MongoDB中批量将时间戳转变通用日期格式
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1,官网提供的MongoDB遍历脚本:
官方文档地址:https://docs.mongodb.org/manual/tutorial/remove-documents/ >var arr = ["ab","cd","ef"] >var show = function(value,index,ar){ print(value) } >arr.forEach(show) ab cd ef
2,MongoDB的模糊查询
本帖下載内容已隐藏,请登入以查看隐藏内容!脚本,这么说来,直接在窗口里面写js脚本来实现就ok了,然后准备拿一条数据来验证是否正确,结果成功了,验证脚本如下: – 单独一条集合数据中,将时间戳变成日期字符串:db.paymentinfo.find({"_id":ObjectId("55d56cbbe4b0c1f89b5356a4")}).forEach(function (a) { #这个函数是在月、日、时分秒的个位数字前面补0操作的 function tran_val(val){ if(parseInt(val)<10){ val="0" +val; } return val; } # 这里是paymentTime为时间戳 var datenew = new Date(parseInt(paymentTime)); # 获取年月日 var year=datenew.getFullYear(); var month=tran_val(datenew.getMonth()+1); var date=tran_val(datenew.getDate()); # 获取时分秒 var hour=tran_val(datenew.getHours()); var minute=tran_val(datenew.getMinutes()); var second=tran_val(datenew.getSeconds()); # 组装成标准的日期格式yyyy-mm-dd hh:mm:ss var datastr=year+"-"+month+"-"+date+" "+hour+":"+minute+":"+second; a["paymentTime"]=datastr print(paymentTime); printjson(a) } );
上面的例子表明直接用js脚本可以实现时间戳到日期格式转变,那么下面就开始for循环批量修改:
db.paymentinfo.update({"_id": ObjectId("55d56fdbe4b0c1f89b5356ae")},{$set:{"paymentTime" : "14400511608049527"}},true); # 使用遍历数组的方式来操作144开头的时间戳 var ds= db.paymentinfo.find({"paymentTime": {$regex: '144', $options:'i'}}); var dschilds=ds.toArray(); for (var i = 0;i <dschilds.length ; i++) { var child=dschilds[i]; var id=child._id; var paymentTime=child.paymentTime; print(paymentTime); function tran_val(val){ if(parseInt(val)<10){ val="0" +val; } return val; } var datenew = new Date(parseInt(paymentTime)); var year=datenew.getFullYear(); var month=tran_val(datenew.getMonth()+1); var date=tran_val(datenew.getDate()); var hour=tran_val(datenew.getHours()); var minute=tran_val(datenew.getMinutes()); var second=tran_val(datenew.getSeconds()); var datestr=year+"-"+month+"-"+date+" "+hour+":"+minute+":"+second; # 这里开始进行修改操作 db.paymentinfo.update({"_id": id},{$set:{"paymentTime" :datestr}},true); db.paymentinfo.find({"_id": id}); }
# 使用遍历数组的方式来操作145开头的时间戳 var ds= db.paymentinfo.find({"paymentTime": {$regex: '145', $options:'i'}}); var dschilds=ds.toArray(); for (var i = 0;i <dschilds.length ; i++) { var child=dschilds[i]; var id=child._id; var paymentTime=child.paymentTime; print(paymentTime); function tran_val(val){ if(parseInt(val)<10){ val="0" +val; } return val; } var datenew = new Date(parseInt(paymentTime)); var year=datenew.getFullYear(); var month=tran_val(datenew.getMonth()+1); var date=tran_val(datenew.getDate()); var hour=tran_val(datenew.getHours()); var minute=tran_val(datenew.getMinutes()); var second=tran_val(datenew.getSeconds()); var datestr=year+"-"+month+"-"+date+" "+hour+":"+minute+":"+second; db.paymentinfo.update({"_id": id},{$set:{"paymentTime" :datestr}},true); db.paymentinfo.find({"_id": id}); }
6,碰到新的问题,统一日期格式,将斜杠变成横杠
–批量修改日期 斜杠变成横杠
var ds= db.paymentinfo.find({"paymentTime": {$regex: '/', $options:'i'}}); var dschilds=ds.toArray(); for (var i = 0;i <dschilds.length; i++) { var child=dschilds[i]; var id=child._id; var paymentTime=child.paymentTime; var paymentTime2=paymentTime.replace(/\//g,"-"); db.paymentinfo.update({"_id": id},{$set:{"paymentTime" :paymentTime2}},true); print(paymentTime);print(paymentTime2); db.paymentinfo.find({"_id": id}); }
-- insert data insert into t1 select 1,'a' from db1.t2; call db1.proc_get_fints
OK,到此圆满解决