我有两个字符串（它们实际上是版本号，可以是任何版本号）
a := "1.05.00.0156"
b := "1.0.221.9289"
我想比较一下哪个更大。在golang中怎么做？
问问题
17740 次
12 回答
24
Hashicorp 有一个很好的解决方案 - https://github.com/hashicorp/go-version
import github.com/hashicorp/go-version
v1, err := version.NewVersion("1.2")
v2, err := version.NewVersion("1.5+metadata")
// Comparison example. There is also GreaterThan, Equal, and just
// a simple Compare that returns an int allowing easy >=, <=, etc.
if v1.LessThan(v2) {
fmt.Printf("%s is less than %s", v1, v2)
}
于 2016-06-20T12:13:14.277 回答
13
前段时间我创建了一个版本比较库：https ://github.com/mcuadros/go-version
version.CompareSimple("1.05.00.0156", "1.0.221.9289")
//Returns: 1
好好享受！
于 2013-11-12T22:48:37.747 回答
9
这是一个通用的解决方案。
package main
import "fmt"
func VersionOrdinal(version string) string {
// ISO/IEC 14651:2011
const maxByte = 1<<8 - 1
vo := make([]byte, 0, len(version)+8)
j := -1
for i := 0; i < len(version); i++ {
b := version[i]
if '0' > b || b > '9' {
vo = append(vo, b)
j = -1
continue
}
if j == -1 {
vo = append(vo, 0x00)
j = len(vo) - 1
}
if vo[j] == 1 && vo[j+1] == '0' {
vo[j+1] = b
continue
}
if vo[j]+1 > maxByte {
panic("VersionOrdinal: invalid version")
}
vo = append(vo, b)
vo[j]++
}
return string(vo)
}
func main() {
versions := []struct{ a, b string }{
{"1.05.00.0156", "1.0.221.9289"},
// Go versions
{"1", "1.0.1"},
{"1.0.1", "1.0.2"},
{"1.0.2", "1.0.3"},
{"1.0.3", "1.1"},
{"1.1", "1.1.1"},
{"1.1.1", "1.1.2"},
{"1.1.2", "1.2"},
}
for _, version := range versions {
a, b := VersionOrdinal(version.a), VersionOrdinal(version.b)
switch {
case a > b:
fmt.Println(version.a, ">", version.b)
case a < b:
fmt.Println(version.a, "<", version.b)
case a == b:
fmt.Println(version.a, "=", version.b)
}
}
}
输出：
1.05.00.0156 > 1.0.221.9289
1 < 1.0.1
1.0.1 < 1.0.2
1.0.2 < 1.0.3
1.0.3 < 1.1
1.1 < 1.1.1
1.1.1 < 1.1.2
1.1.2 < 1.2
于 2013-08-23T21:15:49.930 回答
2
根据杰里米·沃尔的回答：
func compareVer(a, b string) (ret int) {
as := strings.Split(a, ".")
bs := strings.Split(b, ".")
loopMax := len(bs)
if len(as) > len(bs) {
loopMax = len(as)
}
for i := 0; i < loopMax; i++ {
var x, y string
if len(as) > i {
x = as[i]
}
if len(bs) > i {
y = bs[i]
}
xi,_ := strconv.Atoi(x)
yi,_ := strconv.Atoi(y)
if xi > yi {
ret = -1
} else if xi < yi {
ret = 1
}
if ret != 0 {
break
}
}
return
}
于 2014-11-04T07:23:41.597 回答
1
这取决于你所说的更大的意思。
一种天真的方法是：
package main
import "fmt"
import "strings"
func main() {
a := strings.Split("1.05.00.0156", ".")
b := strings.Split("1.0.221.9289", ".")
for i, s := range a {
var ai, bi int
fmt.Sscanf(s, "%d", &ai)
fmt.Sscanf(b[i], "%d", &bi)
if ai > bi {
fmt.Printf("%v is bigger than %v\n", a, b)
break
}
if bi > ai {
fmt.Printf("%v is bigger than %v\n", b, a)
break
}
}
}
于 2013-08-23T19:24:35.830 回答
1
力求清晰和简单：
func intVer(v string) (int64, error) {
sections := strings.Split(v, ".")
