谷歌地图 API 自动完成

Question

我正在使用Google Maps API Autocomplete(Places API(和JavaScript。

输入3个字符后如何运行自动完成？

这是我的代码：

function initAutocomplete() {
var input = $('#autocomplete')[0];
var autocomplete = new google.maps.places.Autocomplete(input);
autocomplete.addListener('place_changed', {
gPlace = autocomplete.getPlace();
});
var _addEventListener = (input.addEventListener) ? input.addEventListener : input.attachEvent;
function addEventListenerWrapper(type, listener) {
if (type == "keydown") {
var orig_listener = listener;
listener = function(event) {
var suggestion_selected = $(".pac-item-selected").length > 0;
if (event.which == 13 && !suggestion_selected) {
var simulated_downarrow = $.Event("keydown", {keyCode:40, which:40})
orig_listener.apply(input, [simulated_downarrow]);
}
orig_listener.apply(input, [event]);
};
}
_addEventListener.apply(input, [type, listener]);
}
if (input.addEventListener) {
input.addEventListener = addEventListenerWrapper;
} else if (input.attachEvent) {
input.attachEvent = addEventListenerWrapper;
}
}

我希望我清楚地解释了我的问题。 提前谢谢你。

Answer 1

您可以参考以下方法在 3 个字符后使用 Google 地图 API 自动完成功能：

function initAutocomplete() {
var input = $('#autocomplete')[0];
var autocomplete = new google.maps.places.Autocomplete(input);
autocomplete.addListener('place_changed', {
gPlace = autocomplete.getPlace();
});
var _addEventListener = (input.addEventListener) ? input.addEventListener : input.attachEvent;
function addEventListenerWrapper(type, listener) {
if (type == "keydown") {
var orig_listener = listener;
listener = function(event) {
var suggestion_selected = $(".pac-item-selected").length > 0;
if (event.which == 13 && !suggestion_selected) {
var simulated_downarrow = $.Event("keydown", {keyCode:40, which:40})
orig_listener.apply(input, [simulated_downarrow]);
}
orig_listener.apply(input, [event]);
};
}
_addEventListener.apply(input, [type, listener]);
}
if (input.addEventListener) {
input.addEventListener = addEventListenerWrapper;
} else if (input.attachEvent) {
input.attachEvent = addEventListenerWrapper;
}
return false;
}