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0

I want to create a log document for each task that is late.

const dte = new Date();
 Task.find()
 .where("status")
 .ne("Done", "Stuck", "Late")
 .where("date")
 .lt(dte)
 .updateMany( status: "Late" )
 .map(task => 
 const newLog = new Log(
 message: `$moment(dte).format("LT"): The task: $
 department.name
 is delayed`,
 type: "late"
 );

 newLog.save();
 );

I expect to have the logs created at the db, the error message that I got is .map() is not a function.

node.js mongodb mongoose

share|improve this question

asked Mar 28 at 0:31

Luis DorizLuis Doriz

511 bronze badge

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  updateMany() returns a Promise and a Promise had no map() function. you should do then() after updatedMany().
  
  – Gerald
  Mar 28 at 0:37
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I already tried that, it returns: "TypeError: tasks.map is not a function"
  
  – Luis Doriz
  Mar 28 at 5:37
  

  
  
  
  

add a comment | 

0

I want to create a log document for each task that is late.

const dte = new Date();
 Task.find()
 .where("status")
 .ne("Done", "Stuck", "Late")
 .where("date")
 .lt(dte)
 .updateMany( status: "Late" )
 .map(task => 
 const newLog = new Log(
 message: `$moment(dte).format("LT"): The task: $
 department.name
 is delayed`,
 type: "late"
 );

 newLog.save();
 );

I expect to have the logs created at the db, the error message that I got is .map() is not a function.

node.js mongodb mongoose

share|improve this question

asked Mar 28 at 0:31

Luis DorizLuis Doriz

511 bronze badge

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  updateMany() returns a Promise and a Promise had no map() function. you should do then() after updatedMany().
  
  – Gerald
  Mar 28 at 0:37
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I already tried that, it returns: "TypeError: tasks.map is not a function"
  
  – Luis Doriz
  Mar 28 at 5:37
  

  
  
  
  

add a comment | 

I want to create a log document for each task that is late.

const dte = new Date();
 Task.find()
 .where("status")
 .ne("Done", "Stuck", "Late")
 .where("date")
 .lt(dte)
 .updateMany( status: "Late" )
 .map(task => 
 const newLog = new Log(
 message: `$moment(dte).format("LT"): The task: $
 department.name
 is delayed`,
 type: "late"
 );

 newLog.save();
 );

I expect to have the logs created at the db, the error message that I got is .map() is not a function.

node.js mongodb mongoose

share|improve this question

asked Mar 28 at 0:31

Luis DorizLuis Doriz

511 bronze badge

const dte = new Date();
 Task.find()
 .where("status")
 .ne("Done", "Stuck", "Late")
 .where("date")
 .lt(dte)
 .updateMany( status: "Late" )
 .map(task => 
 const newLog = new Log(
 message: `$moment(dte).format("LT"): The task: $
 department.name
 is delayed`,
 type: "late"
 );

 newLog.save();
 );

share|improve this question

asked Mar 28 at 0:31

Luis DorizLuis Doriz

511 bronze badge

share|improve this question

asked Mar 28 at 0:31

Luis DorizLuis Doriz

511 bronze badge

asked Mar 28 at 0:31

Luis DorizLuis Doriz

511 bronze badge

asked Mar 28 at 0:31

Luis DorizLuis Doriz

511 bronze badge

1 Answer
1

active

oldest

votes

0

You can do this.

const dte = new Date();
Task.find(
 status: 
 $ne: ["Done", "Stuck", "Late"]
,
 date: 
 $lt: dte
 
).then((tasks) => 

 // loop through tasks
 tasks.forEach(async (task) => 
 const newLog = new Log(
 message: `$moment(dte).format("LT"): The task: $
 department.name
 is delayed`,
 type: "late"
 );

 await newLog.save();

 // update task
 task.set( status: "Late" );
 await task.save();
 );
)

share|improve this answer

edited Mar 28 at 6:10

answered Mar 28 at 6:05

GeraldGerald

13388 bronze badges

add a comment | 

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0

You can do this.

const dte = new Date();
Task.find(
 status: 
 $ne: ["Done", "Stuck", "Late"]
,
 date: 
 $lt: dte
 
).then((tasks) => 

 // loop through tasks
 tasks.forEach(async (task) => 
 const newLog = new Log(
 message: `$moment(dte).format("LT"): The task: $
 department.name
 is delayed`,
 type: "late"
 );

 await newLog.save();

 // update task
 task.set( status: "Late" );
 await task.save();
 );
)

share|improve this answer

edited Mar 28 at 6:10

answered Mar 28 at 6:05

GeraldGerald

13388 bronze badges

add a comment | 

0

You can do this.

const dte = new Date();
Task.find(
 status: 
 $ne: ["Done", "Stuck", "Late"]
,
 date: 
 $lt: dte
 
).then((tasks) => 

 // loop through tasks
 tasks.forEach(async (task) => 
 const newLog = new Log(
 message: `$moment(dte).format("LT"): The task: $
 department.name
 is delayed`,
 type: "late"
 );

 await newLog.save();

 // update task
 task.set( status: "Late" );
 await task.save();
 );
)

share|improve this answer

edited Mar 28 at 6:10

answered Mar 28 at 6:05

GeraldGerald

13388 bronze badges

add a comment | 

You can do this.

const dte = new Date();
Task.find(
 status: 
 $ne: ["Done", "Stuck", "Late"]
,
 date: 
 $lt: dte
 
).then((tasks) => 

 // loop through tasks
 tasks.forEach(async (task) => 
 const newLog = new Log(
 message: `$moment(dte).format("LT"): The task: $
 department.name
 is delayed`,
 type: "late"
 );

 await newLog.save();

 // update task
 task.set( status: "Late" );
 await task.save();
 );
)

share|improve this answer

edited Mar 28 at 6:10

answered Mar 28 at 6:05

GeraldGerald

13388 bronze badges

const dte = new Date();
Task.find(
 status: 
 $ne: ["Done", "Stuck", "Late"]
,
 date: 
 $lt: dte
 
).then((tasks) => 

 // loop through tasks
 tasks.forEach(async (task) => 
 const newLog = new Log(
 message: `$moment(dte).format("LT"): The task: $
 department.name
 is delayed`,
 type: "late"
 );

 await newLog.save();

 // update task
 task.set( status: "Late" );
 await task.save();
 );
)

share|improve this answer

edited Mar 28 at 6:10

answered Mar 28 at 6:05

GeraldGerald

13388 bronze badges

answered Mar 28 at 6:05

GeraldGerald

13388 bronze badges

answered Mar 28 at 6:05

GeraldGerald

13388 bronze badges