使用golang解析Perl正则表达式

huangapple go评论89阅读模式
英文:

Parsing Perl regex with golang

问题

这是我的代码和代码示例。

func insert_comma(input_num int) string {
	temp_str := strconv.Itoa(input_num)
	var validID = regexp.MustCompile(`\B(?=(\d{3})+$)`)
	return validID.ReplaceAllString(temp_str, ",")
}

func main() {
	fmt.Println(insert_comma(1000000000))
}

基本上,我想要的输入是 1,000,000,000。
这个正则表达式在JavaScript中有效,但我不知道如何在Go中使用这个Perl正则表达式。非常感谢。谢谢!

英文:

http://play.golang.org/p/GM0SWo0qGs

This is my code and playground.

func insert_comma(input_num int) string {
 	temp_str := strconv.Itoa(input_num)
  	var validID = regexp.MustCompile(`\B(?=(\d{3})+$)`)
 	return validID.ReplaceAllString(temp_str, ",")
}

 func main() {
 	fmt.Println(insert_comma(1000000000))
 }

Basically, my desired input is 1,000,000,000.
And the regular expression works in Javascript but I do not know how to make this Perl regex work in Go. I would greatly appreciate it. Thanks,

答案1

得分: 2

由于前瞻断言似乎不被支持,我为您提供了一种不使用正则表达式的不同算法:

Perl 代码

sub insert_comma {
    my $x=shift;
    my $l=length($x);
    for (my $i=$l%3==0?3:$l%3;$i<$l;$i+=3) {
        substr($x,$i++,0)=',';
    }
    return $x;
}
print insert_comma(1000000000);

Go 代码免责声明:我对 Go 语言没有任何经验,所以如果有错误,请谅解并随意编辑我的帖子!

func insert_comma(input_num int) string {
    temp_str := strconv.Itoa(input_num)
    var result []string
    i := len(temp_str)%3;
    if i == 0 { i = 3 }
    for index,element := range strings.Split(temp_str, "") {
        if i == index {
            result = append(result, ",");
            i += 3;
        }
        result = append(result, element)
    }
    return strings.Join(result, "")
}

func main() {
    fmt.Println(insert_comma(1000000000))
}

您可以在此链接中查看运行结果:http://play.golang.org/p/7pvo7-3G-s

英文:

Since lookahead assertion seems to be not supported, I'm providing you a different algorithm with no regexp:

Perl code:

sub insert_comma {
    my $x=shift;
    my $l=length($x);
    for (my $i=$l%3==0?3:$l%3;$i&lt;$l;$i+=3) {
        substr($x,$i++,0)=&#39;,&#39;;
    }
    return $x;
}
print insert_comma(1000000000);

Go code: Disclaimer: I have zero experience with Go, so bear with me if I have errors and feel free to edit my post!

func insert_comma(input_num int) string {
    temp_str := strconv.Itoa(input_num)
    var result []string
    i := len(temp_str)%3;
    if i == 0 { i = 3 }
    for index,element := range strings.Split(temp_str, &quot;&quot;) {
        if i == index {
            result = append(result, &quot;,&quot;);
            i += 3;
        }
        result = append(result, element)
    }
    return strings.Join(result, &quot;&quot;)
}

func main() {
    fmt.Println(insert_comma(1000000000))
}

http://play.golang.org/p/7pvo7-3G-s

huangapple
  • 本文由 发表于 2013年9月28日 16:45:15
  • 转载请务必保留本文链接:https://go.coder-hub.com/19065093.html
匿名

发表评论

匿名网友

:?: :razz: :sad: :evil: :!: :smile: :oops: :grin: :eek: :shock: :???: :cool: :lol: :mad: :twisted: :roll: :wink: :idea: :arrow: :neutral: :cry: :mrgreen:

确定