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One-one communication

How to find feature in OpenLayers exsisting layer

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

2

I have a vector layer in my OpenLayers applicaiton.

 var layer = new ol.layer.Vector(
 source: new ol.source.Vector(
 url: "my geojson url"
 )
 )

my GeoJSON data fills the vector layer.

I have an AJAX query that gets single feature from server. But I need to check my layer to control if feature is in layer.

$.ajax(settings.url).then(function(response)
 how to check if response feature in my layer?
)

Response is GeoJSON formatted data.

openlayers geojson

share|improve this question

edited Apr 12 at 11:20

Vince

14.8k32850

asked May 11 '17 at 21:25

bartelomabarteloma

6861226

add a comment | 

2

I have a vector layer in my OpenLayers applicaiton.

 var layer = new ol.layer.Vector(
 source: new ol.source.Vector(
 url: "my geojson url"
 )
 )

my GeoJSON data fills the vector layer.

I have an AJAX query that gets single feature from server. But I need to check my layer to control if feature is in layer.

$.ajax(settings.url).then(function(response)
 how to check if response feature in my layer?
)

Response is GeoJSON formatted data.

openlayers geojson

share|improve this question

edited Apr 12 at 11:20

Vince

14.8k32850

asked May 11 '17 at 21:25

bartelomabarteloma

6861226

add a comment | 

I have a vector layer in my OpenLayers applicaiton.

 var layer = new ol.layer.Vector(
 source: new ol.source.Vector(
 url: "my geojson url"
 )
 )

my GeoJSON data fills the vector layer.

I have an AJAX query that gets single feature from server. But I need to check my layer to control if feature is in layer.

$.ajax(settings.url).then(function(response)
 how to check if response feature in my layer?
)

Response is GeoJSON formatted data.

openlayers geojson

share|improve this question

edited Apr 12 at 11:20

Vince

14.8k32850

asked May 11 '17 at 21:25

bartelomabarteloma

6861226

 var layer = new ol.layer.Vector(
 source: new ol.source.Vector(
 url: "my geojson url"
 )
 )

$.ajax(settings.url).then(function(response)
 how to check if response feature in my layer?
)

share|improve this question

edited Apr 12 at 11:20

Vince

14.8k32850

asked May 11 '17 at 21:25

bartelomabarteloma

6861226

share|improve this question

edited Apr 12 at 11:20

Vince

14.8k32850

asked May 11 '17 at 21:25

bartelomabarteloma

6861226

edited Apr 12 at 11:20

Vince

14.8k32850

edited Apr 12 at 11:20

Vince

14.8k32850

asked May 11 '17 at 21:25

bartelomabarteloma

6861226

asked May 11 '17 at 21:25

bartelomabarteloma

6861226

2 Answers
2

active

oldest

votes

1

I'd use something like:

 features = vectorSource.getFeatures();

 found = false;
 for (i = 0; i < features.length && !found; i++) 
 if (features[i].getId() === id) 
 found = true;
 //Here I'm removing it, but you could do anything you like
 features.splice(i, 1);
 
 

share|improve this answer

answered May 12 '17 at 9:54

Ian Turton♦Ian Turton

50.4k548119

add a comment | 

0

You can easily do that by using the hasFeature() method of ol.source.Vector

var mySource = new ol.source.Vector(url: "geoJsonUrl");
var myLayer = new ol.layer.Vector(source: mySource);

$.ajax(settings.url).then(function(response)
 var feature = the response feature
 //note the feature must be of type ol.Feature and i think it sholud have an id too
 if(mySource.hasFeature(feature))
 //do logic here
 
)

check the doc at OpenLayers VectorSource hasFeature method

share|improve this answer

answered Apr 12 at 11:00

Obi Dennis ChizoluObi Dennis Chizolu

1

add a comment | 

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1

I'd use something like:

 features = vectorSource.getFeatures();

 found = false;
 for (i = 0; i < features.length && !found; i++) 
 if (features[i].getId() === id) 
 found = true;
 //Here I'm removing it, but you could do anything you like
 features.splice(i, 1);
 
