Question

I have these if conditions, but it has compile errors. How can I fix it?

if [ $DEVICE_ID == "" ]; then

I get error:

line 63: [: ==: unary operator expected


if [ 'ls -l Mytest*.log | wc -l' -eq 1 ]; then

i get error:

line 68: [: ls -l Kernel*.log | wc -l: integer expression expected

Answer 1

Quote the variable:

if [ "$DEVICE_ID" == "" ]; then

But it would be better to do:

if [ -z "$DEVICE_ID" ];

The second error is that you need to use backquotes:

if [ $(ls -l Mytest*.log | wc -l) -eq 1 ]; then