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?

Accepted 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.

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

Solution

Problem