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我遇到了一個奇怪的問題。我確定它與變量的範圍有關,但我無法弄清楚原因。這是在谷歌鉻31.0.1650.57和Firefox 24.0被做了下面的代碼: 第一template.js
(以下部分開始對template.js
第28行)是由一個下拉觸發而改變對象在發送到函數時給出了未定義的類型
$("select[name=class]").change(function() {
var subclassSQL = new Array();
subclassSQL[0] = "`class`, `type`, `subclass`";
subclassSQL[1] = "subclass";
subclassSQL[2] = "class";
subclassSQL[3] = $("select[name=class] option:selected").html();
splitArr = new Array();
splitArr[0] = split;
splitArr[1] = "type";
call(subclassSQL, "subClass", "div.subClass", (function(){split(); Handlebars.registerHelper("top", function(object) {return new Handlebars.SafeString("<span>Choose your " + object[0].type + ":</span>\n<select name=\"subClass\">");});Handlebars.registerHelper("bottom", function() {return new Handlebars.SafeString("</select>");});}), splitArr, false);
});
則調用在ajaxCalls.js功能被觸發
function call(SQLarray, source, container, helper, function_, run) {
$.ajax({
url: "includes/ajaxHandler.php",
type: 'post',
dataType: 'json',
data: {info: SQLarray},
success: function(json) {
var template = Handlebars.templates[source],
tempData = {base: json}
console.log(tempData.base);
if (helper) {helper();}
if (function_) {
$(container).html(template(function_[0](function_.slice(1), tempData)));
} else {
$(container).html(template(tempData)); }
if (run) {update(jQuery);}
}
});
};
function split(field, tempData) {
console.log(tempData.base);
return tempData;
};
當在被稱爲控制檯檢查,第一console.log(tempData.base);
返回預期的JSON對象。但第二個返回
Uncaught TypeError: Cannot read property 'base' of undefined ajaxCalls.js:22
split ajaxCalls.js:22
(anonymous function) template.js:37
$.ajax.success ajaxCalls.js:11
c jquery.js:3048
p.fireWith jquery.js:3160
k jquery.js:8235
r jquery.js:8778
呼籲function_
的語法是:
split("type", tempData);
這是爲什麼返回一個未定義的?