Linux 通过 BASH 用下划线替换空格
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Replace spaces with underscores via BASH
提问by Ayush Mishra
Suppose i have a string, $str. I want $str to be edited such that all the spaces in it are replaced by underscores.
假设我有一个字符串，$str。我希望 $str 被编辑，以便其中的所有空格都被下划线替换。
Example
例子
a="hello world"
I want the final output of
我想要最终的输出
echo "$a"
to be hello_world
成为你好世界
采纳答案by William Hay
You could try the following:
您可以尝试以下操作：
str="${str// /_}"
回答by fedorqui 'SO stop harming'
With sedreading directly from a variable:
与sed直接从阅读变量：
$ sed 's/ /_/g' <<< "$a"
hello_world
And to store the result you have to use the var=$(command)syntax:
要存储结果，您必须使用以下var=$(command)语法：
a=$(sed 's/ /_/g' <<< "$a")
For completeness, with awkit can be done like this:
为了完整起见，awk可以这样做：
$ a="hello my name is"
$ awk 'BEGIN{OFS="_"} {for (i=1; i<NF; i++) printf "%s%s",$i,OFS; printf "%s\n", $NF}' <<< "$a"
hello_my_name_is
回答by falsetru
$ a="hello world"
$ echo ${a// /_}
hello_world
According to bash(1):
根据 bash(1)：
${parameter/pattern/string}Pattern substitution. The pattern is expanded to produce a pattern just as in pathname expansion. Parameter is expanded and the longest match of pattern against its value is replaced with string. If pattern begins with /, all matches of pattern are replaced
with string. Normally only the first match is replaced. If pattern begins with #, it must match at the beginning of the expanded value of parameter. If pattern begins with %, it must match at the end of the expanded value of parameter. If string is null, matches of pattern are deleted and the / following pattern may be omitted. If parameter is @ or *, the substitution operation is applied to each positional parameter in turn, and the expansion is the resultant list. If parameter is an array variable subscripted with @ or *, the substitution operation is applied to each member of the array in turn, and the expansion is the resultant list.
${parameter/pattern/string}模式替换。模式被扩展以产生一个模式，就像在路径名扩展中一样。参数被扩展，并且模式与其值的最长匹配被替换为字符串。如果模式以 / 开头，则替换所有匹配的模式
带字符串。通常只替换第一个匹配项。如果模式以# 开头，则它必须匹配参数扩展值的开头。如果模式以 % 开头，则它必须匹配参数扩展值的末尾。如果 string 为空，则删除模式的匹配项，并且可以省略 / 后面的模式。如果参数为@或*，则对每个位置参数依次进行替换操作，展开后即为结果列表。如果参数是一个带有@或*下标的数组变量，则对数组中的每个成员依次进行替换操作，展开的就是结果列表。
回答by anubhava
Pure bash:
纯猛击：
a="hello world"
echo "${a// /_}"
OR tr:
或 tr：
tr -s ' ' '_' <<< "$a"
