bash 如何使用linux限制一行中存在的字符串长度

how to restrict length of string present in a line using linux

提问by Jannat Arora

I have data of the following form:

我有以下形式的数据：

num1    This is a string
num2    This is another string

I want to limit length of all strings which are after the first tab..such that length(string)<4. Therefore, the output which I get is:

我想限制第一个选项卡之后的所有字符串的长度..这样长度（字符串）<4。因此，我得到的输出是：

num1    This is a string
num2    This is another 

I can do this using python. But I am trying to find a linux equivalent in order to achieve the same.

我可以使用 python 来做到这一点。但我试图找到一个 linux 等价物以实现相同的目标。

回答by jramirez

In bash, you can use the following to limit the string, in this case, from index 0 to index 17.

在 bash 中，您可以使用以下内容来限制字符串，在这种情况下，从索引 0 到索引 17。

$ var="this is a another string"

$ echo ${var:0:17}

this is a another

回答by Gilles Quenot

Using awk, by columns :

使用awk，按列：

$ awk '{print , , , }' file

or with sed:

或使用sed：

sed -r 's@^(\S+\s+\S+\s+\S+\s+\S+).*@@' file

or by length using cut:

或按长度使用cut：

$ cut -c 1-23 file

回答by Cole Tierney

If you'd like to truncate strings on word boundaries, you could use foldwith the -s option:

如果您想在单词边界上截断字符串，可以使用fold-s 选项：

awk -F"\t" '{
    printf "%s\t", ; system(sprintf("fold -sw 17 <<< \"%s\" | sed q", ))
}'

The drawback is foldand sedneed to be called for each line (sed qis the same as tail -n1).

缺点是fold并且sed需要为每一行调用（sed q与 相同tail -n1）。