;D:\\test&quot;;
        CopyFile(Input.c_str(), string(OutputFolder+CopiedFile).c_str(), TRUE);
     
        return 0;
    }

so after executing this, it shows me in the `D:`HDD a file `testEmploi NAm.docx`
but I want him to create the test folder if it doesn&#39;t exist.

I want to do that without using the Boost library.

Create a directory if it doesn&#39;t exist

Is there any way to automatically delete all files or folders with few R command lines?
I am aware of the `unlink()` or `file.remove()` functions, but for those you need to define a character vector with exactly all the names of the files you want to delete.  I am looking more for something that lists all the files or folders within a specific path (e.g. &#39;C:/Temp&#39;) and then delete all files with a certain name (regardless of its extension).

Any help is very much appreciated!



Automatically Delete Files/Folders

Let&#39;s say you have a module which contains

    myfile = open(&#39;test.txt&#39;, &#39;r&#39;)

And the &#39;test.txt&#39; file is in the same folder. If you&#39;ll run the module, the file will be opened successfully.

Now, let&#39;s say you import that module from another one which is in another folder. The file won&#39;t be searched in the same folder as the module where that code is.

**So how to make the module search files with relative paths in the same folder first?**

There are various solutions by using &quot;`__file__`&quot; or &quot;`os.getcwd()`&quot;, but I&#39;m hoping there&#39;s a cleaner way, like same special character in the string you pass to open() or file().

How to make an &quot;always relative to current module&quot; file path?

I want to construct a WebSocket URI relative to the page URI at the browser side. Say, in my case convert HTTP URIs like

    http://example.com:8000/path
    https://example.com:8000/path

to

    ws://example.com:8000/path/to/ws
    wss://example.com:8000/path/to/ws

What I&#39;m doing currently is replace the first 4 letters &quot;http&quot; by &quot;ws&quot;, and append &quot;/to/ws&quot; to it. Is there any better way for that?

How to construct a WebSocket URI relative to the page URI?

So, an *absolute path* is a way to get to a certain file or location describing the full route to it, the full path, and it&#39;s OS dependent (the absolute paths for Windows and Linux, for example, are different). A *relative path*, on the other hand, is a route to a file or location which is described from the current location `..` (two dots) indicating a superior level in the directories tree. That has been clear to me for several years now.

When searching I&#39;ve even seen that there are canonicalized files too!
All I know is that CANONICAL means something like &quot;according to the rules&quot; or something.

Can somebody enlighten me in therms of theory about canonical stuff?

What&#39;s a &quot;canonical path&quot;?

I have the following file structure:
&lt;pre&gt;
rootDIR
    dir1
        subdir1
           file0.php
           file1.php
    dir2
       file2.php
       file3.php
       file4.php   
&lt;/pre&gt;

`file1.php` requires `file3.php` and `file4.php` from dir2 like this :

    require(&#39;../../dir2/file3.php&#39;)

`file2.php` requires `file1.php` like this:

    require(&#39;../dir1/subdir1/file1.php&#39;)

But then require in `file1.php` fails to open `file3.php` and `file4.php` ( maybe due to the path relativeness)

However, what is the reason and what can I do for `file2.php` so `file1.php` properly require `file3.php` and `file4.php`?

How to require PHP files relatively (at different directory levels)?

I&#39;ve been here:

* http://www.python.org/dev/peps/pep-0328/
* http://docs.python.org/2/tutorial/modules.html#packages
* https://stackoverflow.com/questions/10059002/python-packages-relative-imports
* https://stackoverflow.com/questions/9123062/python-relative-import-example-code-does-not-work
* https://stackoverflow.com/questions/8299270/ultimate-answer-to-relative-python-imports
* https://stackoverflow.com/questions/4175534/relative-imports-in-python
* https://stackoverflow.com/questions/13233931/python-disabling-relative-import?rq=1

and plenty of URLs that I did not copy, some on SO, some on other sites, back when I thought I&#39;d have the solution quickly.

The forever-recurring question is this: how do I solve this &quot;Attempted relative import in non-package&quot; message?

    ImportError: attempted relative import with no known parent package

I built an exact replica of the package on pep-0328:

    package/
        __init__.py
        subpackage1/
            __init__.py
            moduleX.py
            moduleY.py
        subpackage2/
            __init__.py
            moduleZ.py
        moduleA.py


The imports were done from the console.

