说明如何添加复杂参数

Question

我需要得到url与查询参数，但我不知道如何。我需要这个->"entityTypeId = 172,过滤器(id) = 1,过滤器(id) = 3,过滤(id) = 5"。

在JS中我可以这样做

var httpBuildQuery = require('http-build-query');
var params = {
entityTypeId: 172,
filter: {
id: [1, 3, 5]
}};
const url = url + "?" + httpBuildQuery(params);
console.log(httpBuildQuery(params));

在PHP中

$params = array(
'filter' => array ('ID' => array('1', '3', '5'),),
'entityTypeId' => 172,
);
http_build_query($params);

In dart I try this

var uri = Uri(
scheme: 'http',
host: 'b24-ybr1v4.bitrix24.ru',
path: '/rest/1/token/crm.item.list.json',
queryParameters: {
'entityTypeId': '172',
'filter': [
{'id': '1'}
],
},
);

但是在这种情况下，我得到错误:

The following TypeErrorImpl was thrown while handling a gesture:
Expected a value of type 'String', but got one of type 'IdentityMap<String, String>'

如何获取"filter[id]"之类的参数?

Answer 1

您可以尝试将[id]部分移动到查询参数键中。

var uri = Uri(
scheme: 'http',
host: 'b24-ybr1v4.bitrix24.ru',
path: '/rest/1/token/crm.item.list.json',
queryParameters: {
'entityTypeId': '172',
'filter[id]': ['1', '3', '5'],
},
);