英文:
How to remove string pattern and all the string behind that pattern?
问题
例如：
package main
import "fmt"
func main() {
pattern := "helloworld."
myString := "foo.bar.helloworld.qwerty.zxc.helloworld.asd"
fmt.Println(removeFromPattern(pattern, myString))
}
func removeFromPattern(p, ms string) string {
// 我在这里感到困惑（以高效的方式）
}
期望的输出：
qwerty.zxc.helloworld.asd
我如何获得期望的输出，还有如何从myString中删除第一个pattern及其后面的所有字符串？
英文:
For Example :
package main
import "fmt"
func main() {
pattern := "helloworld."
myString := "foo.bar.helloworld.qwerty.zxc.helloworld.asd"
fmt.Println(removeFromPattern(pattern, myString))
}
func removeFromPattern(p, ms string) string {
// I confused here (in efficient way)
}
Wanted output :
qwerty.zxc.helloworld.asd
How do I get that wanted output, also how to remove the first pattern and all the strings behind that pattern from myString ?
答案1
得分: 1
1- 使用_, after, _ = strings.Cut(ms, p)，尝试这个：
func removeFromPattern(p, ms string) (after string) {
_, after, _ = strings.Cut(ms, p) // 分隔符前后的文本
return
}
其中使用了strings.Index：
// Cut函数在s中找到第一个sep的实例，并将s分割成sep之前和之后的文本。
// found表示sep是否出现在s中。
// 如果sep不在s中，cut返回s、""和false。
func Cut(s, sep string) (before, after string, found bool) {
if i := Index(s, sep); i >= 0 {
return s[:i], s[i+len(sep):], true
}
return s, "", false
}
2- 使用strings.Index，尝试这个：
func removeFromPattern(p, ms string) string {
i := strings.Index(ms, p)
if i == -1 {
return ""
}
return ms[i+len(p):]
}
3- 使用strings.Split，尝试这个：
func removeFromPattern(p, ms string) string {
a := strings.Split(ms, p)
if len(a) != 2 {
return ""
}
return a[1]
}
4- 使用regexp，尝试这个：
func removeFromPattern(p, ms string) string {
a := regexp.MustCompile(p).FindStringSubmatch(ms)
if len(a) < 2 {
return ""
}
return a[1]
}
英文:
1- Using _, after, _ = strings.Cut(ms, p), try this:
func removeFromPattern(p, ms string) (after string) {
_, after, _ = strings.Cut(ms, p) // before and after sep.
return
}
Which uses strings.Index :
// Cut slices s around the first instance of sep,
// returning the text before and after sep.
// The found result reports whether sep appears in s.
// If sep does not appear in s, cut returns s, "", false.
func Cut(s, sep string) (before, after string, found bool) {
if i := Index(s, sep); i >= 0 {
return s[:i], s[i+len(sep):], true
}
return s, "", false
}
2- Using strings.Index, try this:
func removeFromPattern(p, ms string) string {
i := strings.Index(ms, p)
if i == -1 {
return ""
}
return ms[i+len(p):]
}
3- Using strings.Split, try this:
func removeFromPattern(p, ms string) string {
a := strings.Split(ms, p)
if len(a) != 2 {
return ""
}
return a[1]
}
4- Using regexp, try this
func removeFromPattern(p, ms string) string {
a := regexp.MustCompile(p).FindStringSubmatch(ms)
if len(a) < 2 {
return ""
}
return a[1]
}
答案2
得分: 1
func main() {
pattern := "helloworld."
myString := "foo.bar.helloworld.qwerty.zxc"
res := removeFromPattern(pattern, myString)
fmt.Println(res)
}
func removeFromPattern(p, ms string) string {
parts := strings.Split(ms, p)
if len(parts) > 1 {
return parts[1]
}
return ""
}
<details>
<summary>英文:</summary>
[strings.Split][1] is enough
```go
func main() {
pattern := "helloworld."
myString := "foo.bar.helloworld.qwerty.zxc"
res := removeFromPattern(pattern, myString)
fmt.Println(res)
}
func removeFromPattern(p, ms string) string {
parts := strings.Split(ms, p)
if len(parts) > 1 {
return parts[1]
}
return ""
}
答案3
得分: 0
func removeFromPattern(p, ms string) string {
return strings.ReplaceAll(ms, p, "")
}
func main() {
pattern := "helloworld."
myString := "foo.bar.helloworld.qwerty.zxc"
res := removeFromPattern(pattern, myString)
fmt.Println(res)
}
func removeFromPattern(p, ms string) string {
return strings.ReplaceAll(ms, p, "")
}
func main() {
pattern := "helloworld."
myString := "foo.bar.helloworld.qwerty.zxc"
res := removeFromPattern(pattern, myString)
fmt.Println(res)
}
英文:
func removeFromPattern(p, ms string) string {
return strings.ReplaceAll(ms, p, "")
}
func main() {
pattern := "helloworld."
myString := "foo.bar.helloworld.qwerty.zxc"
res := removeFromPattern(pattern, myString)
fmt.Println(res)
}
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论