awk输出不在bash中工作

I've tested the following line in terminal and it works perfectly:

我在终端测试了以下一行,它完美无缺:

zcat file_2016_01_01.gz | awk 'BEGIN { FS= "|"}; {if ($1 =="START" && $7 == "1234") {flag=1}} flag ; {if ($1=="STOP"){flag=0}}'> exit.txt

but when I try to automate the extraction for several files in a bash script the output files are empty:

但是当我尝试在bash脚本中自动提取多个文件时,输出文件为空:

Updated Code: the for loops were commented for showing that it wasn't the problem

更新的代码:对for循环进行了评论,表明它不是问题所在

#!/bin/bash
REF="ref"
CODE="code"
REP="path_to_file"
#for m in 0{1..3}
#do
#  for d in 0{1..9} {10..31};
#  do
#    DATE="2016_${m}_${d}"
DATE="2016_01_04"

EXIT=${REF}_${CODE}_${DATE}.gz
if [ -e $REP/file_$DATE.gz ]
    then
            zcat $REP/file_$DATE.gz | awk 'BEGIN { FS= "|"}; {if ($1 =="START" && $7 == "'"$CODE"'" && $18== "'"$REF"'") {flag=1}} flag ; {if ($1=="STOP"){flag=0}} ' > $EXIT
    else
            echo "File not found!"
fi
#  done
#done
exit 0

could anybody help me? Thanks a lot!

有人能帮助我吗?非常感谢!

1 个解决方案

#1

Problem seems to be here:

问题似乎在这里:

DATE="2016_$m_$d"

Since _ is considered part of variable name it is effectively making it:

由于_被认为是变量名称的一部分,因此它实际上是:

DATE="2016_${m_}${d}"

Since you don't have a variable named $m_, effectively you will be getting the value of "2016_$d" in your DATE variable.

由于您没有名为$ m_的变量,因此您将在DATE变量中获得“2016_ $ d”的值。

You can fix it by using:

您可以使用以下方法修复它:

DATE="2016_${m}_${d}"

Also recommended to check your script on shellcheck.net to find and fix other obvious mistakes.

还建议检查shellcheck.net上的脚本以查找并修复其他明显的错误。

#1