传递外部参数保持隐式参数

Question

我正在为qgis 3编写一个python插件

基本上，当用户点击一个特性时，我会尝试获取它

mapTool.featureIdentified.connect(onFeatureIdentified)

因此，在函数中确定的特性

def onFeatureIdentified(feature):
        print("feature selected : "+ str(feature.id()))

已识别的方法功能传递隐式参数

void QgsMapToolIdentifyFeature::featureIdentified ( const QgsFeature & ) void QgsMapToolIdentifyFeature::featureIdentified ( QgsFeatureId )

我的问题是，我想传递一个其他参数的功能（我想关闭我的窗口时，一个功能是确定的） 就像这样：

mapTool.featureIdentified.connect(onFeatureIdentified(window))

def onFeatureIdentified(feature,window):
            print("feature selected : "+ str(feature.id()))
            window.quit()

这样，window参数就会覆盖本机方法的隐式参数

我该怎么办

Answer 1

有两种方法：

• 使用lambda函数（归功于acw1668）传递第二个参数

  mapTool.featureIdentified.connect(lambda feature: onFeatureIdentified(feature, window))
  

  那么

  def onFeatureIdentified(feature,window):
  

• 如果你使用class（像我一样）：

  在类的\uu init\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu

  mapTool.featureIdentified.connect(self.onFeatureIdentified)
  

  那么

  def onFeatureIdentified(self,feature):
  print("feature selected : "+ str(feature.id()))
  self.window.quit()
  

  第一个参数将是self，然后它将传递函数的本机参数