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如何删除 csv 文件中第七列的第一个值和尾随逗号

tag-icon • tag-icon text-processing sed csv
如何删除 csv 文件中第七列的第一个值和尾随逗号

我正在尝试删除 csv 文件中第 31 列的第一个值。这些列用引号引起来并用逗号分隔，如下所示。

"first_name","last_name","home_address"
"Frank",Russel","123 Hampton road"

第31列是图像列，图像之间用逗号分隔

"http://ddd.com/124dfr3f.jpg,http://ddd.com/hjasgx37trg.jpg"

必须从第 31 列中删除第一个图像和尾部逗号

添加- 这是一个例子。

"one","two","three","four","five","six","seven","eight","nine","ten","eleven","twelve","three","four","five","six","seven","eight","nine","ten","twenty-one","twenty-two","three","four","five","six","seven","eight","nine","ten","thirty-one","thirty-two","three","four","five","six","seven","eight","nine","forty"
"one","two","three","four","five","six","seven","eight","nine","ten","eleven","twelve","three","four","five","six","seven","eight","nine","ten","twenty-one","twenty-two","three","four","five","six","seven","eight","nine","http://pictures.domain.com/e/josequervo/1183/eeeffcc7927sssssssssss5d26c0eb19e1dc53548x.jpg","http://pictures.domain.com/e/josequervo/0083/eeef8cc7927bf1245d26c0eb19e1dc53545x.jpg,http://pictures.domain.com/e/josequervo/0099/eeefscc7927bf1245d26c0eb19e1dc53546x.jpg,http://pictures.domain.com/e/josequervo/0012/eeefdcc7927bf1245d26c0eb19e1dc53547x.jpg,http://pictures.domain.com/e/josequervo/1183/eeeffcc7927bf1245d26c0eb19e1dc53548x.jpg,http://pictures.domain.com/e/josequervo/1234/eeefgcc7927bf1245d26c0eb19e1dc53549x.jpg,http://pictures.domain.com/e/josequervo/asee/eeefhcc7927bf1245d26c0eb19e1dc53540x.jpg,http://pictures.domain.com/e/josequervo/2341/eeefjcc7927bf1245d26c0eb19e1dc53544x.jpg","two",three","four","five","six","seven","eight","nine","forty"

必须删除此网址和逗号

http://pictures.domain.com/e/josequervo/0083/eeef8cc7927bf1245d26c0eb19e1dc53545x.jpg,

我只发布了 csv 文件的前 2 行，但该文件有 300 - 400 行，并且必须从第 2 行到最后一行删除此图像。我们无法更改第一行。

答案1

使用 GNUsed我们可以如图所示解决这个问题：

sed -e '
     s/","/&\n/30
     s/\n[^,]*,//
' file.csv

可以理解为：

  • ","在当前行中第 30 次出现的字符串处放置一个占位符。

  • 然后，从占位符一直到第一个逗号，我们删除沿途的所有内容。

另一种方法给出如下：

 perl -F'/","/' -lane 'my $i;
      print join q[","], map { $i++ == 30 ? substr($_, 1+index($_, ",")) : $_ } @F;
 ' input-file.csv

方法：

  • perl 将逐行读取 csv 文件，并且对于每一行，它将围绕字符串拆分 ","并将拆分元素存储在数组 @F 中。

  • 然后，对于第 31 个元素，我们删除第一个逗号之前的所有内容。

答案2

根据您的问题，图像存在于以逗号分隔的第 31 列中，我们需要从第 31 列中删除第一个字符

awk -F "," '{$31=substr($31,2);print $0}' file.csv

经过测试且工作正常，如有任何疑问请告知我们

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