如何從awk中提取特定字符串的最終字符並將其附加到列？

Question

我有一個看起來像這樣許多數據文件： ,8/9/2015 Timezone,-6 , Serial No.,19000000395CCE41 Location:,LS_trap_9u High temperature limit (�C),20.12 Low temperature limit (�C),0.05 Date - Time,Temperature (�C) 5/28/2015 6:00,20 5/28/2015 8:00,22.6 5/28/2015 10:00,27.1 5/28/2015 12:00,26.1 5/28/2015 14:00,27.1 5/28/2015 16:00,26.1 5/28/2015 18:00,24.6 5/28/2015 20:00,23.6 5/28/2015 22:00,22.6 5/29/2015 0:00,22.1 我分析這個腳本這些文件：

awk -vFS=, -vOFS=, \ 
    '{gsub("\"","")} 
    FNR==4{s=$2} 
    FNR==5{l=$2} 
    FNR>8{gsub(" ",OFS);print l,s,FILENAME,$0}' \ 
    *.csv > formatted_log.csv 
printf "\nDone\n" 

我想從「祿」字符串中提取的最後一個字符（在這種情況下，「U 「）並將其附加到另一列。

最後的文件應該是這個樣子：

LS_trap_9c,3.6E+15,trap9c_3600000039654841_150809.csv,u,5/28/2015,5:59,20.1 
LS_trap_9c,3.6E+15,trap9c_3600000039654841_150809.csv,u,5/28/2015,7:59,27.6 
LS_trap_9c,3.6E+15,trap9c_3600000039654841_150809.csv,u,5/28/2015,9:59,30.1 
LS_trap_9c,3.6E+15,trap9c_3600000039654841_150809.csv,u,5/28/2015,11:59,29.6 
LS_trap_9c,3.6E+15,trap9c_3600000039654841_150809.csv,u,5/28/2015,13:59,29.6 
LS_trap_9c,3.6E+15,trap9c_3600000039654841_150809.csv,u,5/28/2015,15:59,28.1 
LS_trap_9c,3.6E+15,trap9c_3600000039654841_150809.csv,u,5/28/2015,17:59,26.1 
LS_trap_9c,3.6E+15,trap9c_3600000039654841_150809.csv,u,5/28/2015,19:59,23.6 

我嘗試迄今爲止是這樣的：

awk -vFS=, -vOFS=, \ 
    '{gsub("\"","")} 
    FNR==4{ser=$2} 
    FNR==5{loc=$2} 
    FNR>8{gsub(" ",OFS);print loc,ser,FILENAME,${loc:(-1)},$0}' \ 
    *.csv > formatted_log.csv 

我收到以下錯誤：

awk: cmd. line:4:  FNR>8{gsub(" ",OFS);print loc,ser,FILENAME,${loc:(-1)},$0} 
awk: cmd. line:4:            ^syntax error 
awk: cmd. line:4:  FNR>8{gsub(" ",OFS);print loc,ser,FILENAME,${loc:(-1)},$0} 
awk: cmd. line:4:               ^syntax error 
awk: cmd. line:4:  FNR>8{gsub(" ",OFS);print loc,ser,FILENAME,${loc:(-1)},$0} 
awk: cmd. line:4:               ^syntax error 

更改對此的腳本：

awk -vFS=, -vOFS=, \ 
     awk -vFS=, -vOFS=, \ 
    '{gsub("\"","")} 
    FNR==4{ser=$2} 
    FNR==5{loc=$2} 
    my_loc="${loc:(-1)}" 
    FNR>8{gsub(" ",OFS);print loc,ser,FILENAME,my_loc,$0}' \ 
    *.csv > formatted_log.CSV 
printf "\nDone1\n" 
awk -vFS=, -vOFS=, \ 
    '{gsub("\"","")} 
    FNR==4{ser=$2} 
    FNR==5{loc=$2} 
    my_loc="${loc:(-1)}" 
    FNR>8{gsub(" ",OFS);print loc,ser,FILENAME,my_loc,$0}' \ 
    *.csv > formatted_log.CSV 
printf "\nDone1\n" 

將不需要的額外行添加到formattted_log.csv文件。看起來像這樣：

LS_trap_9c,3.6E+15,trap9c_3600000039654841_150809.csv,5/28/2015,5:59,20.1 
5/28/2015 7:59,27.6 
LS_trap_9c,3.6E+15,trap9c_3600000039654841_150809.csv,5/28/2015,7:59,27.6 
5/28/2015 9:59,30.1 
LS_trap_9c,3.6E+15,trap9c_3600000039654841_150809.csv,5/28/2015,9:59,30.1 
5/28/2015 11:59,29.6 
LS_trap_9c,3.6E+15,trap9c_3600000039654841_150809.csv,5/28/2015,11:59,29.6 
5/28/2015 13:59,29.6 
LS_trap_9c,3.6E+15,trap9c_3600000039654841_150809.csv,5/28/2015,13:59,29.6 
5/28/2015 15:59,28.1 

如何從awk中提取特定字符串的最終字符？

Answer 1

要提取AWK的最後一個字符，你可以使用：

substr(var,length(var),1) 

該腳本將是：

awk -vFS=, -vOFS=, \ 
    '{gsub("\"","")} 
    FNR==4{ser=$2} 
    FNR==5{loc=$2} 
    FNR>8{gsub(" ",OFS);print loc,ser,FILENAME,substr(loc,length(loc),1),$0}' \ 
    *.csv > formatted_log.csv 

從人AWK：

SUBSTR（S，I [，n]）
返回從i開始的至多n個字符的子字符串。如果省略n，則使用其餘的s。