Question

我正在使用typeahead.js。

当我从apiController获取数据时，它看起来像这样：

["JobName1", "JobName1", "akhsfkh"]

当它通过此代码时：

 $(function() {
    var projectNumbers = $.getJSON("api/Project/GetAllNumbers")
        .done(function(result) {
            console.log(result);
            return result; // ["JobName1", "JobName1", "akhsfkh"] is here
        })
        .fail(function(jqXHR, textStatus, err) {
            alert('Error: ' + err);
        });


    var substringMatcher = function(strs) {
        return function findMatches(q, cb) {
            var matches, substrRegex;

            // an array that will be populated with substring matches
            matches = [];

            // regex used to determine if a string contains the substring `q`
            substrRegex = new RegExp(q, 'i');

            // iterate through the pool of strings and for any string that
            // contains the substring `q`, add it to the `matches` array
            $.each(strs, function(i, str) {
                if (substrRegex.test(str)) {
                    // the typeahead jQuery plugin expects suggestions to a
                    // JavaScript object, refer to typeahead docs for more info
                    matches.push({ value: str });
                }
            });

            cb(matches);
        };
    };

    $('#jobNumber').typeahead({
        hint: true,
        minLength: 2,
        highlight: true,
    },
        {

            name: 'projects',
            source: substringMatcher(projectNumbers.serializeArray()),
        });
});

我的输入框显示前面的类型，但可用数据是：

JobName1, JobName1, akhsfkh

substringMatcher函数没有改变它。

有什么建议吗？

Answer 1

You cannot return results from a getJSON request like that

你必须存储结果，并在回调中完成其余的工作。

var projectNumbers;
$.getJSON("http://jsbin.com/heyobi/1.json")
    .done(function (result) {
    projectNumbers = result;
    initTypeahead();
})
    .fail(function (jqXHR, textStatus, err) {
    alert('Error: ' + err);
});

var substringMatcher = function (strs) {
......
});

function initTypeahead() {
    $('.typeahead').typeahead({
        hint: true,
        minLength: 2,
        highlight: true,
    }, {
        name: 'projects',
        source: substringMatcher(projectNumbers),
    });
}

这是DEMO

TypeaheadJS方法不将Array转换为值

1 个答案: