If you&#39;re using the Bash builtin `time`, set the `TIMEFORMAT` variable to `%R`:

    $ TIMEFORMAT=%R
    $ time sleep 1
    1.022

`time` can take an optional `--format` or `-f` parameter, but you have to use the full path to the time command, `/usr/bin/time`

    /usr/bin/time -f &quot;%e&quot; sleep 3

    3.00

Without the path, it&#39;ll [use the `time` command of the shell][1] which just treats all arguments as the command, so you&#39;ll get an `-f: command not found`.

  [1]: https://superuser.com/questions/112353/how-can-i-give-arguments-to-time-command-ubuntu

If you write `\time` you enforce not to use the bash built in time.
So with `\time -f &#39;%e&#39; command` you are done.



Alternatively, use `/usr/bin/time` or take a look at the similar question  
[Using time command in bash script][1].


  [1]: https://stackoverflow.com/questions/3683434/using-time-command-in-bash-script/

We can define a constant like

    define(&quot;aconstant&#39;,&#39;avalue&#39;);

 Can&#39;t we define array in this fashion like below ?

    define(&quot;months&quot;,array(&quot;January&quot;, &quot;February&quot;, ---); 

Is it possible to declare an array as constant

I am getting the error &quot;Class &#39;Imagick&#39; not found&quot;.  Somehow I need to make this library accessible to php. I am using Php 5.2.6 on Fedora 8. my php_info has no mention of ImageMagick.

I have tried: yum install ImageMagick and restarted apache, which didn&#39;t work.

I also added extension=imagick.ext to my php.ini file and restarted apache, which didn&#39;t work.



php error: Class &#39;Imagick&#39; not found

I would like to make a script that outputs only the real time value from the time command so I can plot the results. For example `time command` outputs

    real	1m0.001s
    user	1m0.000s
    sys	 0m0.001s

I want to write a script that outputs

    60.001 


How do I get just real time value from &#39;time&#39; command in seconds?

How do I get just real time value from &#39;time&#39; command?

I would like to make a script that outputs only the real time value from the time command so I can plot the results. For example <code>time command</code> outputs
<pre><code class="hljs language-sql">real	1m0.001s
user	1m0.000s
sys	 0m0.001s
</code></pre>
I want to write a script that outputs
<pre><code class="hljs">60.001 
</code></pre>
How do I get just real time value from 'time' command in seconds?

Say I have too programs `a` and `b` that I can run with `./a` and `./b`.

Is it possible to diff their outputs without first writing to temporary files?

Diff output from two programs without temporary files

Is it possible to have a bash script automatically handle prompts that would normally be presented to the user with default actions?  Currently I am using a bash script to call an in-house tool that will display prompts to the user (prompting for Y/N) to complete actions, however the script I&#39;m writing needs to be completely &quot;hands-off&quot;, so I need a way to send `Y|N` to the prompt to allow the program to continue execution.  Is this possible?

Have bash script answer interactive prompts

I have two text files, each of them contains an information by line such like that

    file1.txt            file2.txt
    ----------           ---------
    linef11              linef21
    linef12              linef22
    linef13              linef23
     .                    .
     .                    .
     .                    .

I would like to merge theses files lines by lines using a bash script in order to obtain:

    fileresult.txt
    --------------
    linef11     linef21
    linef12     linef22
    linef13     linef23
     .           .
     .           .
     .           .

How can this be done in Bash?

How to merge two files line by line in Bash

I have defined the following variable:

    myVar=true

now I&#39;d like to run something along the lines of this:

    if [ myVar ]
    then
        echo &quot;true&quot;
    else
        echo &quot;false&quot;
    fi

The above code does work, but if I try to set 

    myVar=false

it will still output true.
What might be the problem?

edit: I know I can do something of the form

    if [ &quot;$myVar&quot; = &quot;true&quot; ]; then ...

but it is kinda awkward.

Thanks


How to evaluate a boolean variable in an if block in bash?

Let&#39;s say I have defined a `function abc()` that will handle the logic related to analyzing the arguments passed to my script.

How can I pass all arguments my bash script has received to it? The number of params is variable, so I can&#39;t just hardcode the arguments passed like this:

    abc $1 $2 $3 $4

**Edit**. Better yet, is there any way for my function to have access to the script arguments&#39; variables?

How to pass all arguments passed to my bash script to a function of mine?

I&#39;ve seen Bash scripts test for a non-zero length string in two different ways. Most scripts use the `-n` option:

    #!/bin/bash
    # With the -n option
    if [ -n &quot;$var&quot; ]; then
      # Do something when var is non-zero length
    fi

But the -n option isn&#39;t really needed:

    # Without the -n option
    if [ &quot;$var&quot; ]; then
      # Do something when var is non-zero length
    fi

Which is the better way?

Similarly, which is the better way for testing for zero-length:

    if [ -z &quot;$var&quot; ]; then
      # Do something when var is zero-length
    fi

or

    if [ ! &quot;$var&quot; ]; then
      # Do something when var is zero-length
    fi




Test for non-zero length string in Bash: [ -n &quot;$var&quot; ] or [ &quot;$var&quot; ]

I have a variable in my bash script whose value is something like this:

    ~/a/b/c

Note that it is unexpanded tilde. When I do ls -lt on this variable (call it $VAR), I get no such directory. I want to let bash interpret/expand this variable without executing it. In other words, I want bash to run eval but not run the evaluated command. Is this possible in bash?

