是否可以检查胡子js的特定值{{name}} == "James"?
数据:
json: {
name: "James"
}
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HTML:
{{name}} //Will give me James as output
{{name == "James" }} //Is it possible to check specific value?
Run Code Online (Sandbox Code Playgroud) 如果我的API中的某些东西变坏了我想要返回一个http 500请求.
$app = new Slim();
$app->halt(500);
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它仍然返回一个http 200.
如果我运行此代码:
$status = $app->response()->status();
echo $status; //Here it is 200
$status = $app->response()->status(500);
echo $status; //Here it is 500
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它仍然给我一个http 200
我在javascript敲门中的$ parent有问题.
如果我有型号:
var Person = function () {
var self = this;
self.person_Id = ko.observable();
self.firstName = ko.observable();
self.lastName = ko.observable();
self.age = ko.observable();
self.role = ko.observable();
self.init = function (data) {
self.person_Id(data.person_Id);
self.firstName(data.firstName);
self.lastName(data.lastName);
self.age(data.age);
self.role(data.role);
};
};
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和集合
var PersonCollection = function () {
var self = this;
self.ItemToEditOrToAdd = ko.observable(new Person());
self.persons = ko.observableArray();
self.AddItem = function (item) {
self.persons.push(item);
};
};
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和HTML
<div id="new" data-bind="with: PersonCollectionInstance.ItemToEditOrToAdd">
<p>Firstname: <input type="text" data-bind="value: firstName"/></p>
<p>Lastname: <input type="text" data-bind="value: …Run Code Online (Sandbox Code Playgroud) 如果我单击一个元素"Link1",则该函数中的e.target是节点Link1.我想知道这个节点在ul子节点中的索引是什么,在这种情况下我希望indexOf返回0,因为Link1在位置0上,而ii单击2我希望它为1.
HTML
<div class="link">
<ul>
<li><a>Link1</a></li>
<li><a>Link2</a></li>
</ul>
</div>
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JAVASCRIPT
self.query('.link').forEach(function(linkNode, flikIndex, flikArr) {
dojo.query(linkNode, 'click', function(e) {
var t = e.target; //If i click Link1 this is Link1 and if i click Link2 and so on.
var parent = t.parentNode; //Contains the parent to my a in this case li
var ancestor = t.parentNode.parentNode.childNodes; //Containes 2 li
var index = Array.prototype.indexOf.call(parent, ancestor); //Return -1 but i want it to return 0 because Link1 is on place [0] in the …Run Code Online (Sandbox Code Playgroud)