Answer

POSIX compatible:

command -v <the_command>

Example use:

if ! command -v <the_command> &> /dev/null
then
    echo "<the_command> could not be found"
    exit
fi

For Bash specific environments:

hash <the_command> # For regular commands. Or...
type <the_command> # To check built-ins and keywords

Explanation

Avoid which. Not only is it an external process you're launching for doing very little (meaning builtins like hash, type or command are way cheaper), you can also rely on the builtins to actually do what you want, while the effects of external commands can easily vary from system to system.

Why care?

• Many operating systems have a which that doesn't even set an exit status, meaning the if which foo won't even work there and will always report that foo exists, even if it doesn't (note that some POSIX shells appear to do this for hash too).
• Many operating systems make which do custom and evil stuff like change the output or even hook into the package manager.

So, don't use which. Instead use one of these:

command -v foo >/dev/null 2>&1 || { echo >&2 "I require foo but it's not installed.  Aborting."; exit 1; }

type foo >/dev/null 2>&1 || { echo >&2 "I require foo but it's not installed.  Aborting."; exit 1; }

hash foo 2>/dev/null || { echo >&2 "I require foo but it's not installed.  Aborting."; exit 1; }

(Minor side-note: some will suggest 2>&- is the same 2>/dev/null but shorter – this is untrue. 2>&- closes FD 2 which causes an error in the program when it tries to write to stderr, which is very different from successfully writing to it and discarding the output (and dangerous!))

If your hash bang is /bin/sh then you should care about what POSIX says. type and hash's exit codes aren't terribly well defined by POSIX, and hash is seen to exit successfully when the command doesn't exist (haven't seen this with type yet). command's exit status is well defined by POSIX, so that one is probably the safest to use.

If your script uses bash though, POSIX rules don't really matter anymore and both type and hash become perfectly safe to use. type now has a -P to search just the PATH and hash has the side-effect that the command's location will be hashed (for faster lookup next time you use it), which is usually a good thing since you probably check for its existence in order to actually use it.

As a simple example, here's a function that runs gdate if it exists, otherwise date:

gnudate() {
    if hash gdate 2>/dev/null; then
        gdate "$@"
    else
        date "$@"
    fi
}

Alternative with a complete feature set

You can use scripts-common to reach your need.

To check if something is installed, you can do:

checkBin <the_command> || errorMessage "This tool requires <the_command>. Install it please, and then run this tool again."

I agree with lhunath to discourage use of which, and his solution is perfectly valid for Bash users. However, to be more portable, command -v shall be used instead:

$ command -v foo >/dev/null 2>&1 || { echo "I require foo but it's not installed.  Aborting." >&2; exit 1; }

Command command is POSIX compliant. See here for its specification: command - execute a simple command

Note: type is POSIX compliant, but type -P is not.

It depends on whether you want to know whether it exists in one of the directories in the $PATH variable or whether you know the absolute location of it. If you want to know if it is in the $PATH variable, use

if which programname >/dev/null; then
    echo exists
else
    echo does not exist
fi

otherwise use

if [ -x /path/to/programname ]; then
    echo exists
else
    echo does not exist
fi

The redirection to /dev/null/ in the first example suppresses the output of the which program.

Try using:

test -x filename

or

[ -x filename ]

From the Bash manpage under Conditional Expressions:

 -x file
          True if file exists and is executable.

There are a ton of options here, but I was surprised no quick one-liners. This is what I used at the beginning of my scripts:

[[ "$(command -v mvn)" ]] || { echo "mvn is not installed" 1>&2 ; exit 1; }
[[ "$(command -v java)" ]] || { echo "java is not installed" 1>&2 ; exit 1; }

This is based on the selected answer here and another source.

Command -v works fine if the POSIX_BUILTINS option is set for the <command> to test for, but it can fail if not. (It has worked for me for years, but I recently ran into one where it didn't work.)

I find the following to be more failproof:

test -x "$(which <command>)"

Since it tests for three things: path, existence and execution permission.

I never did get the previous answers to work on the box I have access to. For one, type has been installed (doing what more does). So the builtin directive is needed. This command works for me:

if [ `builtin type -p vim` ]; then echo "TRUE"; else echo "FALSE"; fi

I wanted the same question answered but to run within a Makefile.

install:
    @if [[ ! -x "$(shell command -v ghead)" ]]; then \
        echo 'ghead does not exist. Please install it.'; \
        exit -1; \
    fi

It could be simpler, just:

#!/usr/bin/env bash                                                                
set -x                                                                             

# if local program 'foo' returns 1 (doesn't exist) then...                                                                               
if ! type -P foo; then                                                             
    echo 'crap, no foo'                                                            
else                                                                               
    echo 'sweet, we have foo!'                                                    
fi                                                                                 

Change foo to vi to get the other condition to fire.

The which command might be useful. man which

It returns 0 if the executable is found and returns 1 if it's not found or not executable:

NAME

       which - locate a command

SYNOPSIS

       which [-a] filename ...

DESCRIPTION

       which returns the pathnames of the files which would
       be executed in the current environment, had its
       arguments been given as commands in a strictly
       POSIX-conformant shell. It does this by searching
       the PATH for executable files matching the names
       of the arguments.

OPTIONS

       -a     print all matching pathnames of each argument

EXIT STATUS

       0      if all specified commands are 
              found and executable

       1      if one or more specified commands is nonexistent
              or not executable

       2      if an invalid option is specified

The nice thing about which is that it figures out if the executable is available in the environment that which is run in - it saves a few problems...

Use Bash builtins if you can:

which programname

...

type -P programname

zsh only, but very useful for zsh scripting (e.g. when writing completion scripts):

The zsh/parameter module gives access to, among other things, the internal commands hash table. From man zshmodules:

THE ZSH/PARAMETER MODULE
       The zsh/parameter module gives access to some of the internal hash  ta‐
       bles used by the shell by defining some special parameters.


[...]

       commands
              This  array gives access to the command hash table. The keys are
              the names of external commands, the values are the pathnames  of
              the  files  that would be executed when the command would be in‐
              voked. Setting a key in this array defines a new entry  in  this
              table  in the same way as with the hash builtin. Unsetting a key
              as in `unset "commands[foo]"' removes the entry  for  the  given
              key from the command hash table.

Although it is a loadable module, it seems to be loaded by default, as long as zsh is not used with --emulate.

example:

martin@martin ~ % echo $commands[zsh]
/usr/bin/zsh

To quickly check whether a certain command is available, just check if the key exists in the hash:

if (( ${+commands[zsh]} ))
then
  echo "zsh is available"
fi

Note though that the hash will contain any files in $PATH folders, regardless of whether they are executable or not. To be absolutely sure, you have to spend a stat call on that:

if (( ${+commands[zsh]} )) && [[ -x $commands[zsh] ]]
then
  echo "zsh is available"
fi

For those interested, none of the methodologies in previous answers work if you wish to detect an installed library. I imagine you are left either with physically checking the path (potentially for header files and such), or something like this (if you are on a Debian-based distribution):

dpkg --status libdb-dev | grep -q not-installed

if [ $? -eq 0 ]; then
    apt-get install libdb-dev
fi

As you can see from the above, a "0" answer from the query means the package is not installed. This is a function of "grep" - a "0" means a match was found, a "1" means no match was found.

This will tell according to the location if the program exist or not:

    if [ -x /usr/bin/yum ]; then
        echo "This is Centos"
    fi

I second the use of "command -v". E.g. like this:

md=$(command -v mkdirhier) ; alias md=${md:=mkdir}  # bash

emacs="$(command -v emacs) -nw" || emacs=nano
alias e=$emacs
[[ -z $(command -v jed) ]] && alias jed=$emacs

#!/bin/bash
a=${apt-cache show program}
if [[ $a == 0 ]]
then
echo "the program doesn't exist"
else
echo "the program exists"
fi

#program is not literal, you can change it to the program's name you want to check

Assuming you are already following safe shell practices:

set -eu -o pipefail
shopt -s failglob

./dummy --version 2>&1 >/dev/null

This assumes the command can be invoked in such a way that it does (almost) nothing, like reporting its version or showing help.

If the dummy command is not found, Bash exits with the following error...

./my-script: line 8: dummy: command not found

This is more useful and less verbose than the other command -v (and similar) answers because the error message is auto generated and also contains a relevant line number.

Late answer but this is what I ended up doing.

I just check if the command I execute returns an error code. If it returns 0 it means program is installed. Moreover you can use this to check the output of a script. Take for instance this script.

foo.sh

#!/bin/bash
echo "hello world"
exit 1 # throw some error code

Examples:

# outputs something bad... and exits
bash foo.sh $? -eq 0 || echo "something bad happened. not installed" ; exit 1

# does NOT outputs nothing nor exits because dotnet is installed on my machine
dotnet --version $? -eq 0 || echo "something bad happened. not installed" ; exit 1

Basically all this is doing is checking the exit code of a command run. the most accepted answer on this question will return true even if the command exit code is not 0.

I couldn't get one of the solutions to work, but after editing it a little I came up with this. Which works for me:

dpkg --get-selections | grep -q linux-headers-$(uname -r)

if [ $? -eq 1 ]; then
        apt-get install linux-headers-$(uname -r)
fi

I would just try and call the program with for example --version or --help and check if the command succeeded or failed

Used with set -e, the script will exit if the program is not found, and you will get a meaningful error message:

#!/bin/bash
set -e
git --version >> /dev/null