 

share|improve this answer

answered May 12 '17 at 9:54

Ian Turton♦Ian Turton

50.4k548119

add a comment | 

1

I'd use something like:

 features = vectorSource.getFeatures();

 found = false;
 for (i = 0; i < features.length && !found; i++) 
 if (features[i].getId() === id) 
 found = true;
 //Here I'm removing it, but you could do anything you like
 features.splice(i, 1);
 
 

share|improve this answer

answered May 12 '17 at 9:54

Ian Turton♦Ian Turton

50.4k548119

add a comment | 

I'd use something like:

 features = vectorSource.getFeatures();

 found = false;
 for (i = 0; i < features.length && !found; i++) 
 if (features[i].getId() === id) 
 found = true;
 //Here I'm removing it, but you could do anything you like
 features.splice(i, 1);
 
 

share|improve this answer

answered May 12 '17 at 9:54

Ian Turton♦Ian Turton

50.4k548119

 features = vectorSource.getFeatures();

 found = false;
 for (i = 0; i < features.length && !found; i++) 
 if (features[i].getId() === id) 
 found = true;
 //Here I'm removing it, but you could do anything you like
 features.splice(i, 1);
 
 

share|improve this answer

answered May 12 '17 at 9:54

Ian Turton♦Ian Turton

50.4k548119

answered May 12 '17 at 9:54

Ian Turton♦Ian Turton

50.4k548119

answered May 12 '17 at 9:54

Ian Turton♦Ian Turton

50.4k548119

0

You can easily do that by using the hasFeature() method of ol.source.Vector

var mySource = new ol.source.Vector(url: "geoJsonUrl");
var myLayer = new ol.layer.Vector(source: mySource);

$.ajax(settings.url).then(function(response)
 var feature = the response feature
 //note the feature must be of type ol.Feature and i think it sholud have an id too
 if(mySource.hasFeature(feature))
 //do logic here
 
)

check the doc at OpenLayers VectorSource hasFeature method

share|improve this answer

answered Apr 12 at 11:00

Obi Dennis ChizoluObi Dennis Chizolu

1

add a comment | 

0

You can easily do that by using the hasFeature() method of ol.source.Vector

var mySource = new ol.source.Vector(url: "geoJsonUrl");
var myLayer = new ol.layer.Vector(source: mySource);

$.ajax(settings.url).then(function(response)
 var feature = the response feature
 //note the feature must be of type ol.Feature and i think it sholud have an id too
 if(mySource.hasFeature(feature))
 //do logic here
 
)

check the doc at OpenLayers VectorSource hasFeature method

share|improve this answer

answered Apr 12 at 11:00

Obi Dennis ChizoluObi Dennis Chizolu

1

add a comment | 

You can easily do that by using the hasFeature() method of ol.source.Vector

var mySource = new ol.source.Vector(url: "geoJsonUrl");
var myLayer = new ol.layer.Vector(source: mySource);

$.ajax(settings.url).then(function(response)
 var feature = the response feature
 //note the feature must be of type ol.Feature and i think it sholud have an id too
 if(mySource.hasFeature(feature))
 //do logic here
 
)

check the doc at OpenLayers VectorSource hasFeature method

share|improve this answer

answered Apr 12 at 11:00

Obi Dennis ChizoluObi Dennis Chizolu

1

var mySource = new ol.source.Vector(url: "geoJsonUrl");
var myLayer = new ol.layer.Vector(source: mySource);

$.ajax(settings.url).then(function(response)
 var feature = the response feature
 //note the feature must be of type ol.Feature and i think it sholud have an id too
 if(mySource.hasFeature(feature))
 //do logic here
 
)

share|improve this answer

answered Apr 12 at 11:00

Obi Dennis ChizoluObi Dennis Chizolu

1

answered Apr 12 at 11:00

Obi Dennis ChizoluObi Dennis Chizolu

1

answered Apr 12 at 11:00

Obi Dennis ChizoluObi Dennis Chizolu

1