如何从javascript对象中删除时间

Question

我试图从我的对象属性之一删除时间。实际上，我用:index检查时间戳是否存在于属性值中。实际上我有这个

输入

[
  {
    "S": "Charge Interest Against Past Due Account",
    "C": "Mon Apr 13 10:38:05 GMT 2015"
  },
  {
    "S": "Charge Interest Against Past Due Account",
    "C": "Mon Apr 13 10:35:05 GMT 2015"
  }
]

输出

[
  {
    "S": "Charge Interest Against Past Due Account",
    "C": "Mon Apr 13  2015"
  },
  {
    "S": "Charge Interest Against Past Due Account",
    "C": "Mon Apr 13  2015"
  }
]

Javascript

var object = ary.map(function(o){
    var result={}
   // console.log(o)
    for(i in o){
        console.log(i+":"+o[i]) ;
        if(o[i].indexOf(":")!=-1){
            console.log("timestam present")
        }
    }
})

原始数组应保持不变，结果应在单独的数组中捕获。

Answer 1

var ary=[
  {
    "S": "Charge Interest Against Past Due Account",
    "C": "Mon Apr 13 10:38:05 GMT 2015"
  },
  {
    "S": "Charge Interest Against Past Due Account",
    "C": "Mon Apr 13 10:35:05 GMT 2015"
  }]
ary.map(function(o){
        var d = new Date(o.C);
        o.C = d.toDateString();
})
console.dir(ary)

C: "Mon Apr 13 2015"
S: "Charge Interest Against Past Due Account"
C: "Mon Apr 13 2015"
S: "Charge Interest Against Past Due Account"

保留原始对象，并在其他对象中获取结果，使用以下代码:

var ary=[
  {
    "S": "Charge Interest Against Past Due Account",
    "C": "Mon Apr 13 10:38:05 GMT 2015"
  },
  {
    "S": "Charge Interest Against Past Due Account",
    "C": "Mon Apr 13 10:35:05 GMT 2015"
  }]
var a = ary.map(function(o){
        var result = {};
        var d = new Date(o.C);
        result.S = o.S;
        result.C = d.toDateString();
        return result;
})
console.dir(a)

Answer 2

最好的方法是将日期解析为date对象，然后重新格式化。您可以按照自己的喜好设置格式，有或没有时间。有许多库可用于格式化日期。你说你使用的是jQuery，有一个日期格式的插件。

我不建议您尝试将字符串分成片段。有许多日期格式(您不指定该日期来自何处)，并且您的解决方案可以很容易地在将来停止工作。

Answer 3

您可以用空格分隔C属性，然后重新连接您想要保留的部分。试试这个:

for (var i = 0; i < ary.length; i++) {
    var dateValues = ary[i].C.split(' ');
    ary[i].C = dateValues[0] + ' ' + dateValues[1] + ' ' + dateValues[2] + ' ' + dateValues[5]    
};

更新小提琴

没有丑陋的字符串连接的替代方案:

for (var i = 0; i < ary.length; i++) {
    var dateValues = ary[i].C.split(' ')
    dateValues.splice(3, 2);
    ary[i].C = dateValues.join(' ');   
};

Answer 4

var ary = [
{
"S": "Charge Interest Against Past Due Account",
"C": "Mon Apr 13 10:38:05 GMT 2015"
},
{
"S": "Charge Interest Against Past Due Account",
"C": "Mon Apr 13 10:35:05 GMT 2015"
}];
for (var i = 0; i < ary.length; i++) {
    ary[i].C = ary[i].C.replace(/d+:d+:d+s*GMTs*/g, '');
}
console.info(ary);