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调试 shell script 方法
wade@V1088:~$ cat b.sh
#!/bin/bashdir=`pwd`dir=$dir‘/‘for f in `ls *.png`do echo $dir$fdone
看每一行代码的执行：
wade@V1088:~$ bash -x b.sh++ pwd+ dir=/home/wade+ dir=/home/wade/++ ls chrome_1407299385726.png chrome_1427299385726.png+ for f in ‘`ls *.png`‘+ echo /home/wade/chrome_1407299385726.png/home/wade/chrome_1407299385726.png+ for f in ‘`ls *.png`‘+ echo /home/wade/chrome_1427299385726.png
进一步，看具体执行的代码与行号对应：
wade@V1088:~$ export PS4=‘+${BASH_SOURCE}:${LINENO}:${FUNCNAME[0]}: ‘wade@V1088:~$ bash -x b.sh++b.sh:2:: pwd+b.sh:2:: dir=/home/wade+b.sh:3:: dir=/home/wade/++b.sh:8:: ls chrome_1407299385726.png chrome_1427299385726.png+b.sh:4:: for f in ‘`ls *.png`‘+b.sh:7:: echo /home/wade/chrome_1407299385726.png/home/wade/chrome_1407299385726.png+b.sh:4:: for f in ‘`ls *.png`‘+b.sh:7:: echo /home/wade/chrome_1427299385726.png
指定脚本的部分去跑详细的行号，其余的忽略：
wade@V1088:~$ cat b.sh #!/bin/bashdir=`pwd`dir=$dir‘/‘#this line above won‘t log out set -xfor f in `ls *.png`do echo $dir$fdoneset +x#the follow line won‘t log out echo ‘end‘abc=‘new var‘
运行输出：
wade@V1088:~$ ./b.sh++./b.sh:10:: ls chrome_1407299385726.png chrome_1427299385726.png+./b.sh:6:: for f in ‘`ls *.png`‘+./b.sh:9:: echo /home/wade/chrome_1407299385726.png/home/wade/chrome_1407299385726.png+./b.sh:6:: for f in ‘`ls *.png`‘+./b.sh:9:: echo /home/wade/chrome_1427299385726.png/home/wade/chrome_1427299385726.png+./b.sh:11:: set +xend
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