在Go中,哪一种有效的方法来修剪字符串变量的前导和尾随空白?


在go中有很多修剪字符串的函数。

看到了吧:修剪

下面是一个例子,改编自文档,删除了前导和后面的空白:

fmt.Printf("[%q]", strings.Trim(" Achtung  ", " "))

strings.TrimSpace (s)

例如,

package main

import (
    "fmt"
    "strings"
)

func main() {
    s := "\t Hello, World\n "
    fmt.Printf("%d %q\n", len(s), s)
    t := strings.TrimSpace(s)
    fmt.Printf("%d %q\n", len(t), t)
}

输出:

16 "\t Hello, World\n "
12 "Hello, World"

为了修剪你的字符串,Go的“strings”包有TrimSpace(), Trim()函数来修剪前导和尾随空格。

有关更多信息,请查看文档。


就像@Kabeer提到的,你可以使用TrimSpace,这里有一个来自golang文档的例子:

package main

import (
    "fmt"
    "strings"
)

func main() {
    fmt.Println(strings.TrimSpace(" \t\n Hello, Gophers \n\t\r\n"))
}

package main

import (
    "fmt"
    "strings"
)

func main() {
    fmt.Println(strings.TrimSpace(" \t\n Hello, Gophers \n\t\r\n"))
}

输出: 你好,打地鼠

点击这个链接——https://golang.org/pkg/strings/#TrimSpace


@peterSO有正确答案。我在这里添加了更多的例子:

package main

import (
    "fmt"
    strings "strings"
)

func main() { 
    test := "\t pdftk 2.0.2  \n"
    result := strings.TrimSpace(test)
    fmt.Printf("Length of %q is %d\n", test, len(test))
    fmt.Printf("Length of %q is %d\n\n", result, len(result))

    test = "\n\r pdftk 2.0.2 \n\r"
    result = strings.TrimSpace(test)
    fmt.Printf("Length of %q is %d\n", test, len(test))
    fmt.Printf("Length of %q is %d\n\n", result, len(result))

    test = "\n\r\n\r pdftk 2.0.2 \n\r\n\r"
    result = strings.TrimSpace(test)
    fmt.Printf("Length of %q is %d\n", test, len(test))
    fmt.Printf("Length of %q is %d\n\n", result, len(result))

    test = "\r pdftk 2.0.2 \r"
    result = strings.TrimSpace(test)
    fmt.Printf("Length of %q is %d\n", test, len(test))
    fmt.Printf("Length of %q is %d\n\n", result, len(result))   
}

你也可以在围棋场上找到它。


一个快速字符串“GOTCHA”与JSON Unmarshall将添加包装引号字符串。

(例如:字符串值{"first_name":" I have whitespace "}将转换为"\" I have whitespace \"")

在你可以修剪任何东西之前,你需要先删除额外的引号:

操场上的例子

// ScrubString is a string that might contain whitespace that needs scrubbing.
type ScrubString string

// UnmarshalJSON scrubs out whitespace from a valid json string, if any.
func (s *ScrubString) UnmarshalJSON(data []byte) error {
    ns := string(data)
    // Make sure we don't have a blank string of "\"\"".
    if len(ns) > 2 && ns[0] != '"' && ns[len(ns)] != '"' {
        *s = ""
        return nil
    }
    // Remove the added wrapping quotes.
    ns, err := strconv.Unquote(ns)
    if err != nil {
        return err
    }
    // We can now trim the whitespace.
    *s = ScrubString(strings.TrimSpace(ns))

    return nil
}

我对性能很感兴趣,所以我做了一个比较,只修剪左边 的一面:

package main

import (
   "strings"
   "testing"
)

var s = strings.Repeat("A", 63) + "B"

func BenchmarkTrimLeftFunc(b *testing.B) {
   for n := 0; n < b.N; n++ {
      _ = strings.TrimLeftFunc(s, func(r rune) bool {
         return r == 'A'
      })
   }
}

func BenchmarkIndexFunc(b *testing.B) {
   for n := 0; n < b.N; n++ {
      i := strings.IndexFunc(s, func(r rune) bool {
         return r != 'A'
      })
      _ = s[i]
   }
}

func BenchmarkTrimLeft(b *testing.B) {
   for n := 0; n < b.N; n++ {
      _ = strings.TrimLeft(s, "A")
   }
}

TrimLeftFunc和IndexFunc是一样的,TrimLeft要慢一些:

BenchmarkTrimLeftFunc-12        10325200               116.0 ns/op
BenchmarkIndexFunc-12           10344336               116.6 ns/op
BenchmarkTrimLeft-12             6485059               183.6 ns/op