如何打破以下的python行

Question

我遇到了几行与此类似的代码，但我不确定如何打破它：

blueprint = Blueprint(self.blueprint_map[str(self.ui.blueprint_combo.currentText())], runs=self.ui.runs_spin.text(), me=self.ui.me_spin.text(), pe=self.ui.pe_skill_combo.currentIndex())

提前致谢

Answer 1

blueprint = Blueprint(
    self.blueprint_map[str(self.ui.blueprint_combo.currentText())],
    runs=self.ui.runs_spin.text(), 
    me=self.ui.me_spin.text(),
    pe=self.ui.pe_skill_combo.currentIndex(),
)

Answer 2

这个怎么样

blueprint_item = self.blueprint_map[str(self.ui.blueprint_combo.currentText())]
blueprint = Blueprint(blueprint_item,
                      runs=self.ui.runs_spin.text(),
                      me=self.ui.me_spin.text(),
                      pe=self.ui.pe_skill_combo.currentIndex())

Answer 3

我这样做：

blueprint = Blueprint(
              self.blueprint_map[str(self.ui.blueprint_combo.currentText())],
              runs=self.ui.runs_spin.text(),
              me=self.ui.me_spin.text(),
              pe=self.ui.pe_skill_combo.currentIndex())

Answer 4

括号内的任何地方都应该有效，例如：

blueprint = Blueprint(self.blueprint_map[str(self.ui.blueprint_combo.currentText())],
        runs=self.ui.runs_spin.text(), me=self.ui.me_spin.text(),
        pe=self.ui.pe_skill_combo.currentIndex())

Answer 5

blueprint = Blueprint(self.blueprint_map[str(self.ui.blueprint_combo.currentText())], 
                      runs=self.ui.runs_spin.text(), me=self.ui.me_spin.text(),
                      pe=self.ui.pe_skill_combo.currentIndex())