2 回答

TA貢獻(xiàn)1780條經(jīng)驗(yàn) 獲得超5個(gè)贊
即使 JSON 很大,您也只需定義您實(shí)際關(guān)心的字段
僅當(dāng)密鑰不是有效的 Go 標(biāo)識(shí)符(在本例中為密鑰是標(biāo)識(shí))時(shí),才需要使用 JSON 標(biāo)記,即使這樣,有時(shí)也可以使用
map[string]something
除非您需要子s用于某些功能或其他功能,否則您不需要定義它們
struct
除非您需要重用該類型,否則您甚至不必定義它,您只需在聲明時(shí)定義結(jié)構(gòu)即可
例:
package main
import (
"encoding/json"
"fmt"
)
const s = `
{
"results": [
{
"times": [1, 2, 3, 4]
}
]
}
`
func main() {
var t struct {
Results []struct {
Times []int
}
}
json.Unmarshal([]byte(s), &t)
fmt.Printf("%+v\n", t) // {Results:[{Times:[1 2 3 4]}]}
}

TA貢獻(xiàn)1995條經(jīng)驗(yàn) 獲得超2個(gè)贊
[...]嘗試僅訪問一個(gè)字段(在眾多可用字段中)。
對(duì)于這個(gè)具體的用例,我將使用一個(gè)庫來查詢和訪問已知路徑中的單個(gè)值,例如:
https://github.com/jmespath/go-jmespath
另一方面,如果您正在練習(xí)如何訪問JSON中的嵌套值,我建議您嘗試編寫一個(gè)遞歸函數(shù),該函數(shù)以與go-jmespath相同(但簡單)的方式遵循未知結(jié)構(gòu)中的路徑。
好吧,我挑戰(zhàn)了自己,花了一個(gè)小時(shí)寫這篇文章。它的工作原理。不確定性能或錯(cuò)誤,它真的很有限:)
https://play.golang.org/p/dlIsmG6Lk-p
package main
import (
"encoding/json"
"errors"
"fmt"
"strings"
)
func main() {
// I Just added a bit more of data to the structure to be able to test different paths
fileContent := []byte(`
{"results": [
{"times": [
1,
2,
3,
4
]},
{"times2": [
5,
6,
7,
8
]},
{"username": "rosadabril"},
{"age": 42},
{"location": [41.5933262, 1.8376757]}
],
"more_results": {
"nested_1": {
"nested_2":{
"foo": "bar"
}
}
}
}`)
var content map[string]interface{}
if err := json.Unmarshal(fileContent, &content); err != nil {
panic(err)
}
// some paths to test
valuePaths := []string{
"results.times",
"results.times2",
"results.username",
"results.age",
"results.doesnotexist",
"more_results.nested_1.nested_2.foo",
}
for _, p := range valuePaths {
breadcrumbs := strings.Split(p, ".")
value, err := search(breadcrumbs, content)
if err != nil {
fmt.Printf("\nerror searching '%s': %s\n", p, err)
continue
}
fmt.Printf("\nFOUND A VALUE IN: %s\n", p)
fmt.Printf("Type: %T\nValue: %#v\n", value, value)
}
}
// search is our fantastic recursive function! The idea is to search in the structure in a very basic way, for complex querying use jmespath
func search(breadcrumbs []string, content map[string]interface{}) (interface{}, error) {
// we should never hit this point, but better safe than sorry and we could incurr in an out of range error (check line 82)
if len(breadcrumbs) == 0 {
return nil, errors.New("ran out of breadcrumbs :'(")
}
// flag that indicates if we are at the end of our trip and whe should return the value without more checks
lastBreadcrumb := len(breadcrumbs) == 1
// current breadcrumb is always the first element.
currentBreadcrumb := breadcrumbs[0]
if value, found := content[currentBreadcrumb]; found {
if lastBreadcrumb {
return value, nil
}
// if the value is a map[string]interface{}, go down the rabbit hole, recursion!
if aMap, isAMap := value.(map[string]interface{}); isAMap {
// we are calling ourselves popping the first breadcrumb and passing the current map
return search(breadcrumbs[1:], aMap)
}
// if it's an array of interfaces the thing gets complicated :(
if anArray, isArray := value.([]interface{}); isArray {
for _, something := range anArray {
if aMap, isAMap := something.(map[string]interface{}); isAMap && len(breadcrumbs) > 1 {
if v, err := search(breadcrumbs[1:], aMap); err == nil {
return v, nil
}
}
}
}
}
return nil, errors.New("woops, nothing here")
}
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