bash Shell 脚本错误：[: 0: 应为一元运算符
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Shell script error: [: 0: unary operator expected
提问by JHarder51
Using this if statment in bash:
使用这个 if 语句bash：
if [ "$TOTAL_LOAD" >= "2" ];then
RESULT=$STATE_WARNING
msg_text="The system load on $HOST is greater than 200% please investigate {$TOTAL_LOAD}"
fi
Getting the error:
得到错误：
line 28: [: 0: unary operator expected
Not seeing the error in my ways. Anyone able to help?
没有以我的方式看到错误。任何人都可以提供帮助？
回答by Alex Howansky
Bash uses different operators for string vs arithmetic comparison. You want -geinstead of >=:
Bash 使用不同的运算符进行字符串与算术比较。你想要-ge而不是>=：
if [ "$TOTAL_LOAD" -ge "2" ];then
回答by Tom Tromey
If I run this by hand I see the error:
如果我手动运行它，我会看到错误：
pokyo. if [ "0" >= "2" ]; then echo hi; fi
bash: [: 0: unary operator expected
The problem is that ">=" is not a valid operator for test. Instead you must write -ge:
问题是 ">=" 不是 的有效运算符test。相反，你必须写-ge：
pokyo. if [ "0" -ge "2" ]; then echo hi; fi
