Question

我已经阅读了源代码，并查看了示例但尚未找到答案。

我需要设置鼠标光标下方修改叠加层上显示的图像的样式。

我正在使用自定义样式函数将中点和自定义端点添加到ol.interaction.Modify使用的图层。 ol.interaction.Modify将样式应用于鼠标光标附近的点，以指示可以修改该特征。这很好，除了光标样式落在自定义端点之下。我找不到改变z-index的方法。

Answer 1

所以，我正在为自己回答我的问题。我猜这就是互联网的精彩之处。我不是狗。

// we'd normally pass feature & resolution parameters to the function, but we're going to
// make this dynamic, so we'll return a style function for later use which will take those params.

DynamicStyleFunction = ( function( /* no feat/res yet!*/ ) {

  /**
   you really only get style are rendered upon simple geometries, not features. features are made of different geometry types, and styleFunctions are passed a feature that has its geometries rendered.  in terms of styling vector geometries, you have only a few options. side note: if there's some feature you expect to see on the the map and it's not showing up, you probably haven't properly styled it. Or, maybe it hasn't been put it in a collection that is included in the source layer... which is a hiccup for a different day.
  */

  // for any geometry that you want to be rendered, you'll want a style. 
  var styles = {};
  var s = styles;

 /**
   an ol.layer.Vector or FeatureOverlay, renders those features in its source by applying Styles made of Strokes, Fills, and Images (made of strokes and fills) on top of the simple geometries which make up the features

   Stroke styles get applied to ol.geom.GeometryType.LINE_STRING

   MULTI_LINE_STRING can get different styling if you want

  */

  var strokeLinesWhite =   new ol.style.Stroke({ 
      color:  [255, 255, 255, 1], // white
      width:  5,
    })

  var whiteLineStyle  new ol.style.Style({
    stroke: strokeLinesWhite
  })

  styles[ol.geom.GeometryType.LINE_STRING] = whiteLineStyle

  /** 
    Polygon styles get applied to ol.geom.GeometryType.POLYGON

    Polygons are gonna get filled. They also have Lines... so they can take stroke

  */

  var fillPolygonBlue = new ol.style.Style({
    fill:  new ol.style.Fill({
      color: [0, 153, 255, 1], // blue
    })
  })

  var whiteOutlinedBluePolygon = new ol.style.Style({
    stroke: strokeLinesWhite,
    fill:   fillPolygonBlue,
  })

  styles[ol.geom.GeometryType.POLYGON] = fillPolygonBlue

  /** 
    Circle styles get applied to ol.geom.GeometryType.POINT

    They're made with a radius and a fill, and the edge gets stroked... 

  */

  var smallRedCircleStyle = new ol.style.Style({ 
    image: new ol.style.Circle({
      radius: 5,
      fill: new ol.style.Fill({
        color: '#FF0000', // red... but i had to look it up
      })
    })
  })

  var whiteBigCircleWithBlueBorderStyle = new ol.style.Style({ 
    image: new ol.style.Circle({
      radius: 10,
      fill: new ol.style.Fill({
        color: '#FFFFFF' // i guessed it
      })
    }),
    stroke: new.ol.style.Stroke({
      color: '#0000FF',  // blue
      width: 5
    })
  })

  // render all points as small red circles
  styles[ol.geom.GeometryType.POINT] = smallRedCircleStyle


  // if you pass an array as the style argument, every rendering of the feature will apply every defined style style rendered with the geometry as the argument. that can be a whole lot of rendering in a FeatureOverlay... 

  smallRedCircleStyle.setZIndex(Infinity)
  whiteBigCircleWithBlueBorderStyle.setZIndex(Infinity -1) // that prob wouldn't work, but i hope it's instructive that you can tinker with styles

  // so... 
  var bullseyePointStyle = [ smallRedCircleStyle, whiteBigCircleWithBlueBorderStyle ];

  return function dynamicStyleFunction (feature, resolution){

    // this is the actual function getting invoked on each function call
    // do whatever you want with the feature/resolution.

    if (Array.indexOf(feature.getKeys('thisIsOurBullseyeNode') > -1) {
      return bullseyePointStyle
    } else if (feature.getGeometryName('whiteBlueBox')){
      return whiteOutlinedBluePolygon
    } else {
      return styles[feature.getGeometryName()]
    }
  }
})()

Answer 2

ol.interaction.Modify，ol.interaction.Select和ol.interaction.Draw使用style参数来更改草绘功能的外观。

在哪里可以找到在openlayers 3中实现自定义ol.style.Style的详细描述？

2 个答案: