當我嘗試回傳.Name我的JSON.
這是我的 javascript(console.log 用于測驗):
var roomsCodes = [
['232','DDSB'],
['232','DDMB'],
['232','SJJB'],
['232','SJJS'],
];
var jsonMin =
{
"DDSB": {
"Name": "Doble Estándar",
},
"DDMB": {
"Name": "Doble con vista mar",
},
"SJJB": {
"Name": "Jr. Suite",
},
"SJJS": {
"Name": "Jr. Suite Superior",
},
}
for(var p = 0; p < roomsCodes.length; p ){
if(232 == roomsCodes[p][0]){
console.log(jsonMin.roomsCodes[p][1].Name);
}
}
uj5u.com熱心網友回復:
您可以通過使用 [] 訪問物件來修復它,如下所示:
var roomsCodes = [
["232", "DDSB"],
["232", "DDMB"],
["232", "SJJB"],
["232", "SJJS"]
];
var jsonMin = {
DDSB: {
Name: "Doble Estándar"
},
DDMB: {
Name: "Doble con vista mar"
},
SJJB: {
Name: "Jr. Suite"
},
SJJS: {
Name: "Jr. Suite Superior"
}
};
for (var p = 0; p < roomsCodes.length; p) {
if (232 == roomsCodes[p][0]) {
console.log(jsonMin[roomsCodes[p][1]].Name);
}
}
當您想使用動態變數訪問屬性時,您需要使用方括號,否則您可以使用點。
uj5u.com熱心網友回復:
var roomsCodes = [
['232','DDSB'],
['232','DDMB'],
['232','SJJB'],
['232','SJJS'],
];
var jsonMin =
{
"DDSB": {
Name: "Doble Estándar",
},
"DDMB": {
Name: "Doble con vista mar",
},
"SJJB": {
Name: "Jr. Suite",
},
"SJJS": {
Name: "Jr. Suite Superior",
},
}
for(var p = 0; p < roomsCodes.length; p ){
if(232 == roomsCodes[p][0]){
const item = roomsCodes[p][1]
console.log(jsonMin[item].Name)
}
}
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標籤:javascript 数组 json
