Javascript 检查字典对象的长度
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Checking length of dictionary object
提问by PositiveGuy
I'm trying to check the length here. Tried count. Is there something I'm missing?
我想在这里检查长度。试过数。有什么我想念的吗？
var dNames = {};
dNames = GetAllNames();
for (var i = 0, l = dName.length; i < l; i++)
{
alert("Name: " + dName[i].name);
}
dNames holds name/value pairs. I know that dNames has values in that object but it's still completely skipping over that and when I alert out even dName.length obviously that's not how to do this...so not sure. Looked it up on the web. Could not find anything on this.
dNames 保存名称/值对。我知道 dNames 在那个对象中有值，但它仍然完全跳过那个，当我警告甚至 dName.length 显然这不是如何做到这一点......所以不确定。在网上查了一下。找不到任何关于此的信息。
回答by Greg Hornby
What I do is use Object.keys() to return a list of all the keys and then get the length of that
我所做的是使用 Object.keys() 返回所有键的列表，然后获取它的长度
Object.keys(dictionary).length
回答by sberry
var c = {'a':'A', 'b':'B', 'c':'C'};
var count = 0;
for (var i in c) {
if (c.hasOwnProperty(i)) count++;
}
alert(count);
回答by meder omuraliev
This question is confusing. A regular object, {}doesn't have a lengthproperty unless you're intending to make your own function constructor which generates custom objects which do have it ( in which case you didn't specify ).
这个问题令人困惑。常规对象{}没有length属性，除非您打算制作自己的函数构造函数来生成具有它的自定义对象（在这种情况下，您没有指定）。
Meaning, you have to get the "length" by a for..instatement on the object, since lengthis not set, and increment a counter.
意思是，您必须通过for..in对象上的语句获取“长度” ，因为length未设置，并增加计数器。
I'm confused as to why you need the length. Are you manually setting 0on the object, or are you relying on custom string keys? eg obj['foo'] = 'bar';. If the latter, again, why the need for length?
我很困惑为什么你需要length. 您是手动设置0对象，还是依赖自定义字符串键？例如obj['foo'] = 'bar';。如果是后者，为什么需要长度？
Edit #1: Why can't you just do this?
编辑#1：你为什么不能这样做？
list = [ {name:'john'}, {name:'bob'} ];
Then iterate over list? The lengthis already set.
然后遍历列表？在length已经设置。
回答by Cees Timmerman
Count and show keys in a dictionary (run in console):
计算并显示字典中的键（在控制台中运行）：
o=[];count=0; for (i in topicNames) { ++count; o.push(count+": "+ i) } o.join("\n")
Sample output:
示例输出：
"1: Phase-out Left-hand
2: Define All Top Level Taxonomies But Processes
3: 987
4: 16:00
5: Identify suppliers"
Simple count function:
简单的计数功能：
function size_dict(d){c=0; for (i in d) ++c; return c}
