unary operator expected in shell script when comparing null value with string

Question

I have two variables  var    var1 abcd   Here is my shell script code  if    var     var1    then   do something else   do something fi   If I run this code it will prompt a warning         unary operator expected   How can I solve this

User · Answer

Why all people want to use      instead of simple        It is bad habit  It used only in       expression  And in       too  But you may use just   too  It work well in any case  If you use numbers  not strings use not parcing to strings and then compare like strings but compare numbers  like that  let -i i 5   garantee that i is nubmber test  i -eq 5  amp  amp  echo   i is equal 5     echo   i not equal 5    It s match better and quicker  I m expert in C C    Java  JavaScript  But if I use bash i never use      instead      Why you do so

User · Answer

Since the value of  var is the empty string  this   if    var     var1    then   expands to this   if      abcd    then   which is a syntax error   You need to quote the arguments   if     var       var1     then   You can also use   rather than     that s the original syntax  and it s a bit more portable   If you re using bash  you can use the    syntax  which doesn t require the quotes   if     var    var1     then   Even then  it doesn t hurt to quote the variable reference  and adding quotes   if      var      var1      then   might save a future reader a moment trying to remember whether           requires them

[shell] unary operator expected in shell script when comparing null value with string

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