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[英]How can I populate a html select element with the contents of an array using vanilla javascript?
[英]How Can I Populate Dropdown Select Element using getJSON
我想填充下拉 select 菜单。 我需要从此 JSON 获取 bankName 和 iINNo
我的 JSON:
{"status":true,"message":"Request Completed","data":[{"id":1,"activeFlag":1,"bankName":"Union Bank of India","details":"Union Bank of India","iINNo":"607161","remarks":"","timestamp":"21/11/2016 16:00:00"},{"id":2,"activeFlag":1,"bankName":"Bank of India","details":"Bank of India","iINNo":"508505","remarks":"","timestamp":"21/11/2016 16:00:00"},],"statusCode":0}
我的 Java 脚本:
let dropdown = $('#Bank');
dropdown.empty();
dropdown.append('<option selected="true" disabled>Select Bank</option>');
dropdown.prop('selectedIndex', 0);
const url = 'http://cors-anywhere.herokuapp.com/';
// Populate dropdown with list of provinces
$.getJSON(url, function (data) {
$.each(data, function (res, code) {
console.info(res);
console.info(code);
dropdown.append($('<option></option>').attr('value', value.iINNo).text(index.bankName));
})
});
带有银行数据的数组位于 JSON 内的属性data
下。 所以诀窍是遍历该属性 - 现在您遍历整个 JSON 响应,包括状态和消息属性。
您将获得个别银行 object 作为第二个参数item
,然后您可以访问它的属性并将 append 访问到您的下拉列表中:
$.each(data.data, function (index, item) {
console.info('Bank Data: ', item);
dropdown.append($('<option></option>').text(item.bankName));
});
这应该有效
const banks = document.getElementById('banks'); banks.innerHTML = '<option selected="true" disabled>Select Bank</option>'; fetch('https://cors-anywhere.herokuapp.com/http://www.allindiaapi.com/HMESEVAAEPS/GetBankDetails?tokenkey=KzhnBIUi1wuM97yizKJ07WB3YwPSykyX9CjB6C6O5LRyClZ9YZl9NcIk5f6Fjna4').then(response => response.json()).then(response => { if(response.status){ response.data.forEach(bank => { const option = document.createElement('option'); option.value = bank.iINNo; option.innerContent = bank.bankName; }); } });
<select id="banks"></select>
问题是因为你需要的数据在data
object 的data
属性中,所以你需要遍历data.data
。 此外,每个 object 的属性将在您的示例中的code
处理程序 function 的第二个参数中。 例如code.bankName
。
然而值得注意的是,代码可以被整理,并且还可以通过使用map()
构建单个option
元素字符串来提高其性能,您只需 append 到 DOM 一次。 尝试这个:
let $dropdown = $('#Bank'); $dropdown.html('<option value="" disabled>Select Bank</option>').val(''); // AJAX commented for example only: //$.getJSON(url, function(data) { let data = { "status": true, "message": "Request Completed", "data": [{ "id": 1, "activeFlag": 1, "bankName": "Union Bank of India", "details": "Union Bank of India", "iINNo": "607161", "remarks": "", "timestamp": "21/11/2016 16:00:00" }, { "id": 2, "activeFlag": 1, "bankName": "Bank of India", "details": "Bank of India", "iINNo": "508505", "remarks": "", "timestamp": "21/11/2016 16:00:00" }, ], "statusCode": 0 } let html = data.data.map(bank => `<option value="${bank.iINNo}">${bank.bankName}</option>`); $dropdown.append(html); //});
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script> <select id="Bank"></select>
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