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Curious Sam
Asked: 2018-06-16 17:49:47 +0800 CST

如何从 csv 文件的第 7 列中删除第一个值和尾随逗号

  • 772

我正在尝试从 csv 文件的第 31 列中删除第一个值。这些列用引号括起来，并用逗号分隔，就像这样。

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

第 31 列是图像列，图像用逗号分隔

"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- 这是一个示例。 

"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"

此网址和逗号必须删除

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

我只发布了 csv 文件的前 2 行，但该文件是 300 - 400 行，并且必须从第 2 行删除该图像到最后一行。我们不能改变第一行。

text-processing sed

2 个回答

  1. Best Answer

    使用 GNU sed，我们可以这样处理：

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

    可以理解为：

    • ","在当前行中字符串的第 30 次出现处放置一个占位符。

    • 然后，从占位符一直到第一个逗号，我们一路删除所有内容。

    其他方法如下：

     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，我们删除直到第一个逗号的所有内容。

    • 2

  2. 根据您的问题，第 31 列中存在以逗号分隔的图像，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

    经过测试并且工作正常让我们知道任何混淆

    • 0

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