Reactjs 为什么条件渲染测试没有失败?
我有一个typescript react组件,如果未定义或为零,则不应显示数字:Reactjs 为什么条件渲染测试没有失败?,reactjs,enzyme,Reactjs,Enzyme,我有一个typescript react组件,如果未定义或为零,则不应显示数字: import * as React from "react"; export class NoZero extends React.Component<{ number?: number }, {}> { render() { return ( <div> {this.props.number &&am
import * as React from "react";
export class NoZero extends React.Component<{ number?: number }, {}> {
render() {
return (
<div>
{this.props.number && this.props.number > 0 && <div className="number">{this.props.number}</div>}
</div>
);
}
}
这些测试都是成功的。如果浏览器显示0,为什么第二次测试没有失败?这就是
&&
的工作原理。它返回第一个假操作数
console.log(0&&true)
log(true&&0)
这就是&&
的工作原理。它返回第一个假操作数
console.log(0&&true)
console.log(true和&0)
import { expect } from "chai";
import "mocha";
import * as React from "react";
import { shallow } from "enzyme";
import { NoZero } from "./NoZero";
describe("NoZero", () => {
it("should show not undefined", () => {
const wrapper = shallow(<NoZero number={undefined} />);
expect(wrapper.find(".number")).to.have.length(0);
});
it("should not show '0'", () => {
const wrapper = shallow(<NoZero number={0}/>);
expect(wrapper.find(".number")).to.have.length(0);
});
it("should show other numbers", () => {
const wrapper = shallow(<NoZero number={2} />);
expect(wrapper.find(".number")).to.have.length(1);
});
});