如何在JSON對象的鍵刪除空間

我有一個輸出象下面這樣：如何在JSON對象的鍵刪除空間

output = { 
    "New Classroom": [{ 
    "Name": "Apple", 
    "Age": "6", 
    "Percentage": "24.00%" 
    }, { 
    "Name": "Orange", 
    "Age": "5", 
    "Percentage": "9.88%" 
    }, { 
    "Name": "Green", 
    "Age": "2", 
    "Percentage": "27.27%" 
    }, { 
    "Name": "Grey", 
    "Age": "6", 
    "Percentage": "12.63%" 
    }] 
}

如何與NewClassroom取代New Classroom和新教室並不總是一個「NewClassroom」。它可能是不同的文字

ob = JSON.parse(output); 

alert(Object.keys(ob))

我這樣做的時候，我得到Newclassroom爲重點

來源

2016-11-25 chandra sekhar juthuga

您可以通過對象的頂級屬性名稱循環您RECE如果有空格，請檢查任何空格，並刪除空格。（你不需要，他們是完全有效的屬性名稱，但你可以如果你想。）

var output = { "New Classroom": [{"Name": "Apple","Age": "6","Percentage": "24.00%"},{"Name": "Orange","Age": "5","Percentage": "9.88%"},{"Name": "Green","Age": "2","Percentage": "27.27%"},{"Name": "Grey","Age": "6","Percentage": "12.63%"}]}; 
 
var name, newName; 
 
// Loop through the property names 
 
for (var name in output) { 
 
    // Get the name without spaces 
 
    newName = name.replace(/ /g, ""); 
 
    // If that's different... 
 
    if (newName != name) { 
 
    // Create the new property 
 
    output[newName] = output[name]; 
 
    // Delete the old one 
 
    delete output[name]; 
 
    } 
 
} 
 
console.log(output);

注意，在對象上使用delete可以降低性能後續的財產查詢。 99.99％的時間，沒關係。如果你的情況的問題，創建一個新對象，而不是在地方修改它：

var output = { "New Classroom": [{"Name": "Apple","Age": "6","Percentage": "24.00%"},{"Name": "Orange","Age": "5","Percentage": "9.88%"},{"Name": "Green","Age": "2","Percentage": "27.27%"},{"Name": "Grey","Age": "6","Percentage": "12.63%"}]}; 
 
var name, newName; 
 
var newOutput = {}; 
 
// Loop through the property names 
 
for (var name in output) { 
 
    // Get the name without spaces 
 
    newName = name.replace(/ /g, ""); 
 
    
 
    // Copy the property over 
 
    newOutput[newName] = output[name]; 
 
} 
 
console.log(newOutput);

來源

2016-11-25 11:55:36

所以如果'Percentage'已經得到了一些空間，這將不起作用？ – Mahi

@馬希：我的印象不是問題。不，它不會，因爲我們只處理頂層對象。當然，可以將相同的主體應用於從屬數組中的對象... –

使用Object.keys獲取對象
使用String#replace到的所有鍵替換字符String

var obj = { 
 
    "New Classroom": [{ 
 
    "Name": "Apple", 
 
    "Age": "6", 
 
    "Percentage": "24.00%" 
 
    }, { 
 
    "Name": "Orange", 
 
    "Age": "5", 
 
    "Percentage": "9.88%" 
 
    }, { 
 
    "Name": "Green", 
 
    "Age": "2", 
 
    "Percentage": "27.27%" 
 
    }, { 
 
    "Name": "Grey", 
 
    "Age": "6", 
 
    "Percentage": "12.63%" 
 
    }] 
 
}; 
 

 
Object.keys(obj).forEach(function(key) { 
 
    var replaced = key.replace(' ', ''); 
 
    if (key !== replaced) { 
 
    obj[replaced] = obj[key]; 
 
    delete obj[key]; 
 
    } 
 
}); 
 
console.log(obj);

注：的空間只有單一的出現被認爲，RegEx可如果space發生是不止一次被使用！

來源

2016-11-25 11:55:46 Rayon

使用正則表達式進行替換，否則它只會替換第一個空格。 – MrCode

循環在json的每個鍵中，然後解析。

嘗試正則表達式

var word = "New Classroom" 
 
word = word.replace(/\s/g, ''); 
 
console.log(word)

來源

2016-11-25 11:56:09 vistajess

如何在JSON對象的鍵刪除空間

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