Question

我有一个组合数组的函数，如果数组有一些值。这是我的阵列：

var emailAddr = [{
    customerID : "C20061102105257_678",
    emails : ["fandi@gmail.com", "fanjavaid@gmail.com"]
}];

var dataEmail2 = {
    customerID : "C20061102105257_678",
    emails : ["weisk@gmail.com"]
};

var dataEmail3 = {
    customerID : "C20061102105257_600",
    emails : ["eko@gmail.com"]
};

var dataEmail4 = {
    customerID : "C20061102105257_678",
    emails : ["1@gmail.com", "156@gmail.com"]
};

这是我的功能：

function addEmail(emailAddresses) {
    if(emailAddr.length == 0) {
        emailAddr.push(emailAddresses);
    } else {
        // Checking
        for (var i = 0; i < emailAddr.length; i++) {            
            if (emailAddr[i].customerID == emailAddresses.customerID) {
                for (var a = 0; a < emailAddresses.emails.length; a++) {
                    emailAddr[i]['emails'].push(emailAddresses.emails[a]);
                }

            } else {
                emailAddr.push(emailAddresses);
            }
        }

    }
} // End function

addEmail(dataEmail2);
addEmail(dataEmail3);
addEmail(dataEmail4);

console.log(emailAddr);

但是当我执行以下语句中的else语句时，我的浏览器崩溃了：

emailAddr.push(emailAddresses);

如何处理？谢谢。

Answer 1

这可能有用。我想你正试图拥有一系列独特的customerID，其中包含所有电子邮件。

＆＃13;

function addEmail(emailAddresses) {
    if(emailAddr.length == 0) {
        emailAddr.push(emailAddresses);
    } else {
        // Checking
        var tempCheck = true;
        for (var i = 0; i < emailAddr.length; i++) {            
            if (emailAddr[i].customerID == emailAddresses.customerID) {
                for (var a = 0; a < emailAddresses.emails.length; a++) {
                    emailAddr[i]['emails'].push(emailAddresses.emails[a]);
                }
                tempCheck = false;

            } 
        }
        if(tempCheck)
        {          
            emailAddr.push(emailAddresses);
        }

    }
} // End function

＆＃13;

Answer 2

这里有两个问题。

1 - 通过添加到当前正在阅读的数组，您将获得无限循环。这将导致它扩展，你将继续反复阅读这个扩展。

2 - 您实际上并没有将NEW对象推送到数组中，只需添加对现有对象的引用即可。这里会发生的是，当您添加到对象时，您还要添加到刚刚添加的对象。换句话说，你在这里有另一个无限循环

for (var a = 0; a < emailAddresses.emails.length; a++) {
    emailAddr[i]['emails'].push(emailAddresses.emails[a]);
}

当你推送到你的数组时，你实际上正在推动引用，这意味着它也将无限扩展。要解决此问题，请使用.slice（0）来复制数组，如下所示。

 var emailAddr = [{
        customerID: "C20061102105257_678",
        emails: ["fandi@gmail.com", "fanjavaid@gmail.com"]
    }];

    var dataEmail2 = {
        customerID: "C20061102105257_678",
        emails: ["weisk@gmail.com"]
    };

    var dataEmail3 = {
        customerID: "C20061102105257_600",
        emails: ["eko@gmail.com"]
    };

    var dataEmail4 = {
        customerID: "C20061102105257_678",
        emails: ["1@gmail.com", "156@gmail.com"]
    };

    function addEmail(emailAddresses) {
        if (emailAddr.length == 0) {
            emailAddr.push({ customerID: emailAddresses.customerID, emails: emailAddresses.emails.slice(0) });
        } else {
            // Checking
            var success = false;
            for (var i = 0; i < emailAddr.length; i++) {
                if (emailAddr[i].customerID == emailAddresses.customerID) {
                    success = true;
                    for (var a = 0; a < emailAddresses.emails.length; a++) {
                        console.log("adding : " + emailAddresses.emails[a]);
                        emailAddr[i]['emails'].push(emailAddresses.emails[a]);
                    }
                }
            }

            if (!success) {
                emailAddr.push({ customerID: emailAddresses.customerID, emails: emailAddresses.emails.slice(0) });
            }
        }
    } // End function

    addEmail(dataEmail2);
    addEmail(dataEmail3);
    addEmail(dataEmail4);

    console.log(emailAddr);

Answer 3

function addEmail(emailAddresses) {
    if(emailAddr.length == 0) {
        emailAddr.push(emailAddresses);
    } else {
        // Checking
        for (var i = 0; i < emailAddr.length; i++) {            
            if (emailAddr[i].customerID == emailAddresses.customerID) {
                for (var a = 0; a < emailAddresses.emails.length; a++) {
                   // emailAddr[a]['emails'].push(emailAddresses.emails[a]);

                   //replaced i with a on above line
                   //replaced i with a on above line

                    emailAddr[a]['emails'].push(emailAddresses.emails[a]);

                }

            } else {
                emailAddr.push(emailAddresses);
            }
        }

    }
} // End function

addEmail(dataEmail2);
addEmail(dataEmail3);
addEmail(dataEmail4);

console.log(emailAddr);

合并两个有一些价值的数组，在浏览器中崩溃

3 个答案: