列出当前工作目录中当前日期的所有文件到新文件中
List all the files for current date from a present working directory into a new file
我正在尝试获取当前目录中列出的当前日期的所有文件的文件名,并将它们列在同一路径中的新文件中。
#!/bin/bash
for file in /queues/intermediate/outbound/*EXP_GP_$(date %Y-%m-%d)* do f1=`basename $file`
if [ -f "$file" ];then
cat >> All_Name $f1;
else
echo "no files to collect"
done
这有效
find -maxdepth 1 -mtime -1 | grep -v bash | grep / | sed 's/\.\///' > FileList.txt
试试这个:
#!/bin/bash
for file in /queues/intermediate/outbound/\*EXP_GP_$(date +%Y-%m-%d)\*
do
f1=`basename $file`
if [ -f "$file" ]; then
cat >> All_Name $f1;
else
echo "no files to collect"
fi
done
还缺少 'fi' 个。
我正在尝试获取当前目录中列出的当前日期的所有文件的文件名,并将它们列在同一路径中的新文件中。
#!/bin/bash
for file in /queues/intermediate/outbound/*EXP_GP_$(date %Y-%m-%d)* do f1=`basename $file`
if [ -f "$file" ];then
cat >> All_Name $f1;
else
echo "no files to collect"
done
这有效
find -maxdepth 1 -mtime -1 | grep -v bash | grep / | sed 's/\.\///' > FileList.txt
试试这个:
#!/bin/bash
for file in /queues/intermediate/outbound/\*EXP_GP_$(date +%Y-%m-%d)\*
do
f1=`basename $file`
if [ -f "$file" ]; then
cat >> All_Name $f1;
else
echo "no files to collect"
fi
done
还缺少 'fi' 个。