How do I find the number of arguments passed to a Bash script

Question

How do I find the number of arguments passed to a Bash script   This is what I have currently      bin bash i 0 for var in      do   i i 1 done   Are there other  better  ways of doing this

User · Answer

The number of arguments is     Search for it on this page to learn more  http   tldp org LDP abs html internalvariables html ARGLIST

User · Answer

bin bash echo  The number of arguments is      a      echo  The total length of all arguments is     a     count 0 for var in      do     echo  The length of argument   var  is     var          count             accum       var     done echo  The counted number of arguments is   count  echo  The accumulated length of all arguments is   accum

User · Answer

that value is contained in the variable

User · Answer

to add the original reference   You can get the number of arguments from the special parameter     Value of 0 means  no arguments      is read-only    When used in conjunction with shift for argument processing  the special parameter    is decremented each time Bash Builtin shift is executed    see Bash Reference Manual in section 3 4 2 Special Parameters      The shell treats several parameters specially  These parameters may only be referenced   and in this section for keyword     Expands to the number of positional parameters in decimal

User · Answer

Below is the easy one -   cat countvariable sh  echo       awk   for i 0 i lt  NF i     print i-1      Output       countvariable sh 1 2 3 4 5 6 6    countvariable sh 1 2 3 4 5 6 apple orange 8

[bash] How do I find the number of arguments passed to a Bash script?

Examples related to bash

Examples related to arguments