如何用点“。”精确地在数组中“grep”？

3  

It sounds like you want:

这听起来像你想要的:

• literal string matching

文字字符串匹配

• against entire input lines.

针对整个输入线。

Therefore, use -F (for literal matching) and -x for whole-line matching:

因此,使用-F(用于文字匹配)和-x用于整行匹配:

array=(
  aaa.bbb.ccc
  ccc.ddd.aaa
)

printf '%s\n' "${array[@]}" | grep -Fx "$1"

Now, only literal strings aaa.bbb.ccc and ccc.ddd.aaa will match, not substrings such as aaa.
Since you're matching the entire line, use of -o (to only output the matching part of each line) is unnecessary.

现在,只有文字字符串aaa.bbb.ccc和ccc.ddd.aaa匹配,而不是像aaa这样的子字符串。由于您匹配整行,因此不需要使用-o(仅输出每行的匹配部分)。

Also note how printf '%s\n' "${array[@]}" must be used to print each array element on its own line, whereas echo ${array[@]} prints the elements on a single line (with whitespace normalization due to word-splitting and potential globbing applied, because ${array[@]} is not enclosed in "...").

另请注意printf'%s \ n'“$ {array [@]}”必须如何用于在自己的行上打印每个数组元素,而echo $ {array [@]}在单行上打印元素(带有由于分词和应用潜在的globbing而导致的空白归一化,因为$ {array [@]}未包含在“...”中。

The problem with your approach:

您的方法存在问题:

• grep defaults to (basic) regular expressions, so a $1 value containing . would result in that . being interpreted to mean "any single character".

  grep默认为(基本)正则表达式,因此包含的$ 1值。会导致这种情况。被解释为“任何单个字符”。

• echo ${array[@]} is problematic due to potentially unwanted whitespace normalization and globbing, as stated; echo "${array[@]}" | grep -Fwo "$1" would be an improvement, but bears a slight risk of false positives.

  如上所述,echo $ {array [@]}是有问题的,因为可能不需要的空格规范化和通化。 echo“$ {array [@]}”| grep -Fwo“$ 1”将是一个改进,但具有误报的轻微风险。

0  

Escape the . with a preceding \, and it won't match aa but will match a.a:

逃避。使用前面的\,它将不匹配aa但会匹配a.a:

echo a.a | grep '\.'

Output:

a.a

To show this works with positional parameters, here's a little function that behaves much like the previous code:

为了证明这适用于位置参数,这里有一个小功能,其行为与前面的代码非常相似:

foo() { echo "$2" | grep "$1" ; }
foo '\.' a.a

Output:

a.a

3  

It sounds like you want:

这听起来像你想要的:

• literal string matching

文字字符串匹配

• against entire input lines.

针对整个输入线。

Therefore, use -F (for literal matching) and -x for whole-line matching:

因此,使用-F(用于文字匹配)和-x用于整行匹配:

array=(
  aaa.bbb.ccc
  ccc.ddd.aaa
)

printf '%s\n' "${array[@]}" | grep -Fx "$1"

Now, only literal strings aaa.bbb.ccc and ccc.ddd.aaa will match, not substrings such as aaa.
Since you're matching the entire line, use of -o (to only output the matching part of each line) is unnecessary.

现在,只有文字字符串aaa.bbb.ccc和ccc.ddd.aaa匹配,而不是像aaa这样的子字符串。由于您匹配整行,因此不需要使用-o(仅输出每行的匹配部分)。

Also note how printf '%s\n' "${array[@]}" must be used to print each array element on its own line, whereas echo ${array[@]} prints the elements on a single line (with whitespace normalization due to word-splitting and potential globbing applied, because ${array[@]} is not enclosed in "...").

另请注意printf'%s \ n'“$ {array [@]}”必须如何用于在自己的行上打印每个数组元素,而echo $ {array [@]}在单行上打印元素(带有由于分词和应用潜在的globbing而导致的空白归一化,因为$ {array [@]}未包含在“...”中。

The problem with your approach:

您的方法存在问题:

• grep defaults to (basic) regular expressions, so a $1 value containing . would result in that . being interpreted to mean "any single character".

  grep默认为(基本)正则表达式,因此包含的$ 1值。会导致这种情况。被解释为“任何单个字符”。

• echo ${array[@]} is problematic due to potentially unwanted whitespace normalization and globbing, as stated; echo "${array[@]}" | grep -Fwo "$1" would be an improvement, but bears a slight risk of false positives.

  如上所述,echo $ {array [@]}是有问题的,因为可能不需要的空格规范化和通化。 echo“$ {array [@]}”| grep -Fwo“$ 1”将是一个改进,但具有误报的轻微风险。

0  

Escape the . with a preceding \, and it won't match aa but will match a.a:

逃避。使用前面的\,它将不匹配aa但会匹配a.a:

echo a.a | grep '\.'

Output:

a.a

To show this works with positional parameters, here's a little function that behaves much like the previous code:

为了证明这适用于位置参数,这里有一个小功能,其行为与前面的代码非常相似:

foo() { echo "$2" | grep "$1" ; }
foo '\.' a.a

Output:

a.a