Question

我在每一行都有一个长字符串，如

1000 AS34_59329 RICwdsRSYHSD11-2-IPAAPEK-93 /ifshk5/BC_IP/PROJECT/T11073/T11073_RICekkR/Fq/AS34_59329/111220_I631_FCC0E5EACXX_L4_RICwdsRSYHSD11-2-IPAAPEK-93_1.fq.gz / ifshk5 /BC_IP/PROJECT/T11073/T11073_RICekkR/Fq/AS34_59329/111220_I631_FCC0E5EACXX_L4_RICwdsRSYHSD11-2-IPAAPEK-93_2.fq.gz 的 / ifshk5 / BC_IP / PROJECT / T11073 / T11073_RICekkR / Fq上/ AS34_59329 / clean_111220_I631_FCC0E5EACXX_L4_RICwdsRSYHSD11-2- IPAAPEK-93_1.fq.gz.total.info 11.824 0.981393 43.8283 95.7401确定

此行包含三个文件位置（粗体）。我需要将这些文件scp到另一个位置，例如/ sample。

如何在每行中使用shell来scp文件？

 while read myline  
  do  
    echo "LINE:"$myline 
    scp .......

  done < datafile.list

Answer 1

如果这些是第4,5和6列，您可以执行以下操作：

while read _ _ _ file1 file2 file3 _ 
do  
    scp "$file1" "$file2" "$file3" "user@host:path"
done < datafile.list

Answer 2

使用grep：

# grep each file name
$ grep -o "/[^ ]*" datafile.list
/ifshk5/BC_IP/PROJECT/T11073/T11073_RICekkR/Fq/AS34_59329/111220_I631_FCC0E5EACXX_L4_RICwdsRSYHSD11-2-IPAAPEK-93_1.fq.gz
/ifshk5/BC_IP/PROJECT/T11073/T11073_RICekkR/Fq/AS34_59329/111220_I631_FCC0E5EACXX_L4_RICwdsRSYHSD11-2-IPAAPEK-93_2.fq.gz
/ifshk5/BC_IP/PROJECT/T11073/T11073_RICekkR/Fq/AS34_59329/clean_111220_I631_FCC0E5EACXX_L4_RICwdsRSYHSD11-2-IPAAPEK-93_1.fq.gz.total.info

# Pipe files to scp 
$ grep -o "/[^ ]*" datafile.list | xargs -i% scp % host:/sample

Answer 3

您可以检查字符串中的部分是否为-f的文件，然后再执行scp。

#!/bin/bash
str="1000 AS34_59329 RICwdsRSYHSD11-2-IPAAPEK-93 /ifshk5/BC_IP/PROJECT/T11073/T11073_RICekkR/Fq/AS34_59329/111220_I631_FCC0E5EACXX_L4_RICwdsRSYHSD11-2-IPAAPEK-93_1.fq.gz /ifshk5/BC_IP/PROJECT/T11073/T11073_RICekkR/Fq/AS34_59329/111220_I631_FCC0E5EACXX_L4_RICwdsRSYHSD11-2-IPAAPEK-93_2.fq.gz /ifshk5/BC_IP/PROJECT/T11073/T11073_RICekkR/Fq/AS34_59329/clean_111220_I631_FCC0E5EACXX_L4_RICwdsRSYHSD11-2-IPAAPEK-93_1.fq.gz.total.info 11.824 0.981393 43.8283 95.7401 OK"

for i in $str
do
  if [ -f $i ]; then
    echo $i
    # scp here
  fi
done

Answer 4

如果您的目标是捕获文件名，记录中的单词数量是可变的，并且文件名不理解空格（如您的示例中所示），您可以尝试这样做：

while read -rd ' ' word || [ "$word" ] ; do
    if [ "$word" != "${word#\/}" ] ; then
        echo '$word is the name of a file'
    fi
done < datafile.list

从字符串中提取文件名

4 个答案: