Question

我正在为PopUp组件编写一个小型测试。但是，我收到此错误：方法“文本”应在1个节点上运行。找到0个。我正在学习过程中，任何帮助将不胜感激。谢谢这是我的component.tests.js

        const small = popUp.find('small');
        expect(small).toHaveLength(1);
        expect(small.text()).toBe(messages['en'] . 
        ['segments.create.timeline.isScheduled']);
        expect(warning).toBeDefined();
        expect(popover).toBeDefined();
        expect(message).toBeDefined();       
    });
});

这是我的component.js

class SegmentWarningPopup extends React.Component {
    state = {
        anchorEl: null,
    };

    handleClick = event => {
        this.setState({
            anchorEl: event.currentTarget,
        });
    };

    handleClose = () => {
        this.setState({
            anchorEl: null,
        });
    };

    render() {
        const { anchorEl } = this.state;
        const open = Boolean(anchorEl);

        return (
            <div>
                <Warning size="small" className="duration-has-changed-icon" onClick={this.handleClick}/>
                <Popover
                    id="durationHasChanged"
                    open={open}
                    anchorEl={anchorEl}
                    onClose={this.handleClose}
                    anchorOrigin={{
                        vertical: 'bottom',
                        horizontal: 'left',
                    }}

                    transformOrigin={{
                        vertical: 'top',
                        horizontal: 'left',
                    }}

                    PaperProps={{
                        className: 'warning-popover-paper'
                    }}
                >
                    <FormattedMessage id={"segments.create.timeline.isScheduled"} description="text"/>
                </Popover>
            </div>
        );
    }
}

    SegmentWarningPopup.propTypes = {
    targetEventObj: PropTypes.object
};

    export default SegmentWarningPopup;

Answer 1

您只需要.find <Warning>组件本身即可：

  popUp.find(Warning)

错误：使用酶找不到节点的问题。方法文字

1 个答案: