如何在bash中从“name1 11/22 name2 33 / 44 name3 last3 55/66”中提取名称

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英文:

How to extract the names from a "name1 11/22 name2 33 / 44 name3 last3 55/66" in bash

问题

My input data formatted as the pattern '<name> <a>/<b>'. It is possible occurs zero or many times in the same line. And it is possible exist a extra space between the '/'.

I expect to extract the names as

  1. name1
  2. name2
  3. name3 last3

Here is the wrong code

  1. echo "name1 11/22 name2 33 / 44 name3 last3 55/66" \
  2.  | grep -o -E '\b[a-zA-Z][a-zA-Z0-9. ]+\b'))

will extract

  1. name1 11
  2. name2 33
  3. name3 last3 55

This script should also pass empty line as no output.

英文:

My input data formatted as the pattern '<name> <a>/<b>'. It is possible occurs zero or many times in the same line. And it is possible exist a extra space between the '/'.

I expect to extract the names as

  1. name1
  2. name2
  3. name3 last3

Here is the wrong code

  1. echo "name1 11/22 name2 33 / 44 name3 last3 55/66" \
  2.  | grep -o -E '\b[a-zA-Z][a-zA-Z0-9. ]+\b'))

will extract

  1. name1 11
  2. name2 33
  3. name3 last3 55

This script should also pass empty line as no output.

答案1

得分: 1

name1
name2
name3 last3

英文: 
  1. $ echo "name1 11/22 name2 33 / 44 name3 last3 55/66" | 
  2.     awk -F' *[0-9]* */ *[0-9]* *' '{for(i=1;i<NF;i++) print $i}'
  3. name1
  4. name2
  5. name3 last3
  7. $ echo "name1 11/22 name2 33 / 44 name3 last3 55/66" | 
  8.     awk -F' *[0-9]* */ *[0-9]* *' -v OFS='\n' '{NF-=1}1'
  9. name1
  10. name2
  11. name3 last3

答案2

得分: 0

使用 grep，您可以轻松提取<a>/<b>部分：

  1. $ echo "name1 11/22 name2 33 / 44 name3 last3 55/66" |
  2.   grep -oE '[^[:space:]]+[[:space:]]*/[[:space:]]*[^[:space:]]+'
  3. 11/22
  4. 33 / 44
  5. 55/66

但如果您想要的不是打印这些内容，而是用换行符替换它们，sed或awk可能是更好的选择。使用 sed 的示例：

  1. $ echo "name1 11/22 name2 33 / 44 name3 last3 55/66" |
  2.   sed 's![[:space:]]*[^[:space:]][^[:space:]]*[[:space:]]*/[[:space:]]*[^[:space:]][^[:space:]]*[[:space:]]*!\n!g'
  3. name1
  4. name2
  5. name3 last3

或者，使用 GNU sed：

  1. $ echo "name1 11/22 name2 33 / 44 name3 last3 55/66" |
  2.   sed -E 's!\s*\S+\s*/\s*\S+\s*!\n!g'
  3. name1
  4. name2
  5. name3 last3

请注意，在每行的最后一个名称后也会添加一个换行符，导致输出中有空行。如果这不可接受，我们可以单独处理每行的最后一个名称。使用 GNU sed 的示例：

  1. $ echo "name1 11/22 name2 33 / 44 name3 last3 55/66" |
  2.   sed -E 's!\s*\S+\s*/\s*\S+\s*$!!;s!\s*\S+\s*/\s*\S+\s*!\n!g'
  3. name1
  4. name2
  5. name3 last3

使用 awk，我们可以将字段分隔符定义为您要删除的<a>/<b>部分，并在单独的行上打印所有字段（除了最后一个空字段）：

  1. $ echo "name1 11/22 name2 33 / 44 name3 last3 55/66" |
  2.   awk -F '[[:space:]]*[^[:space:]]+[[:space:]]*/[[:space:]]*[^[:space:]]+[[:space:]]*' '{for(i=1;i<NF;i++) print $i}'
  3. name1
  4. name2
  5. name3 last3

或者，使用 GNU awk：

  1. $ echo "name1 11/22 name2 33 / 44 name3 last3 55/66" |
  2.   awk -F '\\s*\\S+\\s*/\\s*\\S+\\s*' '{for(i=1;i<NF;i++) print $i}'
  3. name1
  4. name2
  5. name3 last3
英文:

With grep you could easily extract the &lt;a&gt;/&lt;b&gt; parts:

  1. $ echo &quot;name1 11/22 name2 33 / 44 name3 last3 55/66&quot; |
  2.   grep -oE &#39;[^[:space:]]+[[:space:]]*/[[:space:]]*[^[:space:]]+&#39;
  3. 11/22
  4. 33 / 44
  5. 55/66

But as what you want is not to print these but replace them with newlines, sed or awk are probably better choices. Example with sed:

  1. $ echo &quot;name1 11/22 name2 33 / 44 name3 last3 55/66&quot; |
  2.   sed &#39;s![[:space:]]*[^[:space:]][^[:space:]]*[[:space:]]*/[[:space:]]*[^[:space:]][^[:space:]]*[[:space:]]*!\n!g&#39;
  3. name1
  4. name2
  5. name3 last3

Or, with GNU sed:

  1. $ echo &quot;name1 11/22 name2 33 / 44 name3 last3 55/66&quot; |
  2.   sed -E &#39;s!\s*\S+\s*/\s*\S+\s*!\n!g&#39;
  3. name1
  4. name2
  5. name3 last3

Note that a newline is also added after the last name of a line, leading to empty lines in the output. If this is not acceptable we can process the last name of a line separately. Example with GNU sed:

  1. $ echo &quot;name1 11/22 name2 33 / 44 name3 last3 55/66&quot; |
  2.   sed -E &#39;s!\s*\S+\s*/\s*\S+\s*$!!;s!\s*\S+\s*/\s*\S+\s*!\n!g&#39;
  3. name1
  4. name2
  5. name3 last3

With awk we can define the field separator as the &lt;a&gt;/&lt;b&gt; parts you want to remove and print all fields (except the last empty field) on a separate line:

  1. $ echo &quot;name1 11/22 name2 33 / 44 name3 last3 55/66&quot; |
  2.   awk -F &#39;[[:space:]]*[^[:space:]]+[[:space:]]*/[[:space:]]*[^[:space:]]+[[:space:]]*&#39; &#39;
  3.     {for(i=1;i&lt;NF;i++) print $i}&#39;
  4. name1
  5. name2
  6. name3 last3

Or, with GNU awk:

  1. $ echo &quot;name1 11/22 name2 33 / 44 name3 last3 55/66&quot; |
  2.   awk -F &#39;\\s*\\S+\\s*/\\s*\\S+\\s*&#39; &#39;{for(i=1;i&lt;NF;i++) print $i}&#39;
  3. name1
  4. name2
  5. name3 last3

答案3

得分: 0

  1. 如果您的 `grep` 支持 `-P`（PCRE）选项，请尝试：
  3. ```shell
  4. echo "name1 11/22 name2 33 / 44 name3 last3 55/66" \
  5.  | grep -oP '\b[a-zA-Z][a-zA-Z0-9. ]+\b(?=\s+\d+\s*/\s*\d+)'

输出：

  1. name1
  2. name2
  3. name3 last3

(?=\s+\d+\s*/\s*\d+) 是前瞻断言，用于匹配以下序列：

  • 一个或多个空白字符
  • 一个或多个数字
  • 零个或多个空白字符
  • 一个斜杠字符
  • 零个或多个空白字符
  • 一个或多个数字

匹配的子字符串不包括在输出中。

  2. <details>
  3. <summary>英文:</summary>
  5. If your `grep` supports `-P` (PCRE) option, would you please try:

echo "name1 11/22 name2 33 / 44 name3 last3 55/66"
| grep -oP '\b[a-zA-Z][a-zA-Z0-9. ]+\b(?=\s+\d+\s*/\s*\d+)'

  1. Output:

name1
name2
name3 last3

  1. `(?=\s+\d+\s*/\s*\d+)` is the lookahead assertion which matches
  2. a sequence of:
  3. - one or more blank character(s)
  4. - one or more digit(s)
  5. - zero or more blank character(s)
  6. - a slash character
  7. - zero or more blank character(s)
  8. - one or more digit(s)
  10. The matched substring is not included in the output.
  12. </details>
  16. # 答案4
  17. **得分**: 0

使用GNU awk来处理多字符的RS、RT和\s/\S，这可能是您想要的：

$ echo "name1 11/22 name2 33 / 44 name3 last3 55/66" |
awk -v RS='\s+\S+\s*/\s*\S+\s*' 'RT'
name1
name2
name3 last3

  2. <details>
  3. <summary>英文:</summary>
  5. Using GNU awk for multi-char `RS`, `RT`, and `\s/\S`, this might be what you want:
  7.     $ echo &quot;name1 11/22 name2 33 / 44 name3 last3 55/66&quot; | 
  8.         awk -v RS=&#39;\\s+\\S+\\s*/\\s*\\S+\\s*&#39; &#39;RT&#39;
  9.     name1
  10.     name2
  11.     name3 last3
  15. </details>

huangapple
  • 本文由 发表于 2023年7月20日 13:30:33
  • 转载请务必保留本文链接：https://go.coder-hub.com/76726920.html
  • bash
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