用-z和-n在bash中进行字符串检查 [英] string check in bash with -z and -n
问题描述
我在bash中的测试语法中犯了一个错误，但现在我想了解使用 -n 和 -z 。
我写了以下几行以从 grep 的行中获取 LINENUM 变量。如果找不到该字符串（肯定在文件中只有一个），则 LINENUM 变量为空。
I made a mistake in test syntax in bash but now I want to understand what really happens in string check with -n and -z.
I wrote the following lines to get in LINENUM variable the line number from a grep. When the string is not found (there is only one in the file, for sure), the LINENUM variable is empty.
$ LINENUM=$(grep -w -n mystring myfile | cut -d: -f1)
$ echo --$LINENUM--
----
$ if [ -n $LINENUM ] ; then echo "Checked -n"; fi
Checked -n
$ if [ -z $LINENUM ] ; then echo "Checked -z"; fi
Checked -z
然后我意识到我忘记了双引号，然后进行了以下检查给我：
Then I realized I forgot the double quotes and then the following check gave to me:
$ if [ -n "$LINENUM" ] ; then echo "Checked -n"; fi
$ if [ -z "$LINENUM" ] ; then echo "Checked -z"; fi
Checked -z
因此，在以前的测试中，我忘记了双引号，而不是进行if测试检查，实际上是因为我从 -n 和 -z 获得了两次肯定的检查？
So, in the former tests, where I forgot the double quotes, versus what did the if test check , really, since I got two positive checks from both -n and -z ?
推荐答案
没有引号的情况下，您的测试语句（使用任一运算符，均用 -X ），如果[-X]，则减少为
Without the quotes, your test statement (with either operator, represented with -X here), reduces to
if [ -X ]; then echo "Checked -X"; fi
根据POSIX标准， test <如果参数为非null，则/ code>（现在在此处具有）为true。由于文字字符串 -X 不为空（不再是运算符），因此其计算结果为true。
According to the POSIX standard, the one-argument form of test (which you now have here) is true if the argument is non-null. Since the literal string -X is non-null (it's not an operator anymore), it evaluates to true.
使用引号，如果[-X]，您将获得
With the quotes, you get
if [ -X "" ]; then echo "Checked -X"; fi
由于引号强制为空的第二个参数，因此您使用的是2个参数形式的 test 和 -X （ -n 或 -z ）被正确识别为对第二个null参数起作用的主要运算符。
Since the quotes force an empty 2nd argument, you have a 2-argument form of test, and -X (whether -n or -z) is properly recognized as a primary operator acting on the 2nd, null, argument.
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