Regex 如何使用grep获取某些文件名？

我在某个目录中有这样的文件：

my@unix:~/kys$ ls
address_modified_20130312.txt   customer_rows_full_20131202.txt
customer_full_20131201.txt      customer_rows_modified_20131202.txt
customer_modified_20131201.txt
my@unix:~/kys$ 

我想使用grep获取某些以单词“customer”开头的文件名。我试过这个

my@unix:~/kys$ ls | grep customer.*
customer_full_20131201.txt
customer_modified_20131201.txt
customer_rows_full_20131202.txt
customer_rows_modified_20131202.txt
my@unix:~/kys$

但这给了我这些我不想要的customer_行。*文件。正确的结果集是

customer_full_20131201.txt
customer_modified_20131201.txt

如何实现这一点？

您可以尝试：

ls customer_[fm]*

或

您可以尝试：

ls customer_[fm]*

或

您可以尝试：

ls customer_[fm]*

或

您可以尝试：

ls customer_[fm]*

或

---

使用

grep-v

过滤掉你不想要的东西

ls customer* | grep -v '^customer_rows'

---

使用

grep-v

过滤掉你不想要的东西

ls customer* | grep -v '^customer_rows'

---

使用

grep-v

过滤掉你不想要的东西

ls customer* | grep -v '^customer_rows'

---

使用

grep-v

过滤掉你不想要的东西

ls customer* | grep -v '^customer_rows'

---

使用

grep

ls -1 | grep "^customer_[^r].*$"

使用

find

命令

find . \! -iname "customer_rows*"

---

使用

grep

ls -1 | grep "^customer_[^r].*$"

使用

find

命令

find . \! -iname "customer_rows*"

---

使用

grep

ls -1 | grep "^customer_[^r].*$"

使用

find

命令

find . \! -iname "customer_rows*"

---

使用

grep

ls -1 | grep "^customer_[^r].*$"

使用

find

命令

find . \! -iname "customer_rows*"

---

使用Bash扩展的globbing，您可以说

ls customer_!(rows)*

for f in customer_*; do
    case $f in customer_rows* ) continue ;; esac
    : something with "$f"
done

或者，更可能的是，类似于

for f in customer_!(rows)*; do
    : something with "$f"
done

对于POSIX shell或传统Bourne，您可以说

ls customer_!(rows)*

for f in customer_*; do
    case $f in customer_rows* ) continue ;; esac
    : something with "$f"
done

---

使用Bash扩展的globbing，您可以说

ls customer_!(rows)*

for f in customer_*; do
    case $f in customer_rows* ) continue ;; esac
    : something with "$f"
done

或者，更可能的是，类似于

for f in customer_!(rows)*; do
    : something with "$f"
done

对于POSIX shell或传统Bourne，您可以说

ls customer_!(rows)*

for f in customer_*; do
    case $f in customer_rows* ) continue ;; esac
    : something with "$f"
done

---

使用Bash扩展的globbing，您可以说

ls customer_!(rows)*

for f in customer_*; do
    case $f in customer_rows* ) continue ;; esac
    : something with "$f"
done

或者，更可能的是，类似于

for f in customer_!(rows)*; do
    : something with "$f"
done

对于POSIX shell或传统Bourne，您可以说

ls customer_!(rows)*

for f in customer_*; do
    case $f in customer_rows* ) continue ;; esac
    : something with "$f"
done

---

使用Bash扩展的globbing，您可以说

ls customer_!(rows)*

for f in customer_*; do
    case $f in customer_rows* ) continue ;; esac
    : something with "$f"
done

或者，更可能的是，类似于

for f in customer_!(rows)*; do
    : something with "$f"
done

对于POSIX shell或传统Bourne，您可以说

ls customer_!(rows)*

for f in customer_*; do
    case $f in customer_rows* ) continue ;; esac
    : something with "$f"
done