Question

我正在使用API，只要您搜索某些内容，就会返回JSON。它基本上是一个自动完成搜索API。每当您开始在框中键入内容时，它都会使用GET请求命中API端点并返回JSON。假设您开始输入＆＃34; lucky＆＃34; ，然后请求是https://autocomplete.clearbit.com/v1/companies/suggest?query=lucky，JSON响应是

[{"name":"Lucky Brand","domain":"luckybrand.com","logo":"https://logo.clearbit.com/luckybrand.com"},{"name":"LuckyVitamin.com","domain":"luckyvitamin.com","logo":"https://logo.clearbit.com/luckyvitamin.com"},{"name":"Lucky Gunner Ammo","domain":"luckygunner.com","logo":"https://logo.clearbit.com/luckygunner.com"},{"name":"Lucky Orange","domain":"luckyorange.com","logo":"https://logo.clearbit.com/luckyorange.com"},{"name":"Lucky's Market","domain":"luckysmarket.com","logo":"https://logo.clearbit.com/luckysmarket.com"}]

它返回名称，域名和徽标。我有一个HTML搜索框，因此，无论何时开始输入，我都要在每个项目的行中显示logo image，name和domain。但它没有正确显示。这是我的代码，

html： -

<input type="text" placeholder="type something ..." id="suggest" />

css： -

body{
    padding: 30px;
}

JS： - （我使用的是jQuery）

$(document).ready(function () {

    $("#suggest").autocomplete({
        delay: 100,
        source: function (request, response) {

            // Suggest URL
            var suggestURL = "https://autocomplete.clearbit.com/v1/companies/suggest?query=%QUERY";
            suggestURL = suggestURL.replace('%QUERY', request.term);

            // JSON Request
            $.ajax({
                method: 'GET',
                dataType: 'json',
                jsonCallback: 'jsonCallback',
                url: suggestURL
            })
            .success(function(data){
                response(data[]); //Here I want to pass all the return 
                                  //items. I can show only one item, 
                                  //like data[1].name but not sure how 
                                  //to go through each item.

            });
        }
    });

});

Answer 1

您可以使用forEach（）方法来完成您的响应，例如

data.forEach(function(item) {
    console.log(item.name, item.logo, item.domain);
});

有关forEach here

的更多信息

使用jQuery显示JSON响应的数据

1 个答案: