Question

我正在尝试从我的对象中的某个属性中删除时间。实际上我检查：index，如果属性值中是否存在时间戳。其实我有这个

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"
  }
]

的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。

这是小提琴的链接。

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)

<强> OUTPUTS

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

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

<强>更新

要保留原始对象并使用以下代码获取其他对象的结果。

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

您可以将C属性拆分为空格，然后重新加入要保留的部分。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]    
};

Updated fiddle

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 3

最好的办法是将日期解析为Date对象，然后重新格式化。您可以根据自己的喜好对其进行格式化，无论是否有时间。有许多库可用于格式化日期。你声明你使用的是jQuery，有一个日期格式的插件。

我不建议您尝试将字符串分解为片段。有许多日期格式（您不必指定此日期的来源），您的解决方案将来很容易停止工作。

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*GMT\s*/g, '');
}
console.info(ary);

如何在javascript中删除对象的时间？

4 个答案: