[GIS] A script on this page may be busy, or it may have stopped responding

javascriptopen-source-gisopenlayers-2

I'm using OpenLayers – 2.11,GeoServer 2.13, ExtJS 3.4, and Java for developing GIS based web applications.

Recently i'm getting below error message in my application

"A script on this page may be busy, or it may have stopped responding. You can stop the script now, or you can continue to see if the script will complete." Mostly it showing OpenLayers.js file.

I've done"about:config" and set "dom.max_script_run_time" to 20 and more and nothing seems to work.

Please guide me to solve this issue..

Best Answer

You probably have an infinite loop in your code or you are trying to do something too heavy for your processor to handle like loading to many vector in your map. We can't help you if you don't provide some code, but even with an example, I suspect your code is really complex so it won't help much...

Solution

Check for infinite loop (or really long loop).
Check for big object cloning (maybe inside a loop).
Check for multiple object creation that may use all your RAM or CPU.
Try to remove some code or part of your application one by one until it start working again to find the problematic part
Build a small example that you can share through jsfiddle.net that reproduce the problem outside of your environment.
Ask again, if you didn't find the problem yet, by showing us your small example.

Related Solutions

[GIS] OpenLayers popup – random losing click event, have to refresh page

as i understand from the codes and your explanation, it's come to my mind:

in OpenLayers.Popup method, there is a parameter which name is closeBox. this works for:

{Boolean} Whether to display a close box inside the popup.

if you make it true, the closing button will show in popup, in this case when you click to closing box, it is going to destroy all feature from map which is irreversible for this popup.

 destroy: function()

 nullify references to prevent circular references and memory leaks

Solution:

make it hidden or unselect for second call.

you can achieve this sth. as:

function onPopupClose(evt) {
            selectControl.unselect(selectedFeature);
        }

   function onFeatureSelect(feature) {
            selectedFeature = feature;
            popup = new OpenLayers.Popup.FramedCloud("chicken", 
                                     feature.geometry.getBounds().getCenterLonLat(),
                                     null,
                                     "<div style='font-size:.8em'>Feature: " + feature.id 
            +"<br>Area: " + feature.geometry.getArea()+"</div>",
                                     null, true, onPopupClose);
            feature.popup = popup;
            map.addPopup(popup);
        }

i hope it helps you...

OpenLayers 2 JavaScript – How to Filter Vector Layer Features

Since you already setup the stylemaps for diplaying labels. You can easily do this by checking the feature attribute value of the required field with the search text

Try the below code to search in attributes

//JS code
    function search() {
    var search_name = document.getElementById("search_text").value;
    var features = layer.features;
    layer.destroyFeatures();
    for(var i=0; i< features.length; i++) {

     //features[i].attributes.name. you have the attribute field "name"

      if(features[i].attributes.name === search_name) {
        layer.addFeatures([features[i]])
      }
    }

    layer.redraw();
    }

<!-- HTML stuff  -->
    <input type="text" id="search_text" />
    <button onclick="search()">Search</button>

Update: hide/show using renderIntent

    //JS code

        function search() {
        var search_name = document.getElementById("search_text").value;
        var features = layer.features;
        for(var i=0; i< features.length; i++) {

         //features[i].attributes.name. you have the attribute field "name"

         //hide  a feature by default
          features[i].renderIntent = "delete"; 
          if(features[i].attributes.name === search_name) {
           //show a feature value of name attribute is search_name 
            features[i].renderIntent = "default" 
          }
        }

        layer.redraw();
        }

The second method becomes faster as renderIntent="delete" will set the css property display: none;(hiding) and renderIntent="default" will use the default rendering from your stylemap obivously showing the features.

Best Answer

Related Solutions

[GIS] OpenLayers popup – random losing click event, have to refresh page

OpenLayers 2 JavaScript – How to Filter Vector Layer Features

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