You need to use double quotes. Single quotes prevent the shell variable from being interpolated by the shell. You use single quotes to prevent the shell from doing interpolation which you may have to do if your regular expression used `$` as part of the pattern. You can also use a backslash to _quote_ a `$` if you&#39;re using double quotes.

Also, you may need to put your variable in curly braces `${var}` in order to help separate it from the rest of the pattern.

Example:

    $ string=&quot;test this&quot;
    $ var=&quot;test&quot;
    $ echo $string | grep &quot;^${var}&quot;

I am about to start a project with **AngularJS** for the client side and **Django** for the server side. 

What are the best practices to make them work like best friends (static files,auth, deployment, etc.) 

What are the best practices to use AngularJS with Django

I have been trying to get the `head` utility to display all but the last line of standard input. The actual code that I needed is something along the lines of `cat myfile.txt | head -n $(($(wc -l)-1))`. But that didn&#39;t work. I&#39;m doing this on Darwin/OS X which doesn&#39;t have the nice semantics of `head -n -1` that would have gotten me similar output.

None of these variations work either.

    cat myfile.txt | head -n $(wc -l | sed -E -e &#39;s/\s//g&#39;)
    echo &quot;hello&quot; | head -n $(wc -l | sed -E -e &#39;s/\s//g&#39;)

I tested out more variations and in particular found this to work:

    cat &lt;&lt;EOF | echo $(($(wc -l)-1))
    &gt;Hola
    &gt;Raul
    &gt;Como Esta
    &gt;Bueno?
    &gt;EOF
    3

Here&#39;s something simpler that also works.

    echo &quot;hello world&quot; | echo $(($(wc -w)+10))

This one understandably gives me an illegal line count error. But it at least tells me that the `head` program is not consuming the standard input before passing stuff on to the subshell/command substitution, a remote possibility, but one that I wanted to rule out anyway.

    echo &quot;hello&quot; | head -n $(cat &amp;&amp; echo 1)

What explains the behavior of `head` and `wc` and their interaction through subshells here? Thanks for your help.

Getting head to display all but the last line of a file: command substitution and standard I/O redirection

I&#39;m trying to use a variable in a grep regex.  I&#39;ll just post an example of the failure and maybe someone can suggest how to make the variable be evaluated while running the grep command.  I&#39;ve tried `${var}` as well.

    $ string=&quot;test this&quot;
    $ var=&quot;test&quot;
    $ echo $string | grep &#39;^$var&#39;
    $ 

Since my regex should match lines which start with &quot;test&quot;, it should print the line echoed thru it.

    $ echo $string
    test this
    $

shell variable in a grep regex

I'm trying to use a variable in a grep regex. I'll just post an example of the failure and maybe someone can suggest how to make the variable be evaluated while running the grep command. I've tried <code>${var}</code> as well.
<pre><code class="hljs language-shell">$ string="test this"
$ var="test"
$ echo $string | grep '^$var'
$ 
</code></pre>
Since my regex should match lines which start with "test", it should print the line echoed thru it.
<pre><code class="hljs language-shell">$ echo $string
test this
$
</code></pre>

I want to subtract &quot;number of days&quot; from a date in bash. I am trying something like this ..

    echo $dataset_date #output is 2013-08-07

    echo $date_diff #output is 2   
 
    p_dataset_date=`$dataset_date --date=&quot;-$date_diff days&quot; +%Y-%m-%d` # Getting Error



subtract days from a date in bash

What is a `login shell` and `interactive shell`, and what is a `.bash_profile` and `.bashrc`?

What are the differences between a login shell and interactive shell?

I have two large files (sets of filenames). Roughly 30.000 lines in each file. I am trying to find a fast way of finding lines in file1 that are not present in file2.

For example, if this is **file1:**

    line1
    line2
    line3

And this is **file2:**

    line1
    line4
    line5

Then my **result/output** should be:

    line2
    line3

This works:

`grep -v -f file2 file1`

But it is very, very slow when used on my large files.

I suspect there is a good way to do this using `diff()`, but the output should be *just* the lines, nothing else, and I cannot seem to find a switch for that.

Can anyone help me find a fast way of doing this, using bash and basic Linux binaries?


**EDIT**: To follow up on my own question, this is the best way I have found so far using `diff()`:

     diff file2 file1 | grep &#39;^&gt;&#39; | sed &#39;s/^&gt;\ //&#39;
Surely, there must be a better way?

