题
注意:没有jQuery
我怎么能这样做:
var array = new Array();
array[name] = "Tom";
array[age] = 15;
foreach(array as key=>value){
alert(key + " = " + value);
}
解决方案
首先,你应该把它叫做 obj
或 person
而不是 array
;数组是一系列相似的元素,而不是单个对象。
你可以这样做:
var person = new Object();
person['name'] = "Tom";
person['age'] = 15;
for (var key in person) {
if(!person.hasOwnProperty(key)) continue; //Skip base props like toString
alert(key + " = " + person[key]);
}
您还可以使用属性初始化对象,如下所示:
person.name = "Tom";
person.age = 15;
您还可以使用 JavaScript对象文字语法:
var person = { name: "Tom", age: 15 };
其他提示
这将在您的简单示例场景中起作用:
for (var key in array) {
alert(key + " = " + array[key]);
}
对于一般用途,建议您进行测试以确保该物业没有'被嫁接到继承链中其他地方的对象上:
for (var key in array) {
if (array.hasOwnProperty(key)) {
alert(key + " = " + array[key]);
}
}
使用javascript对象
var object = {};
object.name = "Tom";
object.age = 15;
for ( var i in object ) {
console.log(i+' = '+ object[i]);
}
首先,你不需要一个数组,你想要一个对象。关于什么构成数组的PHP的想法坦率地说有点奇怪。
var stuff = {
name: "Tom",
age: 15
};
/* Note: you could also have done
var stuff = {};
stuff.name = "Tom";
stuff.age = 15;
// or
var stuff = {};
stuff["name"] = "Tom";
stuff["age"] = 15;
*/
for (var key in stuff) {
alert(key + " = " + stuff[key];
}
key=0;while(key<array.length) {
alert(key + " = " + array.item(key));
key++;
}
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