How would i parse a multiline stdout string with a variable in bash script, i am getting error "unary operator expected"

I have a bash script:

#!/bin/bash
JAVA_VERSION="1.6.0_17"
_STDOUT=`java -version`

if [ $JAVA_VERSION = $_STDOUT ]; then
        echo "Matched"
else
        echo "Not Matched"
fi

i get the result:

java version "1.6.0_17"
OpenJDK Runtime Environment (IcedTea6 1.7.5) (rhel-1.16.b17.el5-x86_64)
OpenJDK 64-Bit Server VM (build 14.0-b16, mixed mode)
t4.sh: line 8: [: 1.6.0_17: unary operator expected
Not Matched

How would i match $JAVA_VERSION with $_STDOUT when $_开发者_开发技巧STDOUT has multiple lines

You have a few problems.

1. It appears java -version puts its output on STDERR, not STDOUT, so you'll have to redirect STDERR to STDOUT to parse it.
2. You need to match the double quotes literally. With JAVA_VERSION="1.6.0_17" the shell will remove the quotes, you can wrap the double quotes in single quotes to make them literal.
3. Finally, if you're going to use bash you should be using [[ ]] and not [ ]. The latter is actually a synonym to the test builtin and the former is native syntax that allows for more capability; one of which is you don't need to quote the variables inside.

.

#!/bin/bash

JAVA_VERSION='"1.6.0_17"'
_STDOUT=$(java -version 2>&1 | awk 'NR==1{print $3}')

if [[ $JAVA_VERSION = $_STDOUT ]]; then
  echo "Matched"
else
  echo "Not Matched"
fi

Proof of Concept

$ java -version
java version "1.6.0_20"
Java(TM) SE Runtime Environment (build 1.6.0_20-b02)
Java HotSpot(TM) Client VM (build 16.3-b01, mixed mode, sharing)

$ JAVA_VERSION='"1.6.0_20"'; _STDOUT=$(java -version 2>&1 | awk 'NR==1{print $3}'); if [[ $JAVA_VERSION = $_STDOUT ]]; then echo "Matched"; else echo "Not Matched"; fi
Matched

$ JAVA_VERSION='"1.6.0_19"'; _STDOUT=$(java -version 2>&1 | awk 'NR==1{print $3}'); if [[ $JAVA_VERSION = $_STDOUT ]]; then echo "Matched"; else echo "Not Matched"; fi
Not Matched

You can use Bash's inbuilt comparison checker to see if a string is contained within another string. So you don't need to pipe into awk or cut.

JAVA_VERSION=1.6.0_17
_STDOUT=`java -version 2>&1`
if [[ $_STDOUT == *$JAVA_VERSION* ]]; then
     echo "Matched"
else
     echo "Not Matched"
fi

Use quotes.

#!/bin/bash
JAVA_VERSION="1.6.0_17"
_STDOUT=`java -version`

if [ "$JAVA_VERSION" = "$_STDOUT" ]; then
        echo "Matched"
else
        echo "Not Matched"
fi

The problem is not of having multiple lines -- even if the output had appeared in a single line, it wouldn't have matched. You made a good try, but what that code does is like comparing an apple with a basket of fruits. So, we need to isolate the apple from that basket :)

This is how we do it:

Note: Corrected the code as per SiegeX's comments

#!/bin/bash
JAVA_VERSION="1.6.0_17"
_STDOUT=`java -version 2>&1 | grep "java version" | cut -d'"' -f2` # Just extract the version

if [[ "$JAVA_VERSION" = "$_STDOUT" ]]; then
        echo "Matched"
else
        echo "Not Matched"
fi