同样，按照Alberto Piras解决方案，如果你不想复制所有的“state”对象:

handleChange(el) {
    let inputName = el.target.name;
    let inputValue = el.target.value;

    let jasperCopy = Object.assign({}, this.state.jasper);
    jasperCopy[inputName].name = inputValue;

    this.setState({jasper: jasperCopy});
  }

创建一个状态对象

this.state = {
  objName: {
    propertyOne: "",
    propertyTwo: ""
  }
};

使用setState更新状态

this.setState(prevState => ({
  objName: {
    ...prevState.objName,
    propertyOne: "Updated Value",
    propertyTwo: "Updated value"
  }
}));

舰队指挥官:样品

   const [formData, setformData] = useState({
           project_admin_permissions: {
              task_forms: false,
              auto_assign_rules: false,
              project_notes: true,
              alerts: false,
              update_criteria: true,
              project_flow: false,
              reports: false,
            }
        
          })
    
     const handleChangeCheckBox = (e) => {
       setformData({
          ...formData, project_admin_permissions: { ...formData.project_admin_permissions, [e.target.name]: e.target.checked }
    
        })
      }

同样，按照Alberto Piras解决方案，如果你不想复制所有的“state”对象:

handleChange(el) {
    let inputName = el.target.name;
    let inputValue = el.target.value;

    let jasperCopy = Object.assign({}, this.state.jasper);
    jasperCopy[inputName].name = inputValue;

    this.setState({jasper: jasperCopy});
  }

试试这个，应该没问题

this.setState(Object.assign(this.state.jasper,{name:'someOtherName'}));

不使用异步和等待使用此…

funCall(){    
     this.setState({...this.state.jasper, name: 'someothername'});
}

如果你使用异步和等待使用这个…

async funCall(){
      await this.setState({...this.state.jasper, name: 'someothername'});
}