再請教一個關于自動代入變量的問題

tasteoftime_90

下午已經寫過這樣的了 不行啊
我的日志目錄如下：
a1-1.log.1
a1-1.log.2
a1-1.log.3
a1-2.log.1
a1-2.log.2
a1-2.log.3
b1-1.log.1
b1-1.log.2
b1-1.log.3
b1-2.log.1
b1-2.log.2
b1-2.log.3

shell內容如下：
a1-1_log=$(ls -lt /try/a1-1.log.* |head -1 |awk '{print $9}')
a1-2_log=$(ls -lt /try/a1-2.log.* |head -1 |awk '{print $9}')
b1-1_log=$(ls -lt /try/b1-1.log.* |head -1 |awk '{print $9}')
b1-2_log=$(ls -lt /try/b1-2.log.* |head -1 |awk '{print $9}')
for file in a1-1.log a1-2_log b1-1_log b1-2_log; do
    test=$(grep -Po '(?<=xxoo )\d+' $file |tail -5 |awk '{print ;}')
    echo "$test" >> /try/try.txt
done

結果是：
test.sh: line 6: a1-1_log=: command not found
test.sh: line 7: a1-2_log=: command not found
test.sh: line 8: b1-1_log=: command not found
test.sh: line 9: b1-2_log=: command not found
grep: a1-1.log: 沒有那個文件或目錄
grep: a1-2_log: 沒有那個文件或目錄
grep: b1-1_log: 沒有那個文件或目錄
grep: b1-2_log: 沒有那個文件或目錄

另外，單獨執(zhí)行：
a1-1_log=$(ls -lt /try/a1-1.log.* |head -1 |awk '{print $9}')
grep -Po '(?<=xxoo )\d+' $file |tail -5 |awk '{print ;}'
就沒問題，應該是for in循環(huán)的問題？
回復 10# Shell_HAT


   

Buring__

test=$(grep -Po '(?<=xxoo )\d+' <<<$file |tail -5 |awk '{print ;}')

Shell_HAT

回復 11# tasteoftime_90

1. a1_1_log=$(ls -1t /try/a1-1.log.* | head -1)
   
2. a1_2_log=$(ls -1t /try/a1-2.log.* | head -1)
   
3. b1_1_log=$(ls -1t /try/b1-1.log.* | head -1)
   
4. b1_2_log=$(ls -1t /try/b1-2.log.* | head -1)
   
5. for file in ${a1_1_log} ${a1_2_log} ${b1_1_log} ${b1_2_log}; do
   
6.     test=$(grep -Po '(?<=xxoo )\d+' $file | tail -5)
   
7.     echo "$test" >> /try/try.txt
   
8. done

復制代碼

tasteoftime_90

我早上也剛剛發(fā)現(xiàn)問題 哈哈哈
想請教下為什么定義a1-1_log的時候不能使用“-”這一個符號
回復 13# Shell_HAT


   

jason680

回復 14# tasteoftime_90

說說你要什么功能...(看起來一個awk就能完成)

別再一直問 X-Y problem
just talk us what you want to do
http://www.72891.cn/forum.php?mod=viewthread&tid=4118053
   

Shell_HAT

回復 14# tasteoftime_90


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