I did make functions named spam and eggs in their appropriate modules.  Naturally, it didn&#39;t work.  The answer is apparently in the 4th URL I listed, but it&#39;s all alumni to me. There was this response on one of the URLs I visited:

&gt; Relative imports use a module&#39;s name attribute to determine that module&#39;s position in the package hierarchy. If the module&#39;s name does not contain any package information (e.g. it is set to &#39;main&#39;) then relative imports are resolved as if the module were a top level module, regardless of where the module is actually located on the file system.

The above response looks promising, but it&#39;s all hieroglyphs to me.  So my question, how do I make Python not return to me &quot;Attempted relative import in non-package&quot;? has an answer that involves -m, supposedly.

Can somebody please tell me why Python gives that error message, what it means by &quot;non-package&quot;, why and how do you define a &#39;package&#39;, and *the precise answer put in terms easy enough for a kindergartener to understand*. 


Relative imports for the billionth time

I am running next command within a crontab to encrypt a file and I don&#39;t want a keyboard interaction

    echo &quot;PASSPHRASE&quot; | gpg --passphrase-fd 0 -r USER --encrypt FILENAME.TXT

but I have this answer:

    gpg: C042XXXX: There is no assurance this key belongs to the named user
    
    pub  40XXX/C042XXXX 2012-01-11 Name LastName. (comment) &lt;user@email.com&gt;
     Primary key fingerprint: XXXX XXXX XXXX XXXX XXXX  XXXX XXXX XXXX XXXX XXXX
          Subkey fingerprint: XXXX XXXX XXXX XXXX XXXX  XXXX XXXX XXXX XXXX XXXX
    
    It is NOT certain that the key belongs to the person named
    in the user ID.  If you *really* know what you are doing,
    you may answer the next question with yes.
    
    Use this key anyway? (y/N) 

gpg encrypt file without keyboard interaction

I am new to scheduling tasks with cron and crontab.  I am trying to schedule execution of a task as if I had logged on, opened a terminal, and executed it myself.

However, I scheduled a task to help me observe what $USER and $PATH a scheduled task is executing with, and this is what I found:

    $ crontab -l
    41 11 * * * echo &quot;USER: $USER&quot; &gt; ~/Desktop/cron_env.log; echo &quot;PATH: $PATH&quot; &gt;&gt; ~/Desktop/cron_env.log
    $ cat ~/Desktop/cron_env.log
    USER:
    PATH: /usr/bin:/bin

It appears as though $USER is not set, and $PATH is something very basic and/or default.  On the contrary, this is what I see when I open a terminal (logged in) and echo this same information:

    USER: aschirma
    PATH: /usr/lib/jvm/java-6-sun/bin:/usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin:/usr/games:/pkg/icetools/bin:/pkg/hwtools/bin:/pkg/netscape/bin:/pkg/gnu/bin

What do I need to do to make my crontab tasks run the way I want?

crontab PATH and USER

Source Wikipedia:   

&gt; **Question mark** (?) is used instead of &#39;*&#39; for leaving either day-of-month or day-of-week blank.

The above statement is not making much sense to me.

So if I write some cron as **`0 0 0 ? * *`** then does it mean first of every month or it means it will execute daily?

It is a bit confusing as Java crons start with seconds while other crons start with minute.

Significance of question mark in Java cron

i want to execute a script twice daily at 00:00 and 13:30 so i write :

    0,30 0,13 * * *

it seems wrong for me, because like this, the script will fire at 00:00 , 00:30 , 13:00 and 13:30. Any idea ?

execute crontab twice daily at 00h and 13:30

I would like to know if there is a way to setup a cronjob/task to execute every minute. Currently any of my instances should be able to run this task.

This is what I have tried to do in the config files without success:

    container_commands:
      01cronjobs:
        command: echo &quot;*/1 * * * * root php /etc/httpd/myscript.php&quot;

I&#39;m not really sure if this is the correct way to do it

Any ideas?


Content Type	Original Author	Original Content on Stackoverflow
Question	user333746	View Question on Stackoverflow
Solution 1 - Unix	Gilles 'SO- stop being evil'	View Answer on Stackoverflow
Solution 2 - Unix	dlamblin	View Answer on Stackoverflow

Crontab - Run in directory

Unix Problem Overview

Unix Solutions

Solution 1 - Unix

Solution 2 - Unix

How can I retrieve the remote git address of a repo?

Can someone explain mappedBy in JPA and Hibernate?

Attributions