所以我查看了包括Finding All Combinations of JavaScript array values和https://codereview.stackexchange.com/questions/52119/calculate-all-possible-combinations-of-an-array-of-arrays-or-strings在内的多个数据源,并且仍然出现空白。在不改变原始数组的情况下迭代多维数组
我将字符串“35 w 35 ave”拆分为" "
的数组,然后将其与数组进行比较。这里是对象:
var addressObject = {
ave : ["av", "aven", "avenu", "avenue", "avn", "avnue"],
w : ["west", "wst", "w"]
};
我添加这var address = txtNode.val().split(' ');
它返回阵列['35','w','35','ave']
,然后我走阵列和把它通过一个嵌套forEach循环像这样:
address.forEach(function(addressElement, indexOfAddressElement) {
var abbreviationMatch = addressObject.hasOwnProperty(addressElement);
if (abbreviationMatch) {
function addSentenceVariationsToAddressLists(arrayToLookAt) {
arrayToLookAt.forEach(function (anItem, a) {
var y = address.splice(indexOfAddressElement, 1, anItem);
addressLists.push(address.join(" "));
})
}
var elementsFromArray = addressObject[addressElement];
addSentenceVariationsToAddressLists(elementsFromArray);
}
});
console.log(addressLists);
这返回["35 west 35 ave", "35 wst 35 ave", "35 w 35 ave", "35 w 35 av", "35 w 35 aven", "35 w 35 avenu", "35 w 35 avenue", "35 w 35 avn", "35 w 35 avnue"]
我的问题是:如何将"ave"
的所有变体与"wst"
和"west"
而不是仅包含原始值的"w"
e的address
数组?
你为什么有' “W”'在'addressObject'的'在首位w'财产。 – Redu
因为当我把它拿出来的时候,它并没有显示在地址列表中,它只显示“[35 35 west 35 ave”,“35 th 35 35 ave”,“35 th 35 35 av”,“35 th 35 35 aven” ,“35岁35岁”,“35岁35岁”,“35岁35岁”,“35岁35岁” –