按鍵名過濾對象

Question

我有一個對象，我想過濾。這是我用：

query = { 
    "teststring-123": "true", 
    "teststring-12344566": "false", 
    test: "true" 
} 

我想，這樣過濾後，我只需要篩選查詢：

query = { 
    "teststring-123": "true", 
    "teststring-12344566": "false" 
} 

$(query).each(function(index, value) { 
    $.each(value, function(i, v) { 
     if(i.indexOf('teststring-')==-1) { 
      // remove part of query object where index is this one  
      console.log(index) 
     } 
    });  
}); 

我如何處理呢？

Answer 1

使用delete操作：

$.each(query, function(key, value) { 
    if(key.indexOf('teststring-') == -1) { 
     delete query[key]; 
    } 
});  

http://jsfiddle.net/NvZyA/（在本演示中，Object.keys()用於顯示所有密鑰）。

Answer 2

您可能正在尋找delete運營商。

Answer 3

您是否試圖刪除所有沒有以「teststring-」開頭的鍵的鍵值對？如果是這樣的......

for(var key in query){ 
    if(query.hasOwnProperty(key) && key.indexOf('teststring-') === -1){ 
     delete query[key]; 
    } 
} 

Answer 4

使用delete操作：

var query = { 
    "teststring-123": "true", 
    "teststring-12344566": "false", 
    test: "true" 
} 
$.each(query, function(sKey) { 
    if (sKey.indexOf("teststring-") < 0) { // or check if it is not on first position: != 0 
     delete query[sKey]; 
    } 
}); 

Answer 5

正如其他人所說，使用delete操作。然而，不需要重複該對象的屬性：

var query = { 
    "teststring-123" : true, 
    "teststring-12344566" : false, 
    test: true 
}; 

delete query['test']; 

Answer 6

喜歡這個？

var query = { 
    "teststring-123": "true", 
    "teststring-12344566": "false", 
    "test": "true" 
} 
var newobj = {}; 

$.each(query, function(i, v) { 
    if(i.indexOf('teststring-') != -1) { 
     // remove part of query object where index is this one   
     console.log(i); 
     newarr[i] = v; 
    } 
}); 
console.log(newobj);