如何從javascript中刪除對象的時間？

Question

我想從我的物體中的一個屬性中刪除時間。實際上，我檢查：索引，如果時間戳是否存在於屬性值中。實際上我有此

輸入

[ 
    { 
    "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" 
    } 
] 

輸出

[ 
    { 
    "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") 
     } 
    } 
}) 

原始陣列應保持不變，並且結果應該在單獨的陣列被捕獲。

這是鏈接到小提琴。

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)

產出

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

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

UPDATE

要保留原始對象並獲取引起其他物體使用下面的代碼。

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

您可以按空格拆分C屬性，然後重新加入要保留的部分。試試這個：

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

替代而不醜字符串連接：

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

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);

Answer 4

最好的辦法是將日期解析爲一個Date對象，然後重新格式化它。無論有沒有時間，您都可以根據自己的喜好進行格式化。有許多庫可用於格式化日期。你說你使用的是jQuery，有一個用於日期格式的插件。

我不會推薦你試圖將字符串分解成碎片。有許多日期格式（您不指定此日期的來源），並且您的解決方案可能很容易在將來停止工作。