How do I escape the wildcard asterisk character in bash

Question

For example   me  FOO  BAR   BAR  me  echo  FOO BAR file1 file2 file3 file4 BAR   and using the   escape character   me  FOO  BAR    BAR  me  echo  FOO BAR    BAR   I m obviously doing something stupid   How do I get the output BAR   BAR

User · Answer

SHORT ANSWER  Like others have said - you should always quote the variables to prevent strange behaviour   So use echo   foo  in instead of just echo  foo   LONG ANSWER  I do think this example warrants further explanation because there is more going on than it might seem on the face of it   I can see where your confusion comes in because after you ran your first example you probably thought to yourself that the shell is obviously doing    Parameter expansion Filename expansion   So from your first example   me  FOO  BAR   BAR  me  echo  FOO   After parameter expansion is equivalent to   me  echo BAR   BAR   And after filename expansion is equivalent to   me  echo BAR file1 file2 file3 file4 BAR   And if you just type echo BAR   BAR into the command line you will see that they are equivalent   So you probably thought to yourself  if I escape the    I can prevent the filename expansion   So from your second example   me  FOO  BAR    BAR  me  echo  FOO   After parameter expansion should be equivalent to   me  echo BAR    BAR   And after filename expansion should be equivalent to   me  echo BAR    BAR   And if you try typing  echo BAR    BAR  directly into the command line it will indeed print  BAR   BAR  because the filename expansion is prevented by the escape   So why did using  foo not work   It s because there is a third expansion that takes place - Quote Removal  From the bash manual quote removal is      After the preceding expansions  all   unquoted occurrences of the characters                     and         that did not result   from one of the above expansions are   removed    So what happens is when you type the command directly into the command line  the escape character is not the result of a previous expansion so BASH removes it before sending it to the echo command  but in the 2nd example  the      was the result of a previous Parameter expansion  so it is NOT removed   As a result  echo receives      and that s what it prints   Note the difference between the first example -     is not included in the characters that will be removed by Quote Removal   I hope this makes sense   In the end the conclusion in the same - just use quotes   I just thought I d explain why escaping  which logically should work if only Parameter and Filename expansion are at play  didn t work   For a full explanation of BASH expansions  refer to   http   www gnu org software bash manual bashref html Shell-Expansions

User · Answer

If you don t want to bother with weird expansions from bash you can do this me  FOO  quot BAR  x2A BAR quot      2A is hex code for   me  echo -e  FOO BAR   BAR me    Explanation here why using -e option of echo makes life easier  Relevant quote from man here  SYNOPSIS    echo  SHORT-OPTION      STRING        echo LONG-OPTION  DESCRIPTION    Echo the STRING s  to standard output      -n     do not output the trailing newline     -e     enable interpretation of backslash escapes     -E     disable interpretation of backslash escapes  default      --help display this help and exit     --version           output version information and exit     If -e is in effect  the following sequences are recognized             backslash              0NNN  byte with octal value NNN  1 to 3 digits       xHH   byte with hexadecimal value HH  1 to 2 digits   For the hex code you can check man ascii page  first line in octal  second decimal  third hex      051   41    29                                151   105   69    i    052   42    2A                                152   106   6A    j    053   43    2B                                153   107   6B    k

User · Answer

echo   FOO

User · Answer

It may be worth getting into the habit of using printf rather then echo on the command line   In this example it doesn t give much benefit but it can be more useful with more complex output   FOO  BAR   BAR  printf  s   FOO

User · Answer

FOO  BAR   BAR  echo   FOO

User · Answer

Quoting when setting  FOO is not enough   You need to quote the variable reference as well   me  FOO  BAR   BAR  me  echo   FOO  BAR   BAR

User · Answer

I ll add a bit to this old thread   Usually you would use    echo   FOO    However  I ve had problems even with this syntax  Consider the following script      bin bash curl opts  -s --noproxy   -O  curl  curl opts   1    The   needs to be passed verbatim to curl  but the same problems will arise  The above example won t work  it will expand to filenames in the current directory  and neither will     You also can t quote  curl opts because it will be recognized as a single  invalid  option to curl   curl  option -s --noproxy   -O  is unknown curl  try  curl --help  or  curl --manual  for more information   Therefore I would recommend the use of the bash variable  GLOBIGNORE to prevent filename expansion altogether if applied to the global pattern  or use the set -f built-in flag      bin bash GLOBIGNORE     curl opts  -s --noproxy   -O  curl  curl opts   1      no filename expansion   Applying to your original example   me  FOO  BAR   BAR   me  echo  FOO BAR file1 file2 file3 file4 BAR  me  set -f me  echo  FOO BAR   BAR  me  set  f me  GLOBIGNORE   me  echo  FOO BAR   BAR

[bash] How do I escape the wildcard/asterisk character in bash?

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