使用awk读取今天日期的bash变量并将其传递回bash脚本

问题描述 投票:-1回答:1

我正在尝试遍历包含三个内容的常规文件的内容:1)文件说明2)Linux服务器中文件的命名约定3)Linux服务器路径

将有大约20行以下格式类似的行,如果文件存在,它将被放入csv文件中以作每日报告。

Test File,TEST.$(date+'%Y-%m-%d'),/received/completed/$(date+'%Y-%m-%d')

上面有类似格式的多行,带有不同的文件路径,文件名等。>

如上所述,常规文件分为这3个部分,我需要在今天使用通配符,因为这就是Linux环境中完整文件的文件夹结构。以下是供参考的代码。它全天运行,以上面的文件作为参数来检查文件路径。

## Frequency in seconds to check the path
RECHECK_WAIT=5
## Start and end time to know when we should start
## and stop checking
START_TIME=0000
END_TIME=2359
LIST=$1
DATE_FORMAT=$(date +'%Y-%m-%d')
DATE_FORMAT1=$(date +'%Y%m%d')
REPORT_FILE="report_$(date +%b-%d-%y).csv"
check_if_report_exists() {
   ## Check to see if a report file already exists
   ## If not, use base template as starting point
   if [[ ! -f $REPORT_FILE ]]
   then
      base_report=template.csv
   else
      base_report=$REPORT_FILE
   fi
}
update_csv_file_recv() {
   echo "$(date): Recieved "$2" in $LIST"
   ## Update .csv field for timestamp of recieved time and mark as Recieved
   awk -v date="$(date +"%b-%d-%y %H:%M:%S")" -v type="$1" -v OFS=, -F, '{ if($2 ~ type) $3=date; if($2 ~ type) $6="Recieved"} 1' $base_report > tmp_$REPORT_FILE
   cp tmp_$REPORT_FILE $REPORT_FILE 2>/dev/null
}
update_csv_file_processed(){
   echo "$(date): "$2" was processed and no longer exists"
   ## Update .csv for timestamp of completed time and mark as Completed
   awk -v date="$(date +"%b-%d-%y %H:%M:%S")" -v type="$1" -v OFS=, -F, '{ if($2 ~ type) $4=date; if($2 ~ type) $6="Completed"} 1' $base_report  > tmp_$REPORT_FILE
   cp tmp_$REPORT_FILE $REPORT_FILE 2>/dev/null
}
run_checks(){
   ## Check if file exists in path
   check_if_exists="$(ls $1 | grep "$2")"
   ## Check .csv file to see if item is already marked as Received
   check_already_recv="$(cat report_$(date +%b-%d-%y).csv 2>/dev/null | grep "$1" | awk -F, '{ print $6 }' | grep "Recieved")"
   ## Check .csv file to see if item is already marked as Completed
   check_completed="$(cat report_$(date +%b-%d-%y).csv 2>/dev/null | grep "$1" | awk -F, '{ print $6 }' | grep "Completed")"
}
check_files() {
   cat $LIST | while read line
   do
      TYPE="$(echo $line | awk -F, '{ print $1 }')"
      EXPECTED="$(echo $line | awk -F, '{ print $2 }')"
      path="$(echo $line | awk -F, '{ print $3 }')"
      check_if_report_exists
      ## Check if files were recieved/processed
      run_checks "$path" "$EXPECTED"
      ## If not tagged as completed in .csv perform
      if [[ -z $check_completed ]]
      then
         ## Check if file does not exist and not already recieved
         if [[ ! -z $check_if_exists ]] && [[ -z $check_already_recv ]]
         then
            update_csv_file_recv "$TYPE" "$EXPECTED"
         fi
         ## Check if file no longer exists, but was previously recieved
         if [[ -z $check_if_exists ]] && [[ ! -z $check_already_recv ]]
         then
            update_csv_file_processed "$TYPE" "$EXPECTED"
         fi
      fi
   done
   ## Wait a set amount of time before checking again
   sleep $RECHECK_WAIT
   ## Update timestamp to see if we should no longer execute
   check_time="$(date +%H%M)"
}
## Loop forever
while true
do
   ## Get current time
   check_time="$(date +%H%M)"
   ## Check if current time is between our parameters
   while [[ $check_time -ge $START_TIME ]] && [[ $check_time -lt $END_TIME ]]
   do
      ### Main function used to execute checks and updates
      check_files
   done
   ### If its not within timeframe it will wait and try again in n-seconds
   sleep $RECHECK_WAIT
done

我在这里有两个问题。1)awk命令不会按照我想要的那样将日期作为变量读取。 check_files中的ls命令无法找到路径(我知道它们存在于今天的日期)。我相信这是因为awk命令实际上并没有按预期运行$(date)变量。

ls: cannot access /done/$DATE_FORMAT: No such file or directory

2)最近,我也开始同时出现此错误:./file_exists.sh: line 75: [[: 0931: value too great for base (error token is "0931")

我这样运行脚本./file_exist regular_text_file

任何建议或指示,将不胜感激!

我正在尝试遍历包含三个内容的常规文件的内容:1)文件说明2)linux服务器中文件的命名约定3)Linux服务器路径...

linux bash shell awk
1个回答
0
投票
行情应该是问题。在bash中,如果您在引号内使用相同类型的引号,则会导致错误。为了避免这种情况,我们需要转义内部引号。就您而言echo "$(date): Recieved "$2" in $LIST"应该是echo "$(date): Recieved \"$2\" in $LIST"
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