所以我们的目标是得到一个像这样的字符串：
somesite.com/-%20Luzon%20-/Bicol%20Region/Albay/Busay%20Falls/Busay_falls_10.jpg
所以我开始构建对象数组的多维事物，我不确定你是如何使用数组的名称构建url的。
var photoArray = {},
islandGroups = ["- Luzon -", " - Visayas -", "-Mindanao-"],
luzonRegions = ["Bicol Region", "Cagayan Valley", "Calabarzon", "CAR", "Central Luzon", "Ilocos Region", "Mimaropa"],
bicolProvinces = ["Albay", "Camarines Norte", "Camarines Sur", "Catanduanes", "Masbate", "Sorsogon"],
albayProvinceTravelDestinations = ["Busay Falls", "Hoyop-Hoyopan Cave", "Lignon Hill", "Malabsay Falls", "Mt Mayon", "Panicuason Hot Spring Resort", "Vera Falls"];
busayFallsPhotoPattern = ["Cover, Busay_falls_"];
busayFallsPhotoPatternID1 = [""];
busayFallsPhotoPatternID2 = ["1", "1_2", "10", "2", "2_2", "3", "3_2", "4", "4_2", "6", "7", "8", "9"];
baseURL = "somesite.com";
/* build sub-array */
var islandGroupsArrayLength = islandGroups.length;
for (var i = 0; i < islandGroupsArrayLength; i++) {
photoArray[islandGroups[i]] = {};
}
var luzonRegionsLength = luzonRegions.length;
for (var i = 0; i < luzonRegionsLength; i++) {
photoArray["- Luzon -"][luzonRegions[i]] = {};
}
var bicolProvincesLength = bicolProvinces.length;
for (var i = 0; i < bicolProvincesLength; i++) {
photoArray["- Luzon -"]["Bicol Region"][bicolProvinces[i]] = {};
}
var albayProvinceTravelDestinationsLength = albayProvinceTravelDestinations.length;
for (var i = 0; i < albayProvinceTravelDestinations; i++) {
photoArray["- Luzon -"]["Bicol Region"]["Albay"][i] = {};
}
/* build string before converting space to %20% */
/*
busayFallsPhotoPattern = ["Cover, Busay_falls_"];
busayFallsPhotoPatternID1 = [""];
busayFallsPhotoPatternID2 = ["1", "1_2", "10", "2", "2_2", "3", "3_2", "4", "4_2", "6", "7", "8", "9"];
*/
var busayFallsPhotoPatternLength = busayFallsPhotoPattern.length,
busayFallsPhotoPatternID2Length = busayFallsPhotoPatternID2.length;
/* setup photoURLs array */
photoArray["- Luzon -"]["Bicol Region"]["Albay"]["Busay Falls"] = {};
photoArray["- Luzon -"]["Bicol Region"]["Albay"]["Busay Falls"]["busayFallsPhotoURLs"] = {};
photoArray["- Luzon -"]["Bicol Region"]["Albay"]["Busay Falls"]["busayFallsPhotoURLs"][0] = "Cover" + ".jpg";
console.log(photoArray.[0].[0].[0].[0].[0].[0]); // this doesn't work
可能很明显，我的思绪目前有点不知所措
我也想知道如何更有效地构建这个/使用指针，所以你没有例如superLongNameLength。
编辑：
我想我知道什么是错的，首先你不要与之相提并论。在PHP的PHP中，我认为我必须在将空格转换为％20后依次将每个部分附加到字符串以获取URL但我仍然不确定这是否是执行此操作的最佳方式
这是更接近但仍然错误/冗长，我得到了最后一个有意义的条目，因为它们都有相同的名称。
var islandGroups = ["- Luzon -", "- Visayas -", "-Mindanao"],
regions = {
"- Luzon -" : "Bicol Region",
"- Luzon -" : "Cagayan Valley",
"- Luzon -" : "Calabarzon",
"- Luzon -" : "CAR",
"- Luzon -" : "Central Luzon",
"- Luzon -" : "Ilocos Region",
"- Luzon -" : "Mimaropa"
},
provinces = {
"Bicol Region" : "Albay",
"Bicol Region" : "Camarines Norte",
"Bicol Region" : "Camarines Sur",
"Bicol Region" : "Catanduanes",
"Bicol Region" : "Masbate",
"Bicol Region" : "Sorsogon"
},
travelDestinations = {
"Albay" : "Busay Falls",
"Albay" : "Hoyop-Hoyopan Cave",
