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I need to get all features' names, and for that I did a listview to put all features' names on it.

So when I click one feature, not on the map but on the listview, the feature must be selected.

Here it is my code:

for(var f=0;f<select.features.length;f++) {

        if(select.features[f].attributes.ADI == "TALAS") {
        alert(select.features[0].attributes.ADI);

        break;
        }
}

When I click a feature on a map, the select.features.length will be 1.

But I don't need to select the feature on the map, I must have the array of features and then I search attributes on this array and select that feature. That's what I would like to get.

I appreciate for any hint to get this working. I try to write some codes about that:

select.removeAllFeatures();     
filter = new OpenLayers.Filter.Comparison({
                type: OpenLayers.Filter.Comparison.EQUAL_TO,
                property: "ADI",
                value: "TALAS"
});

filterStrategy = new OpenLayers.Strategy.Filter({filter: filter});    
groups = new OpenLayers.Control.GetFeature({ 
            strategies: [new OpenLayers.Strategy.Fixed(),filterStrategy],
            protocol: OpenLayers.Protocol.WFS.fromWMSLayer(ilce) 
});

select.addFeatures(groups.feature);

I got this error message: Uncaught TypeError: Cannot set property 'layer' of undefined

1

I think you need something like this:

'featureselected': function(panel, feature) {
                    for (var f = 0; f < features.length; f++) {
                        features[f], 'default';
                    }
                    feature, 'select';
                }

This is perhaps what you have in mind:

http://api.geoext.org/1.1/examples/feature-grid.html

the code is here:

http://api.geoext.org/1.1/examples/feature-grid.js

  • that'is ext also i make this map for mobile phone. this codes are harmonious with mobile – Görkem Karadoğan Dec 11 '13 at 10:07

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