[英]unexpected token '.' when executing a shell command when changing file extension within groovy script
[英]Shell script - Two for loops and changing extension of file
我有这个Shell脚本,我已经设法把它弄糊涂了,我希望我可以被纠正并走上正确的道路,并希望添加一些我没有能力胜任自己的事情。 我在下面的Shell脚本中将我想做的事情作为注释。
#!/bin/bash
#Get all files from dir "videos" and send to processLine function one at a time
cd /home/test/videos/
for file in `dir -d *` ; do
processLine -f $file
done
processLine(){
# I was hoping to have a further for loop that would loop 4 times and change the $ext
#variable to avi, mpg, wmv and mov
#For loop, execute a command on each file
for i in 1 2 3 4 5 6 7 8 9 10
do
START=$(date +%s.%N)
echo "$line"
#The saved file in done dir should have filename as $file + START.
eval "ffmpeg -i $file -ar 44100 /home/test/videos/done/$fileSTART.$ext" > /dev/null 2>&1
END=$(date +%s.%N)
DIFF=$(echo "$END - $START" | bc)
echo "$line, $START, $END, $DIFF" >> file.csv 2>&1
echo "It took $DIFF seconds"
echo $line
done
}
该脚本的基本思想是:从dir中获取所有文件,并对它们运行ffmpeg命令,并查看需要花费多长时间。 我正在尝试收集一些统计数据
感谢您的任何帮助
利用Juliano的脚本并交换循环2和3。我设法在下面获得此输出:
.
.
.
/home/test/videos/done 8 mov took 0.012 seconds
/home/test/videos/done 9 mov took 0.012 seconds
/home/test/videos/video1236104961.flv 0 avi took 0.446 seconds
它在那里暂停。
许多事情是错误的。
另一种尝试,修复了一些问题:
#!/bin/bash
TIMEFORMAT=%6R
for file in /home/test/videos/* ; do
if [ ! -f "$file" ]; then
continue # anything that is not a regular file
fi
for ext in avi mpg wmv mov; do
for (( i = 0; i < 10; i++ )); do
base="${file##*/}"
elapsed=$({ time ffmpeg -i "$file" -ar 44100 -y "${file%/*}/done/${base%.*}-$i.$ext" &>/dev/null; } 2>&1)
echo "$file $ext $i took $elapsed seconds"
done
done
done
1) $line
在哪里定义?
2)您曾经使用$i
吗?
3)您可以通过
for ext in avi mov mpg wmv; do
ffmpg ...
done
4)您可以使用双括号在bash中执行基本算术。 所以$(($x-$y))
而不是管道到bc
我闻到了一个陷阱。
$ touch aaa
$ touch "bbb ccc"
$ ls -1
aaa
bbb ccc
$ for file in `dir -d *`; do echo $file; done
aaa
bbb\
ccc
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