Fast way of finding lines in one file that are not in another?

I&#39;m writing a script that requires root level permissions, and I want to make it so that if the script is not run as root, it simply echoes &quot;Please run as root.&quot; and exits.

Here&#39;s some pseudocode for what I&#39;m looking for:

    if (whoami != root)
      then echo &quot;Please run as root&quot;
     
      else (do stuff)
    fi

    exit

How could I best (cleanly and securely) accomplish this? Thanks!

Ah, just to clarify: the (do stuff) part would involve running commands that in-and-of themselves require root. So running it as a normal user would just come up with an error. This is just meant to cleanly run a script that requires root commands, without using sudo inside the script, I&#39;m just looking for some syntactic sugar. 

How to check if running as root in a bash script

I have a table in my PostgreSQL database which has 3 columns - `c_uid`, `c_defaults` and `c_settings`. `c_uid` simply stores the name of a user and `c_defaults` is a long piece of text which contains a lot of data w.r.t that user.

I have to execute a statement from a bash script which selects the value of the `c_defaults` column based on the `c_uid` value and this needs to be done by the database user &#39;postgres&#39;.

On the CLI I can do the following:

    [mymachine]# su postgres
    bash-4.1$psql
    postgres=#\c database_name
    You are now connected to database &quot;database_name&quot; as user &quot;postgres&quot;.
    database_name=#SELECT c_defaults  FROM user_info WHERE c_uid = &#39;testuser&#39;;
However, how do I achieve this through a bash script?

The aim is to get the information from that column, edit it and write it back into that column - all through a bash script.




PostgreSQL - query from bash script as database user &#39;postgres&#39;

I&#39;d like find lines in files with an occurrence of some pattern and an absence of some other pattern. For example, I need find all files/lines including `loom` except ones with `gloom`. So, I can find `loom` with command:

    grep -n &#39;loom&#39; ~/projects/**/trunk/src/**/*.@(h|cpp)

Now, I want to search `loom` excluding `gloom`. However, both of following commands failed:

    grep -v &#39;gloom&#39; -n &#39;loom&#39; ~/projects/**/trunk/src/**/*.@(h|cpp)
    grep -n &#39;loom&#39; -v &#39;gloom&#39; ~/projects/**/trunk/src/**/*.@(h|cpp)

What should I do to achieve my goal?

***EDIT 1:*** I mean that `loom` and `gloom` are the character sequences (not necessarily the words). So, I need, for example, `bloomberg` in the command output and don&#39;t need `ungloomy`.

***EDIT 2:*** There is sample of my expectations.
Both of following lines are in command output:
&gt; I faced the icons that *loomed* through the veil of incense. 
&gt;
&gt; Arty is *slooming* in a *gloomy* day.

Both of following lines aren&#39;t in command output:
&gt; It’s *gloomyin’* ower terrible — great muckle doolders o’ cloods.
&gt;
&gt; In the south west round of the heigh pyntit hall


How to grep, excluding some patterns?

I want to extract the value pair from a key-value pair syntax but I can not.   
Example I tried:  

    echo employee_id=1234 | sed &#39;s/employee_id=$[0-9]+$/\1/g&#39;

But this gives `employee_id=1234` and not `1234` which is actually the capture group.  

What am I doing wrong here? I also tried:

    echo employee_id=1234| egrep -o employee_id=([0-9]+)

but no success.

Can not extract the capture group with either sed or grep

How do I `grep` a file for `19:55` and get the Line 1,2,3,4,5?

    2013/10/08 19:55:27.471
    Line 1
    Line 2
    Line 3
    Line 4
    Line 5

    2013/10/08 19:55:29.566
    Line 1
    Line 2
    Line 3
    Line 4
    Line 5



How do you grep a file and get the next 5 lines

How do you grep and only return the matching line? i.e. The path/filename is omitted from the results.

In this case I want to look in all .bar files in the current directory, searching for the term FOO

    find . -name &#39;*.bar&#39; -exec grep -Hn FOO {} \;

Content Type	Original Author	Original Content on Stackoverflow
Question	mike_b	View Question on Stackoverflow
Solution 1 - Bash	David W.	View Answer on Stackoverflow

shell variable in a grep regex

Bash Problem Overview

Bash Solutions

Solution 1 - Bash

Getting head to display all but the last line of a file: command substitution and standard I/O redirection

What are the best practices to use AngularJS with Django

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