fmt.Println("Enter position to delete::")
fmt.Scanln(&pos)

new_arr := make([]int, (len(arr) - 1))
k := 0
for i := 0; i < (len(arr) - 1); {
    if i != pos {
        new_arr[i] = arr[k]
        k++
        i++
    } else {
        k++
    }
}

for i := 0; i < (len(arr) - 1); i++ {
    fmt.Println(new_arr[i])
}

我正在使用这个命令从切片中删除一个元素,但它不起作用,请建议。


当前回答

没有必要检查每一个元素,除非你关心的内容,你可以利用切片追加。试试吧

pos := 0
arr := []int{1, 2, 3, 4, 5, 6, 7, 9}
fmt.Println("input your position")
fmt.Scanln(&pos)
/* you need to check if negative input as well */
if (pos < len(arr)){
    arr = append(arr[:pos], arr[pos+1:]...)
} else {
    fmt.Println("position invalid")
}

其他回答

下面是带有指针的操场示例。 https://play.golang.org/p/uNpTKeCt0sH

package main

import (
    "fmt"
)

type t struct {
    a int
    b string
}

func (tt *t) String() string{
    return fmt.Sprintf("[%d %s]", tt.a, tt.b)
}

func remove(slice []*t, i int) []*t {
  copy(slice[i:], slice[i+1:])
  return slice[:len(slice)-1]
}

func main() {
    a := []*t{&t{1, "a"}, &t{2, "b"}, &t{3, "c"}, &t{4, "d"}, &t{5, "e"}, &t{6, "f"}}
    k := a[3]
    a = remove(a, 3)
    fmt.Printf("%v  ||  %v", a, k)
}

从切片中移除一个元素(这称为“重新切片”):

package main

import (
    "fmt"
)

func RemoveIndex(s []int, index int) []int {
    return append(s[:index], s[index+1:]...)
}

func main() {
    all := []int{0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
    fmt.Println(all) //[0 1 2 3 4 5 6 7 8 9]
    all = RemoveIndex(all, 5)
    fmt.Println(all) //[0 1 2 3 4 6 7 8 9]
}

也许这段代码会有所帮助。

它删除具有给定索引的项。

接受数组和要删除的索引,并返回一个新数组,类似于append函数。

func deleteItem(arr []int, index int) []int{
  if index < 0 || index >= len(arr){
    return []int{-1}
  }

    for i := index; i < len(arr) -1; i++{
      arr[i] = arr[i + 1]

    }

    return arr[:len(arr)-1]
}

在这里您可以使用代码:https://play.golang.org/p/aX1Qj40uTVs

找到一条不需要搬迁的路。

更改订单

a := []string{"A", "B", "C", "D", "E"}
i := 2

// Remove the element at index i from a.
a[i] = a[len(a)-1] // Copy last element to index i.
a[len(a)-1] = ""   // Erase last element (write zero value).
a = a[:len(a)-1]   // Truncate slice.

fmt.Println(a) // [A B E D]

维持秩序

a := []string{"A", "B", "C", "D", "E"}
i := 2

// Remove the element at index i from a.
copy(a[i:], a[i+1:]) // Shift a[i+1:] left one index.
a[len(a)-1] = ""     // Erase last element (write zero value).
a = a[:len(a)-1]     // Truncate slice.

fmt.Println(a) // [A B D E]

也许你可以试试这个方法:

// DelEleInSlice delete an element from slice by index
//  - arr: the reference of slice
//  - index: the index of element will be deleted
func DelEleInSlice(arr interface{}, index int) {
    vField := reflect.ValueOf(arr)
    value := vField.Elem()
    if value.Kind() == reflect.Slice || value.Kind() == reflect.Array {
        result := reflect.AppendSlice(value.Slice(0, index), value.Slice(index+1, value.Len()))
        value.Set(result)
    }
}

用法:

arrInt := []int{0, 1, 2, 3, 4, 5}
arrStr := []string{"0", "1", "2", "3", "4", "5"}
DelEleInSlice(&arrInt, 3)
DelEleInSlice(&arrStr, 4)
fmt.Println(arrInt)
fmt.Println(arrStr)

结果:

0, 1, 2, 4, 5
"0", "1", "2", "3", "5"