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Return latest 2 entries from a mongo database?

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0

I have a database recording the unix time, latitude and longitude of a received message from a GPS unit. I need a way to select the latest two entries from this database to work with in php.

Example database entry:
[3] => stdClass Object ( [_id] => MongoDBBSONObjectId Object ( [oid] => 5c7fe3fc1c37210bf96e8182 ) [Latitude] => 53.385360717773 [Longitude] => -6.6032276153564 [Time] => 1551885285 ) )

So far I only needed one and I got it by searching for the largest timestamp. Is there a way to simply return the latest two entries in mongo like in mySQL? I've noticed 'limit' returns the first entries not the last:

 $options=['limit' => 2];
 $filter=[];
 //$cursor = $collection->find($filter, $options);
 $cursor = $collection->find($filter);
 $largest = 0;
 foreach($cursor as $document){
 if ($largest < $document["Time"]) 
 $largest = $document["Time"];
 $longitude = $document["Longitude"];
 $latitude = $document["Latitude"];
 

Thanks for any help.

edit: I've found this question How to get the last N records in mongodb?. The mongo version is different but the principle seems sound: use 'sort' then 'limit'. The syntax is quite different thought how could I combine the two in this version?

"db.foo.find().sort(_id:1).limit(50);"

 $options=['limit' => 2];
 $filter=[];
 $cursor = $collection->find($filter, $options);

mongodb

share|improve this question

edited Mar 27 at 12:06

Jay.F

asked Mar 27 at 11:53

Jay.FJay.F

4188 bronze badges

add a comment | 

0

I have a database recording the unix time, latitude and longitude of a received message from a GPS unit. I need a way to select the latest two entries from this database to work with in php.

Example database entry:
[3] => stdClass Object ( [_id] => MongoDBBSONObjectId Object ( [oid] => 5c7fe3fc1c37210bf96e8182 ) [Latitude] => 53.385360717773 [Longitude] => -6.6032276153564 [Time] => 1551885285 ) )

So far I only needed one and I got it by searching for the largest timestamp. Is there a way to simply return the latest two entries in mongo like in mySQL? I've noticed 'limit' returns the first entries not the last:

 $options=['limit' => 2];
 $filter=[];
 //$cursor = $collection->find($filter, $options);
 $cursor = $collection->find($filter);
 $largest = 0;
 foreach($cursor as $document){
 if ($largest < $document["Time"]) 
 $largest = $document["Time"];
 $longitude = $document["Longitude"];
 $latitude = $document["Latitude"];
 

Thanks for any help.

edit: I've found this question How to get the last N records in mongodb?. The mongo version is different but the principle seems sound: use 'sort' then 'limit'. The syntax is quite different thought how could I combine the two in this version?

"db.foo.find().sort(_id:1).limit(50);"

 $options=['limit' => 2];
 $filter=[];
 $cursor = $collection->find($filter, $options);

mongodb

share|improve this question

edited Mar 27 at 12:06

Jay.F

asked Mar 27 at 11:53

Jay.FJay.F

4188 bronze badges

add a comment | 

I have a database recording the unix time, latitude and longitude of a received message from a GPS unit. I need a way to select the latest two entries from this database to work with in php.

Example database entry:
[3] => stdClass Object ( [_id] => MongoDBBSONObjectId Object ( [oid] => 5c7fe3fc1c37210bf96e8182 ) [Latitude] => 53.385360717773 [Longitude] => -6.6032276153564 [Time] => 1551885285 ) )

So far I only needed one and I got it by searching for the largest timestamp. Is there a way to simply return the latest two entries in mongo like in mySQL? I've noticed 'limit' returns the first entries not the last:

 $options=['limit' => 2];
 $filter=[];
 //$cursor = $collection->find($filter, $options);
 $cursor = $collection->find($filter);
 $largest = 0;
 foreach($cursor as $document){
 if ($largest < $document["Time"]) 
 $largest = $document["Time"];
 $longitude = $document["Longitude"];
 $latitude = $document["Latitude"];
 

Thanks for any help.

edit: I've found this question How to get the last N records in mongodb?. The mongo version is different but the principle seems sound: use 'sort' then 'limit'. The syntax is quite different thought how could I combine the two in this version?