intVerSection := func(v string, n int) string {
if n < len(sections) {
return fmt.Sprintf("%04s", sections[n])
} else {
return "0000"
}
}
s := ""
for i := 0; i < 4; i++ {
s += intVerSection(v, i)
}
return strconv.ParseInt(s, 10, 64)
}
func main() {
a := "3.045.98.0832"
b := "087.2345"
va, _ := intVer(a)
vb, _ := intVer(b)
fmt.Println(va<vb)
}
比较版本意味着解析，所以我相信这两个步骤应该分开以使其健壮。
于 2015-05-27T19:20:13.217 回答
1
以下是一些用于版本比较的库：
- https://github.com/blang/semver
- https://github.com/Masterminds/semver
- https://github.com/hashicorp/go-version
- https://github.com/mcuadros/go-version
我用过blang/semver。例如： https: //play.golang.org/p/1zZvEjLSOAr
import github.com/blang/semver/v4
v1, err := semver.Make("1.0.0-beta")
v2, err := semver.Make("2.0.0-beta")
// Options availabe
v1.Compare(v2) // Compare
v1.LT(v2) // LessThan
v1.GT(v2) // GreaterThan
于 2020-08-20T12:20:24.797 回答
0
import (
"fmt"
"strconv"
"strings"
)
func main() {
j := ll("1.05.00.0156" ,"1.0.221.9289")
fmt.Println(j)
}
func ll(a,b string) int {
var length ,r,l int = 0,0,0
v1 := strings.Split(a,".")
v2 := strings.Split(b,".")
len1, len2 := len(v1), len(v2)
length = len2
if len1 > len2 {
length = len1
}
for i:= 0;i<length;i++ {
if i < len1 && i < len2 {
if v1[i] == v2[i] {
continue
}
}
r = 0
if i < len1 {
if number, err := strconv.Atoi(v1[i]); err == nil {
r = number
}
}
l = 0
if i < len2 {
if number, err := strconv.Atoi(v2[i]); err == nil {
l = number
}
}
if r < l {
return -1
}else if r> l {
return 1
}
}
return 0
}
于 2016-03-22T09:02:34.560 回答
0
在 leetcode 中测试：https ://leetcode.com/problems/compare-version-numbers/
func compareVersion(version1 string, version2 string) int {
len1, len2, i, j := len(version1), len(version2), 0, 0
for i < len1 || j < len2 {
n1 := 0
for i < len1 && '0' <= version1[i] && version1[i] <= '9' {
n1 = n1 * 10 + int(version1[i] - '0')
i++
}
n2 := 0
for j < len2 && '0' <= version2[j] && version2[j] <= '9' {
n2 = n2 * 10 + int(version2[j] - '0')
j++
}
if n1 > n2 {
return 1
}
if n1 < n2 {
return -1
}
i, j = i+1, j+1
}
return 0
}
于 2021-11-16T08:50:52.963 回答
0
如果你能保证版本字符串的格式相同（即SemVer），你可以转换成int并比较int。这是对 SemVer 的切片进行排序的实现：
versions := []string{"1.0.10", "1.0.6", "1.0.9"}
sort.Slice(versions[:], func(i, j int) bool {
as := strings.Split(versions[i], ".")
bs := strings.Split(versions[j], ".")
if len(as) != len(bs) || len(as) != 3 {
return versions[i] < versions[j]
}
ais := make([]int, len(as))
bis := make([]int, len(bs))
for i := range as {
ais[i], _ = strconv.Atoi(as[i])
bis[i], _ = strconv.Atoi(bs[i])
}
//X.Y.Z
// If X and Y are the same, compare Z
if ais[0] == bis[0] && ais[1] == bis[1] {
return ais[2] < bis[2]
}
// If X is same, compare Y
if ais[0] == bis[0] {
return ais[1] < bis[1]
}
// Compare X
return ais[0] < bis[0]
})
fmt.Println(versions)
于 2021-07-12T16:59:13.283 回答
-8
将“1.05.00.0156”转换为“0001”+“0005”+“0000”+“0156”，然后转换为int64。
将“1.0.221.9289”转换为“0001”+“0000”+“0221”+“9289”，然后转换为int64。
比较两个 int64 值。
于 2013-08-23T19:12:36.273 回答