"Albay" : "Lignon Hill",
"Albay" : "Malabsay Falls",
"Albay" : "Mt Mayon",
"Albay" : "Panicuason Hot Spring Resort",
"Albay" : "Vera Falls"
},
photos = {
"Busay Falls" : "Cover.jpg",
"Busay Falls" : "Busay_falls_10.jpg",
"Busay Falls" : "Busay_falls_2.jpg",
"Busay Falls" : "Busay_falls_3.jpg",
"Busay Falls" : "Busay_falls_4.jpg",
"Busay Falls" : "Busay_falls_5.jpg",
"Busay Falls" : "Busay_falls_6.jpg",
"Busay Falls" : "Busay_falls_7.jpg",
"Busay Falls" : "Busay_falls_8.jpg",
"Busay Falls" : "Busay_falls_9.jpg"
};
console.log(islandGroups[0]+regions["- Luzon -"]+provinces["Bicol Region"]+travelDestinations["Albay"]+photos["Busay Falls"]);
答案 0 :(得分：1)
是的，那里肯定是完全错误的，键是相同的...所以不是例如：
photos = {
"Busay Falls" : "Cover.jpg",
"Busay Falls" : "Busay_falls_10.jpg",
"Busay Falls" : "etc..."
}
应该是
photos = {
"busay_falls":[
"Cover.jpg",
"Busay_falls_10.jpg",
"etc..."
]
}
然后你可以通过遍历数组的索引来构建url
喜欢这里的照片案例：
photos.busay_falls[0] // becomes Cover.jpg
答案 1 :(得分：1)
由于你有n个数组，每个数组都有可变数量的项目，你可以做的最好的事情是递归地选择每个数组的一个元素，如下所示：
const lower = ['a', 'b', 'c']
const upper = ['X', 'Y', 'Z']
const numbers = [1, 2, 3]
const all = [lower, upper, numbers]
// all = a 2d array
// current = the i-th array of `all` we're currently operating on
// result = the concatenated string
function combination(all, current = 0, result = '') {
if (current == all.length) {
console.log(result)
return
}
for (let i = 0; i < all[current].length; i += 1) {
// concatenate a new string from the i-th array
combination(all, current + 1, result + all[current][i] + '/')
}
}
combination(all)
将上述算法应用于字符串需要进行一些修改：
result应编码
const baseURL = ["somesite.com"];
const islandGroups = ["- Luzon -", " - Visayas -", "-Mindanao-"];
const luzonRegions = ["Bicol Region", "Cagayan Valley", "Calabarzon", "CAR", "Central Luzon", "Ilocos Region", "Mimaropa"];
const bicolProvinces = ["Albay", "Camarines Norte", "Camarines Sur", "Catanduanes", "Masbate", "Sorsogon"];
const albayProvinceTravelDestinations = ["Busay Falls", "Hoyop-Hoyopan Cave", "Lignon Hill", "Malabsay Falls", "Mt Mayon", "Panicuason Hot Spring Resort", "Vera Falls"];
const busayFallsPhotoPattern = ["Cover", "Busay_falls_"];
const busayFallsPhotoPatternID2 = ["1", "1_2", "10", "2", "2_2", "3", "3_2", "4", "4_2", "6", "7", "8", "9"];
// all = a 2d array
// current = the i-th array of `all` we're currently operating on
// result = the concatenated string
function combination(all, current = 0, result = '') {
if (current == all.length) {
// add extension
console.log(result + '.jpg')
return
}
for (let i = 0; i < all[current].length; i += 1) {
// concatenate a new string from the i-th array
let newString = encodeURI(result + all[current][i])
// empty string shouldn't have /
// when current >= all.length - 2 we shouldn't add /
if (all[current][i] && current < all.length - 2) { newString += '/' }
combination(all, current + 1, newString)
}
}
combination([
baseURL,
islandGroups,
luzonRegions,
bicolProvinces,
albayProvinceTravelDestinations,
busayFallsPhotoPattern,
busayFallsPhotoPatternID2
])