Question

这是我的命令：

awk -v DATE="$(date +"%Y%m%d")" -F"," 'NR>1 { print > "Assignment_"$1"_"DATE".csv"}' Text_01012020.CSV

但它出来了：

Assignment_"A"_01012017

我想删除＆＃34; ___＆＃34;，你能帮助我吗？

我发现了这个：

awk -v DATE="$(date +"%d%m%Y")" -F"," 'NR>1 { gsub(/"/,"",$1); print > "Assignment_"$1"_"DATE".csv"}' Text_01012020.CSV

但是在我运行此命令之后，我的文件可以是assignment_A_01012017但是然后在我的文件中...列没有分隔成列。怎么样？

Answer 1

使用以下awk

awk -F, -v DATE="$(date +'%Y%m%d')" 'NR>1{s=$1; gsub(/"/,"",s);  print > "Assignment_"s"_"DATE".csv"}' Text_01012020.CSV

<强>解释

awk -F, -v DATE="$(date +'%Y%m%d')" '    # Start awk, where field sep being  comma
                                         # and variable DATE with current date
     NR>1{                               # If no records greater than 1 then 
            s=$1;                        # save field1 data to variable s
            gsub(/"/,"",s);              # substitute double quote with null in variable s, so here we remove quote

            # print current record/line to file Assigment_{field1}_{current_date}.csv
            # {field1} = value of variable s after removing double quote
            # {current_date} = value of variable DATE

            print > "Assignment_"s"_"DATE".csv"
         }
     ' Text_01012020.CSV

Answer 2

awk -v DATE="$(date +"%d%m%Y")" -F"," 'BEGIN{OFS=","}NR>1 { gsub(/"/,"",$1); print > "Assignment_"$1"_"DATE".csv"}' Text_01012020.CSV

只需设置“OFS”配置，然后输出将使用逗号。

Answer 3

如果要包含列名，请在NR==1：

时使用大小写进行阅读

awk -v DATE="$(date +"%d%m%Y")" -F"," 'BEGIN{OFS="," } NR==1 {COLUMN_NAME=$1} NR>1 { gsub(/"/,"",$1); print > "Assignment_"$1"_"COLUMN_NAME"_"DATE".csv"}' a.txt

您还可以使用此great tutorial

自学

如何使用awk删除双引号

3 个答案: