JavaScript解析数组并获得唯一值 [英] JavaScript Parsing array and getting the unique value
问题描述
我有以下数组:
array-1:
Array[11]
0:"265"
1:"text-success"
2:"51"
3:"text-warning"
4:"16"
5:"text-warning"
6:"35"
7:"text-warning"
8:"38"
9:"text-warning"
10:"106"
array-2:
Array[11]
0:"265"
1:"text-success"
2:"51"
3:"text-warning"
4:"16"
5:"text-warning"
6:"35"
7:"text-success"
8:"38"
9:"text-warning"
10:"106"
array-3:
Array[11]
0:"265"
1:"muted"
2:"51"
3:"text-warning"
4:"16"
5:"text-warning"
6:"35"
7:"text-success"
8:"38"
9:"text-warning"
10:"106"
array-4:
Array[11]
0:"265"
1:"text-warning"
2:"51"
3:"text-warning"
4:"16"
5:"muted"
6:"35"
7:"text-warning"
8:"38"
9:"text-warning"
10:"106"
第一个将返回FALSE,因为两者都同时存在文本成功"两次和静音"一次
1st one will return FALSE because both exist "text-success"-twice AND "muted"-once
第二个将返回TRUE,因为它曾经存在文本成功"
2nd one will return TRUE because it exist ONCE "text-success"
第三个将返回FALSE,因为两者都存在文本成功"和静音"
3rd one will return FALSE because both exist "text-success" AND "muted"
第四个将返回TRUE,因为它曾经被静音"
4th one will return TRUE because it exist ONCE "muted"
我需要解析数组并获取结果:
I need to parse the array and get the result:
array-1:
Array[11]
0:"265"
1:"text-success"
2:"51"
array-2: Null
array-3: Null
array-4:
Array[11]
0:"265"
1:"muted"
2:"16"
到目前为止,我有这个:
so far I have this:
function singles( array) {
for( var index = 0, single = []; index < array.length; index++ ) {
if(array[index] == "text-success" || array[index] == "muted") {
single.push(array[index]);
}
}
return single;
};
请帮助任何人?
推荐答案
目前尚不清楚您是否包含一个包含多个数组的对象,或者只是一个标准的一维数组,您已经为其提供了一些可能的示例,但是假设后者,我认为这就是您要寻找的东西:
It's not clear whether you have an object that contains multiple arrays, or just a standard one-dimensional array for which you've given several possible examples, but assuming the latter I think this is what you're looking for:
function singles(array) {
var foundIndex = -1;
for(var index = 0; index < array.length; index++) {
if(array[index] === "text-success" || array[index] === "muted") {
if (foundIndex != -1) {
// we've found two, so...
return null;
}
foundIndex = index;
}
}
if (foundIndex != -1) {
return [ array[0], array[foundIndex], array[foundIndex+1] ];
}
return "Some default value for when it wasn't found at all";
}
console.log(singles(["532","text-warning","51","text-warning","16","muted","35","text-warning","38","text-warning","106"]));
console.log(singles(["533","text-warning","51","text-warning","16","muted","35","text-success","38","text-warning","106"]));
这将循环遍历输入array
.第一次找到"text-success"
或"muted"
时,其索引存储在foundIndex
中.
This loops through the input array
. The first time "text-success"
or "muted"
is found its index is stored in foundIndex
.
如果第二次找到"text-success"
或"muted"
,我们将立即返回null
.
If "text-success"
or "muted"
is found a second time we immediately return null
.
如果我们进入循环的结尾,那么我们知道我们发现一次"text-success"
或"muted"
一次还是根本没有.如果找到,则返回一个包含三个元素的数组,其中包括输入数组中的第一项,"text-success"
或"muted"
的值,然后是紧跟该数组中文本的项.
If we get to the end of the loop then we know we found "text-success"
or "muted"
either once or not at all. If it was found once we return an array of three elements included the first item in the input array, the value of "text-success"
or "muted"
, and then the item that immediately followed that text in the array.
(注意:在数组1的示例输出中,在匹配的文本之后返回 ,但在数组4的示例输出中,在返回的之前匹配的文本.我不知道您真正想要的是哪一个,但是您可以轻松地修改我的函数以执行其中一项.
(Note: in your example output from Array 1 you returned the value after the matched text, but in your example output from Array 4 you returned the value before the matched text. I don't know which one you really want, but you can easily modify my function to do one or the other.)
您没有说要在数组中不存在"text-success"
或"muted"
时要返回的内容,所以我刚刚返回了一个占位符字符串.
You didn't say what you wanted to return if "text-success"
or "muted"
wasn't present in the array, so I've just returned a place-holder string.
这篇关于JavaScript解析数组并获得唯一值的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!