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