案例1-合并2个不同文件夹中的csv文件到另外一个目录，相同的文件名进行数据合并，不同的文件名直接移到新文件夹

#!/bin/bash

filenum=30
dir1=./upload_root/gdby
dir2=./upload_root/bzbygz
dir_merge=./download

#删除旧文件
rm -f $dir_merge/*.csv


dir1_select=`ls $dir1 | sort -r | head -$filenum`
dir2_select=`ls $dir2 | sort -r | head -$filenum`


#2个文件夹共有的
echo "两个学校共有的文件名"

for i in ${dir1_select}
do
    for file2 in ${dir2_select}
    do
        if [ "$i" = "$file2" ];then
        echo $i
        num1=$(head -1 $dir1/$i | awk -F "," ‘{print $2}‘)
        num2=$(head -1 $dir2/$i | awk -F "," ‘{print $2}‘)
        money1=$(head -1 $dir1/$i | awk -F "," ‘{print $4}‘)
        money2=$(head -1 $dir2/$i | awk -F "," ‘{print $4}‘)


        num_all=$(($num1+$num2))
        money_all=$(echo $money1 $money2 | awk ‘{print $1+$2}‘)

        cp $dir1/$i $dir_merge
        sed -ie "1 s#`echo $num1`#`echo $num_all`#; 1 s#`echo $money1`#`echo $money_all`#" $dir_merge/$i
        #sed ‘1d‘ $dir2/$i >> $dir_merge/$i
        tail -n +2 $dir2/$i >> $dir_merge/$i

        fi
    done
done



rm -f $dir_merge/*.csve


#只在$dir1中有的：
echo "只在gdby中存在的文件"
for i in ${dir1_select}
do
  for file2 in ${dir2_select}
  do
    if [ "$i" != "$file2" ];then
    cp $dir1/$i $dir_merge
    fi
  done
done
#grep -q $i <<< ${dir2_select};if [[ $? -ne 0 ]];then cp $dir1/$i $dir_merge;fi;done


#只在$dir2中有的：
echo "只在bzbygz存在的文件"
for i in ${dir2_select}
do
  for file1 in ${dir1_select}
  do
    if [ "$i" != "$file1" ];then
    cp $dir2/$i $dir_merge
    fi
  done
done

#!/bin/bash

filenum=30
dir1=./test1
dir2=./test2
dir_merge=./test3

#删除旧文件
rm -f $dir_merge/*.csv


dir1_select=`ls $dir1 | sort -r | head -$filenum`
dir2_select=`ls $dir2 | sort -r | head -$filenum`


#2个文件夹共有的
for i in ${dir1_select}
do grep -q $i <<< ${dir2_select}  #<<<只能在centos中使用, grep -q安静模式，不打印任何标准输出。如果有匹配的内容则立即返回状态值0。
    if [[ $? -eq 0 ]];then
        echo $i
        num1=$(head -1 $dir1/$i | awk -F "," ‘{print $2}‘)
        num2=$(head -1 $dir2/$i | awk -F "," ‘{print $2}‘)
        money1=$(head -1 $dir1/$i | awk -F "," ‘{print $4}‘)
        money2=$(head -1 $dir2/$i | awk -F "," ‘{print $4}‘)


        num_all=$(($num1+$num2))
        money_all=$(echo $money1 $money2 | awk ‘{print $1+$2}‘)

        cp $dir1/$i $dir_merge
        sed -ie "1 s#`echo $num1`#`echo $num_all`#; 1 s#`echo $money1`#`echo $money_all`#" $dir_merge/$i
        #sed ‘1d‘ $dir2/$i >> $dir_merge/$i
        tail -n +2 $dir2/$i >> $dir_merge/$i
    fi
done

rm -f $dir_merge/*.csve


#只在$dir1中有的：
for i in ${dir1_select};do grep -q $i <<< ${dir2_select};if [[ $? -ne 0 ]];then cp $dir1/$i $dir_merge;fi;done

#只在$dir2中有的：
for i in ${dir2_select};do grep -q $i <<< ${dir1_select};if [[ $? -ne 0 ]];then cp $dir2/$i $dir_merge;fi;done