Compare integer in bash unary operator expected

Question

The following code gives         -ge  unary operator expected   when   i 0  if    i -ge 2   then      some code fi   why

User · Answer

Your piece of script works just great  Are you sure you are not assigning anything else before the if to  i    A common mistake is also not to leave a space after and before the square brackets

User · Answer

I need to add my 5 cents  I see everybody use   or     but it worth to mention that they are not part of if syntax  For arithmetic comparisons  use         instead           is an arithmetic command  which returns an exit status of 0 if the expression is nonzero  or 1 if the expression is zero  Also used as a synonym for  quot let quot   if side effects  assignments  are needed  See  ArithmeticExpression

User · Answer

Your piece of script works just great  Are you sure you are not assigning anything else before the if to  i    A common mistake is also not to leave a space after and before the square brackets

User · Answer

Judging from the error message the value of i was the empty string when you executed it  not 0

User · Answer

Your problem arises from the fact that  i has a blank value when your statement fails   Always quote your variables when performing comparisons if there is the slightest chance that one of them may be empty  e g    if     i  -ge 2     then       fi   This is because of how the shell treats variables   Assume the original example   if    i -ge 2     then       The first thing that the shell does when executing that particular line of code is substitute the value of  i  just like your favorite editor s search  amp  replace function would   So assume that  i is empty or  even more illustrative  assume that  i is a bunch of spaces   The shell will replace  i as follows   if       -ge 2     then       Now that variable substitutions are done  the shell proceeds with the comparison and     fails because it cannot see anything intelligible to the left of -gt   However  quoting  i   if     i  -ge 2     then       becomes   if          -ge 2     then       The shell now sees the double-quotes  and knows that you are actually comparing four blanks to 2 and will skip the if   You also have the option of specifying a default value for  i if  i is blank  as follows   if      i -0   -ge 2     then       This will substitute the value 0 instead of  i is  i is undefined   I still maintain the quotes because  again  if  i is a bunch of blanks then it does not count as undefined  it will not be replaced with 0  and you will run into the problem once again   Please read this when you have the time   The shell is treated like a black box by many  but it operates with very few and very simple rules - once you are aware of what those rules are  one of them being how variables work in the shell  as explained above  the shell will have no more secrets for you

User · Answer

Your piece of script works just great  Are you sure you are not assigning anything else before the if to  i    A common mistake is also not to leave a space after and before the square brackets

User · Answer

Your piece of script works just great  Are you sure you are not assigning anything else before the if to  i    A common mistake is also not to leave a space after and before the square brackets

User · Answer

Judging from the error message the value of i was the empty string when you executed it  not 0

User · Answer

Your problem arises from the fact that  i has a blank value when your statement fails   Always quote your variables when performing comparisons if there is the slightest chance that one of them may be empty  e g    if     i  -ge 2     then       fi   This is because of how the shell treats variables   Assume the original example   if    i -ge 2     then       The first thing that the shell does when executing that particular line of code is substitute the value of  i  just like your favorite editor s search  amp  replace function would   So assume that  i is empty or  even more illustrative  assume that  i is a bunch of spaces   The shell will replace  i as follows   if       -ge 2     then       Now that variable substitutions are done  the shell proceeds with the comparison and     fails because it cannot see anything intelligible to the left of -gt   However  quoting  i   if     i  -ge 2     then       becomes   if          -ge 2     then       The shell now sees the double-quotes  and knows that you are actually comparing four blanks to 2 and will skip the if   You also have the option of specifying a default value for  i if  i is blank  as follows   if      i -0   -ge 2     then       This will substitute the value 0 instead of  i is  i is undefined   I still maintain the quotes because  again  if  i is a bunch of blanks then it does not count as undefined  it will not be replaced with 0  and you will run into the problem once again   Please read this when you have the time   The shell is treated like a black box by many  but it operates with very few and very simple rules - once you are aware of what those rules are  one of them being how variables work in the shell  as explained above  the shell will have no more secrets for you

User · Answer

Judging from the error message the value of i was the empty string when you executed it  not 0

User · Answer

I need to add my 5 cents  I see everybody use   or     but it worth to mention that they are not part of if syntax  For arithmetic comparisons  use         instead           is an arithmetic command  which returns an exit status of 0 if the expression is nonzero  or 1 if the expression is zero  Also used as a synonym for  quot let quot   if side effects  assignments  are needed  See  ArithmeticExpression

User · Answer

Judging from the error message the value of i was the empty string when you executed it  not 0

[linux] Compare integer in bash, unary operator expected

Examples related to linux

Examples related to bash

Examples related to shell