Question

因此,我编篡了我的法典,我 st不 st地试图计算某一具体记录中出现的时间。

如果像<in”这样的东西被搜寻,但现在我得到的却是13,19,-1<>。

我非常赞赏关于如何改进我守则的任何方面。

books = [ { title: "Inheritance: Inheritance Cycle, Book 4", author: "Christopher Paolini", }, { title: "The Sense of an Ending", author: "Julian Barnes"}, { title: "Snuff Discworld Novel 39", author: "Sir Terry Pratchett", } ] search = prompt("Title?"); function count(books, pattern) { var num = 0; var result = []; for (i = 0; i < books.length; i++) { var index = books[i].title.toLowerCase().indexOf(pattern.toLowerCase()); do { result[i] = index; index = books[i].title.toLowerCase().indexOf(pattern.toLowerCase(), index + 1); } while (index >= 0) num = 0; } return result; } alert(count(books, search));

Answer 1

你们需要保持这一指数,而不是其结果的指数。只有当指数大于或等于零时,你才能增加计算。注I ve还做了一些优化工作,以减少你改变扼杀所需时间。我也增加了一种使用定期表述的不同版本。

function count(books, pattern) {
    var result = [];
    for (i = 0; i < books.length; i++) {
        var index = -1, // set to -1 so that the first iteration will set it to 0
            title = books[i].title.toLowerCase(),
            realPattern = pattern.toLowerCase();
        result[i] = 0;
        do {
            index = title.indexOf(pattern, index + 1);
            if (index >= 0) {
               result[i] = result[i] + 1;
            }
        }
        while (index >= 0)
    }
    return result;
}

更好的替代选择

RegExp.escape = function(text) {
    return text.replace(/[-[]{}()*+?.,\^$|#s]/g, "\$&");
}

function count(books, pattern) {
    var result = [];
    for (i = 0; i < books.length; i++) {
        var regex = new RegExp(RegExp.escape(pattern), gi ),
            matches = books[i].title.match(regex);
        result[i] = matches ? matches.length : 0;
    }
    return result;
}

Answer 2

You can use a for in loop and push the loop var into your results variable.

rel=“nofollow”>jsfiddle

var arr = [ a ,  b ,  c ,  a ];

function count(books, pattern) {   
    var results = [];
    for (var index in arr) {
        if (arr[index] === pattern) {
          results.push(index);
        } 
    }
    return results;
}

console.log(count(arr,  a ));

Answer 3

www.un.org/Depts/DGACM/index_spanish.htm 引自:

function count(books, pattern) 
{

var num = 0;
var result = Array();
for (i = 0; i < books.length; i++) 
{
    var times = books[i].title.toLowerCase().split(pattern.toLowerCase());
    if(times.length - 1 >= 0)result[i] = times.length - 1;
}
return result;
}

希望你们能找到这一有益的l!

Answer 4

您可使用<代码>match来简化其代码。我在这里做了一个 j:。它希望:

var books = [
    {
        title: "Inheritance: Inheritance Cycle, Book 4",
        author: "Christopher Paolini",
    },
    {
        title: "The Sense of an Ending",
        author: "Julian Barnes"
    },
    {
        title: "Snuff Discworld Novel 39",
        author: "Sir Terry Pratchett",
    }
]
search = prompt("Title?");

function count(pattern)
{
    var results = [];
    var matches = null;
    pattern = new RegExp(pattern, "g")

    for (i = 0; i < books.length; i++) {
        results[i] = 0;
        matches = books[i].title.toLowerCase().match(pattern);
        if(matches != null) {
            results[i] = matches.length;
        }
    }
    return results
}

alert(count(search))

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