"db.foo.find().sort(_id:1).limit(50);"

 $options=['limit' => 2];
 $filter=[];
 $cursor = $collection->find($filter, $options);

mongodb

share|improve this question

edited Mar 27 at 12:06

Jay.F

asked Mar 27 at 11:53

Jay.FJay.F

4188 bronze badges

 $options=['limit' => 2];
 $filter=[];
 //$cursor = $collection->find($filter, $options);
 $cursor = $collection->find($filter);
 $largest = 0;
 foreach($cursor as $document){
 if ($largest < $document["Time"]) 
 $largest = $document["Time"];
 $longitude = $document["Longitude"];
 $latitude = $document["Latitude"];
 

 $options=['limit' => 2];
 $filter=[];
 $cursor = $collection->find($filter, $options);

share|improve this question

edited Mar 27 at 12:06

Jay.F

asked Mar 27 at 11:53

Jay.FJay.F

4188 bronze badges

share|improve this question

edited Mar 27 at 12:06

Jay.F

asked Mar 27 at 11:53

Jay.FJay.F

4188 bronze badges

asked Mar 27 at 11:53

Jay.FJay.F

4188 bronze badges

asked Mar 27 at 11:53

Jay.FJay.F

4188 bronze badges

2 Answers
2

active

oldest

votes

1

Keep extending the query as follows:

$cursor = $collection->find($filter)->sort(array('_id'=>-1))->limit(2)

share|improve this answer

answered Mar 27 at 12:10

Anirudh SimhaAnirudh Simha

24977 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Caused an error, I don't think this is compatible with my version. Figured it it thought thanks for your help!
  
  – Jay.F
  Mar 27 at 12:22
  

  
  
  
  

add a comment | 

0

Got it to work with:

 $options = ['sort' => ['Time' => -1], 'limit' => 2];
 $filter=[];
 $cursor = $collection->find($filter, $options);

share|improve this answer

answered Mar 27 at 12:21

Jay.FJay.F

4188 bronze badges

add a comment | 

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1

Keep extending the query as follows:

$cursor = $collection->find($filter)->sort(array('_id'=>-1))->limit(2)

share|improve this answer

answered Mar 27 at 12:10

Anirudh SimhaAnirudh Simha

24977 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Caused an error, I don't think this is compatible with my version. Figured it it thought thanks for your help!
  
  – Jay.F
  Mar 27 at 12:22
  

  
  
  
  

add a comment | 

1

Keep extending the query as follows:

$cursor = $collection->find($filter)->sort(array('_id'=>-1))->limit(2)

share|improve this answer

answered Mar 27 at 12:10

Anirudh SimhaAnirudh Simha

24977 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Caused an error, I don't think this is compatible with my version. Figured it it thought thanks for your help!
  
  – Jay.F
  Mar 27 at 12:22
  

  
  
  
  

add a comment | 

Keep extending the query as follows:

$cursor = $collection->find($filter)->sort(array('_id'=>-1))->limit(2)

share|improve this answer

answered Mar 27 at 12:10

Anirudh SimhaAnirudh Simha

24977 bronze badges

$cursor = $collection->find($filter)->sort(array('_id'=>-1))->limit(2)

share|improve this answer

answered Mar 27 at 12:10

Anirudh SimhaAnirudh Simha

24977 bronze badges

answered Mar 27 at 12:10

Anirudh SimhaAnirudh Simha

24977 bronze badges

answered Mar 27 at 12:10

Anirudh SimhaAnirudh Simha

24977 bronze badges

0

Got it to work with:

 $options = ['sort' => ['Time' => -1], 'limit' => 2];
 $filter=[];
 $cursor = $collection->find($filter, $options);

share|improve this answer

answered Mar 27 at 12:21

Jay.FJay.F

4188 bronze badges

add a comment | 

0

Got it to work with:

 $options = ['sort' => ['Time' => -1], 'limit' => 2];
 $filter=[];
 $cursor = $collection->find($filter, $options);

share|improve this answer

answered Mar 27 at 12:21

Jay.FJay.F

4188 bronze badges

add a comment | 

Got it to work with:

 $options = ['sort' => ['Time' => -1], 'limit' => 2];
 $filter=[];
 $cursor = $collection->find($filter, $options);

share|improve this answer

answered Mar 27 at 12:21

Jay.FJay.F

4188 bronze badges

 $options = ['sort' => ['Time' => -1], 'limit' => 2];
 $filter=[];
 $cursor = $collection->find($filter, $options);

share|improve this answer

answered Mar 27 at 12:21

Jay.FJay.F

4188 bronze badges

answered Mar 27 at 12:21

Jay.FJay.F

4188 bronze badges

answered Mar 27 at 12:21

Jay.FJay.F

4188 bronze badges