刪除重複和合並Javascript數組的值

Question

我有一個JavaScript陣列，像這樣：

var recipients = [{ 
    name: 'Michael', 
    task: 'programming', 
    contactdetails: 'michael@michael.com' 
}, { 
    name: 'Michael', 
    task: 'designing', 
    contactdetails: 'michael@michael.com' 
}, { 
    name: 'Shane', 
    task: 'designing', 
    contactdetails: 'shane@shane.com' 
}]; 

什麼我做的是一個排班系統，我發出通知，誰是這個星期，所以電子郵件就像「你好，邁克爾你在這周編程」。目前它不是很好，因爲它發送了一個電子郵件給數組中的每個值。所以在上面的例子中，它會發送Michael 2郵件。

我想要做的是在合併任務屬性字符串時刪除重複項。因此，陣列將是：

var recipients = [{ 
    name: 'Michael', 
    task: 'programming, designing', 
    contactdetails: 'michael@michael.com' 
}, { 
    name: 'Shane', 
    task: 'designing', 
    contactdetails: 'shane@shane.com' 
}]; 

這樣，它可以只發送一個消息，如「嗨邁克爾你是編程，這周的設計」。我如何去做這件事？我也在使用Google Apps腳本，因此我需要一個純JavaScript解決方案。我還應該補充說，每個人的姓名和電子郵件地址將永遠是相同的，所以邁克爾永遠不會有一個不同的電子郵件地址等您的幫助非常感謝！

Answer 1

這將是使用reduce函數的好機會。

我們所做的是循環遍歷每個原始收件人列表，看看我們是否已經處理過元素（如果有的話），將當前元素的任務追加到已處理的元素，否則將當前收件人添加到處理過的列表

// original array 
var recipients = [ 
    {name: 'Michael',task:'programming',contactdetails:'michael@michael.com'}, 
    {name: 'Michael',task:'designing',contactdetails:'michael@michael.com'}, 
    {name: 'Shane',task:'designing',contactdetails:'shane@shane.com'} 
]; 
var recipientKeyList = []; // used to store the contacts we've already processed 
// cycle through each recipient element 
var newRecipients = recipients.reduce(function(allRecipients, recipient){ 
    // get the indexOf our processed array for the current recipient 
    var index = recipientKeyList.indexOf(recipient.contactdetails); 
    // if the contact details already exist, append the task 
    if(index >= 0){ 
     allRecipients[index].task = allRecipients[index].task + ', ' + recipient.task; 
     return allRecipients 
    }else{ // otherwise append the recipient 
     recipientKeyList.push(recipient.contactdetails) 
     return allRecipients.concat(recipient); 
    } 

}, []); 

Answer 2

var recipients = [{name: 'Michael',task:'programming',contactdetails:'michael@michael.com'},{name: 'Michael',task:'designing',contactdetails:'michael@michael.com'},{name: 'Shane',task:'designing',contactdetails:'shane@shane.com'}]; 

var tempObj = {}; 
for (i=0; i<recipients.length; i++) { 
    if (!tempObj[recipients[i]['name']]) { 
     tempObj[recipients[i]['name']] = {}; 
     tempObj[recipients[i]['name']]['task'] = []; 
    } 
    tempObj[recipients[i]['name']]['task'].push(recipients[i]['task']); 
    tempObj[recipients[i]['name']]['contactdetails'] = recipients[i]['contactdetails']; 
} 

var new_arr = []; 
Object.keys(tempObj).forEach(function(key) { 
    new_arr.push({name: key, task: tempObj[key]['task'].join(", "), contactdetails: tempObj[key]['contactdetails']}) 
}); 

Answer 3

轉換陣列與鍵name對象（可以是電子郵件也）

// original array 
 
var recipients = [ 
 
    {name: 'Michael',task:'programming',contactdetails:'michael@michael.com'}, 
 
    {name: 'Michael',task:'designing',contactdetails:'michael@michael.com'}, 
 
    {name: 'Shane',task:'designing',contactdetails:'shane@shane.com'} 
 
]; 
 

 
var recipientsObj = {}; 
 
for (var i = 0; i < recipients.length; i++) { 
 
    var element = recipients[i]; 
 
    var recipientInObj = recipientsObj[element.name] 
 
    if (recipientInObj) { 
 
    // If a recipient is repeated with same task, here duplicates will appear 
 
    recipientInObj.task += ', ' + element.task; 
 
    } else { 
 
    recipientsObj[element.name] = element; 
 
    } 
 
} 
 
console.log(recipientsObj)

Answer 4

迭代和尋找相同的對象，如果再附加任務，這樣

var recipients = [{ 
 
    name: 'Michael', 
 
    task: 'programming', 
 
    contactdetails: 'michael@michael.com' 
 
}, { 
 
    name: 'Michael', 
 
    task: 'designing', 
 
    contactdetails: 'michael@michael.com' 
 
}, { 
 
    name: 'Shane', 
 
    task: 'designing', 
 
    contactdetails: 'shane@shane.com' 
 
}]; 
 

 

 
var uniqueR = []; 
 
var copyRecipients = JSON.parse(JSON.stringify(recipients)); 
 

 
copyRecipients .forEach(function(ele){ 
 
    var obj = uniqueR.find(function(e){ 
 
    return (e.name == ele.name && e.contactdetails == ele.contactdetails); 
 
    }); 
 
    if(obj){ 
 
    obj.task = obj.task + ", " + ele.task; 
 
    }else{ 
 
    uniqueR.push(ele); 
 
    } 
 
}); 
 

 
console.log(uniqueR)

Answer 5

var newJSON = {}; （json.contactdetails）= {name：json [「name」]，任務：newJSON [json.contactdetails]！= undefined & & newJSON [json.contactdetails ] [「task」]！= undefined？newJSON [json.contactdetails] [「task」] +「，」+ json [「task」]：json [「task」]}

}）;