我返回了rsltCallback函数,当我调用googleSearchSuggestions函数时,我得到了未定义的值。当我console.log输入参数里面的rsltCallback函数它将输出打印到控制台。如何访问全局局部变量的值
var googleSearchSuggestions = function(search_Keyword , element) {
var parsed;
var uri = 'http://suggestqueries.google.com/complete/search?output=toolbar&hl=en&q=' + search_Keyword;
var xhr = (typeof XMLHttpRequest !== 'undefined') ? new XMLHttpRequest() : new ActiveXObject(Microsoft.XMLHTTP);
xhr.onreadystatechange = function(){
xhr.responseType = 'xml';
if(xhr.status == 200 && xhr.readyState == 4)
{
var response = xhr.responseXML;
var items = response.getElementsByTagName('toplevel')
for (var i = 0 ; i < items[0].childNodes.length;i++)
{
parsed = items[0].childNodes[i].firstElementChild.getAttribute('data');
rsltcallBack(parsed);
}
}
};
xhr.open('GET', decodeURI(uri), true);
xhr.send();
var rsltcallBack = function(input){
return input;
};
return rsltCallBack();
};
http://stackoverflow.com/questions/14220321/how-to-return-the-response-from-an-ajax-call –