如何删除Shell最后一列中的空白空间

Question

我有一个带有此字符串的文件

select chrom,chromStart,chromEnd,name from snp147 where name="rs12414460      ";
select chrom,chromStart,chromEnd,name from snp147 where name="rs12456              ";
select chrom,chromStart,chromEnd,name from snp147 where name="rs12434212334               ";

我想删除name字段中的空间。

我该如何获得：

select chrom,chromStart,chromEnd,name from snp147 where name="rs12414460";
select chrom,chromStart,chromEnd,name from snp147 where name="rs12456";
select chrom,chromStart,chromEnd,name from snp147 where name="rs1243421111111";

请帮助我

我尝试了：cut test.txt | cut -d ' ' -f1,2,3,4,5,6，但它不起作用

Answer 1

这对sed而不是bash。

sed 's/ *";/";/' test.txt

或(作为此处文档的演示(：

sed 's/ *";/";/'  <<EOF
> select chrom,chromStart,chromEnd,name from snp147 where name="rs12414460      ";
> select chrom,chromStart,chromEnd,name from snp147 where name="rs12456              ";
> select chrom,chromStart,chromEnd,name from snp147 where name="rs12434212334               ";
> EOF
select chrom,chromStart,chromEnd,name from snp147 where name="rs12414460";
select chrom,chromStart,chromEnd,name from snp147 where name="rs12456";
select chrom,chromStart,chromEnd,name from snp147 where name="rs12434212334";