【发布时间】:2018-05-23 15:51:32
【问题描述】:
这是我的 cron 作业 api:
var job = new cronJob('* * * * * *', function () {
Draft.find().then(data => {
var finalData = data;
Profsms.create({
phoneno: draftData.contacts.gsm,
sender: senderName,
message: message
}).then(function (data) {
if (data) {
console.log("successfully moved in profsms mysql");
} else {
console.log("failed");
}
})
});
}, function () {
console.log('DelCron Job finished.');
}, true, 'Asia/Calcutta');
在 finalData 变量中,我得到了这种类型的 json 对象:
{
"draftType" : "contactdraft",
"scheduledTime" : null,
"senderdata" : "",
"draftData" : {
"contacts" : [
{
"updatedAt" : "2017-12-07T12:09:10.000Z",
"createdAt" : "2017-12-07T12:09:10.000Z",
"data3" : "",
"data2" : "",
"data1" : "",
"country" : "",
"url" : "",
"company" : "",
"email" : "sameer@ech.com",
"dob" : null,
"postcode" : "",
"region" : "",
"city" : "",
"street" : "",
"lastName" : "Sameer",
"firstName" : "Mohamed",
"gsm" : "123344",
"id" : 12
},
{
"updatedAt" : "2017-12-07T12:09:58.000Z",
"createdAt" : "2017-12-07T12:09:58.000Z",
"data3" : "",
"data2" : "",
"data1" : "",
"country" : "",
"url" : "",
"company" : "",
"email" : "ganesh@ich.com",
"dob" : null,
"postcode" : "",
"region" : "",
"city" : "",
"street" : "",
"lastName" : "Pandiyan",
"firstName" : "Ganesh",
"gsm" : "1233",
"id" : 13
}
]
},
"senderName" : "ifelse",
"message" : "hey...",
"draftName" : "December 9",
}
我想将来自 draftData 和 senderName 的 gsm 号码和消息值插入到 db 中
想要收藏
phoneno: draftData.contacts.gsm,
sender: senderName,
message: message
如何做到这一点?
我正在使用 mysql 的 sequelize ORM。
你能帮我用 for 或 foreach 或 javascript 中的任何其他函数解决这个问题吗?
【问题讨论】:
标签: mysql arrays json node.js object