问题是我使用的是这样的代码:
Well the problem is that I was using code like this:
new Date().toJSON().slice(0, 10)
将我的日期作为 YYYY-MM-DD
字符串,然后我在一些 mysql 查询和一些条件语句中使用它作为参数.在一天结束时,我没有得到正确的日期,因为它仍然在前一天(我的时区偏移量是 +2/3 小时).
to get my date as YYYY-MM-DD
string, then I use it like parameter in some mysql queries and in some condition statements. In the end of the day I wasn't getting the right date since it was still in the previous day (my timezone offset is +2/3 hours).
我没有注意到 toJSON
方法没有考虑到您的时区偏移,所以我最终得到了这个 hacky 解决方案:
I haven't noticed that the toJSON
method does not take into account your timezone offset, so I've ended up with this hacky solution:
var today = new Date();
today.setHours( today.getHours()+(today.getTimezoneOffset()/-60) );
console.log(today.toJSON().slice(0, 10));
有没有更优雅的解决方案?
Is there a more elegant solution?
ECMAScript 中的日期对象在内部是 UTC.时区偏移量用于当地时间.
Date objects in ECMAScript are internally UTC. The timezone offset is used for local times.
Date.prototype.toJSON 的规范说它使用 Date.prototype.toISOString,表示时区始终为 UTC".您的解决方案正在做的是将日期对象的 UTC 时间值偏移时区偏移量.
The specification for Date.prototype.toJSON says that it uses Date.prototype.toISOString, which states that "the timezone is always UTC". What your solution is doing is offsetting the UTC time value of the date object by the timezone offset.
考虑将您自己的方法添加到 Date.prototype,例如
Consider adding your own method to Date.prototype, e.g.
Date.prototype.toJSONLocal = function() {
function addZ(n) {
return (n<10? '0' : '') + n;
}
return this.getFullYear() + '-' +
addZ(this.getMonth() + 1) + '-' +
addZ(this.getDate());
}
如果你想挤出额外的性能,以下应该更快:
If you want to squeeze extra performance, the following should be faster:
Date.prototype.toJSONLocal = (function() {
function addZ(n) {
return (n<10? '0' : '') + n;
}
return function() {
return this.getFullYear() + '-' +
addZ(this.getMonth() + 1) + '-' +
addZ(this.getDate());
};
}())
但这有点过早优化的味道,所以除非你在很短的时间内调用它数千次,否则我不会打扰.
But that smacks of premature optimisation, so unless you are calling it thousands of times in a very short period, I wouldn't bother.
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