How did I manage to pass this into my script without expansion? I passed the argument in surrounding it with double quotes.

Try this command to see what I mean:

    ls -lt &quot;~&quot;

This is exactly the situation I am in. I want the tilde to be expanded. In other words, what should I replace magic with to make these two commands identical:

    ls -lt ~/abc/def/ghi

and

    ls -lt $(magic &quot;~/abc/def/ghi&quot;)

Note that ~/abc/def/ghi may or may not exist.

How to manually expand a special variable (ex: ~ tilde) in bash

I have a script that automates a process that needs access to a password protected system. The system is accessed via a command-line program that accepts the user password as an argument.

I would like to prompt the user to type in their password, assign it to a shell variable, and then use that variable to construct the command line of the accessing program (which will of course produce stream output that I will process).  

I am a reasonably competent shell programmer in Bourne/Bash, but I don&#39;t know how to accept the user input without having it echo to the terminal (or maybe having it echoed using &#39;*&#39; characters).  

Can anyone help with this?

How to get a password from a shell script without echoing

I have two `*.sql` files that I use when creating a new web site database.  The first file creates all the tables.  The second file populates some default records.  I would like to execute these files from PHP.  I also use the Zend_Framework, if that will help accomplish this.

**Additional Info**

  1. I don&#39;t have console access
  1. I&#39;m trying to automate site generation from within our application.

**SOLUTION**

Using `shell_exec()`...

    $command = &#39;mysql&#39;
            . &#39; --host=&#39; . $vals[&#39;db_host&#39;]
            . &#39; --user=&#39; . $vals[&#39;db_user&#39;]
            . &#39; --password=&#39; . $vals[&#39;db_pass&#39;]
            . &#39; --database=&#39; . $vals[&#39;db_name&#39;]
            . &#39; --execute=&quot;SOURCE &#39; . $script_path
    ;
    $output1 = shell_exec($command . &#39;/site_db.sql&quot;&#39;);
    $output2 = shell_exec($command . &#39;/site_structure.sql&quot;&#39;);

...I never did get useful output, but followed [some suggestions][1] on [another thread][2] and finally got it all working.  I switch to the `--option=value` format for the commands and used `--execute=&quot;SOURCE ...&quot;` instead of `&lt;` to execute the file.

Also, I never got a good explanation of the difference between `shell_exec()` and `exec()`.


  [1]: https://stackoverflow.com/questions/814586/calling-mysql-exe-using-php-exec-doesnt-work/814722#814722
  [2]: https://stackoverflow.com/questions/814586/calling-mysql-exe-using-php-exec-doesnt-work

Running MySQL *.sql files in PHP

I have two files:

file 1

    dsf
    sdfsd
    dsfsdf

file 2

    ljljlj 
    lkklk 
    dsf
    sdfsd
    dsfsdf



I want to display what is in file 2 but not in file 1, so file 3 should look like

    ljljlj 
    lkklk 


find difference between two text files with one item per line

Make seems to me simply a shell script with slightly easier handling of command line arguments.

Why is it standard to run make instead of ./make.sh

Why use make over a shell script?

On Linux you can convert a date like &quot;2010-10-02&quot; to a unix timestamp in shell script by

    date -d &quot;2010-10-02&quot; &quot;+%s&quot;

Since Mac OS does not have the equivalent **-d** for **date**. How do you go about converting a date to a unix timestamp in a shell script.

How to convert DATE to UNIX TIMESTAMP in shell script on MacOS

I want to represent multiple conditions like this:
  
    if [ ( $g -eq 1 -a &quot;$c&quot; = &quot;123&quot; ) -o ( $g -eq 2 -a &quot;$c&quot; = &quot;456&quot; ) ]   
    then  
        echo abc;  
    else  
        echo efg;   
    fi  

but when I execute the script, it shows

```lang-none
syntax error at line 15: `[&#39; unexpected,
```
 
where line 15 is the one showing if ....  

What is wrong with this condition? I guess something is wrong with the `()`.


How to represent multiple conditions in a shell if statement?

How would you go about telling whether files of a specific extension are present in a directory, with bash?

Something like

    if [ -e *.flac ]; then 
    echo true; 
    fi 

Content Type	Original Author	Original Content on Stackoverflow
Question	Matthew Sowders	View Question on Stackoverflow
Solution 1 - Bash	Dennis Williamson	View Answer on Stackoverflow
Solution 2 - Bash	eumiro	View Answer on Stackoverflow
Solution 3 - Bash	Martin T.	View Answer on Stackoverflow
Solution 4 - Bash	Arc	View Answer on Stackoverflow

How do I get just real time value from 'time' command?

Bash Problem Overview

Bash Solutions

Solution 1 - Bash

Solution 2 - Bash

Solution 3 - Bash

Solution 4 - Bash

php error: Class 'Imagick' not found

Is it possible to declare an array as constant

Attributions