酒店"并不存在类型'string'当使用makestyles@mui

0

的问题

我有下列代码:

    const useStyles = makeStyles(() => ({
  dialog: {
    root: {
      position: 'absolute'
    },
    backdrop: {
      position: 'absolute'
    },
    paperScrollPaper: {
      overflow: 'visible'
    },
    paper: {
      background: 'none',
      boxShadow: 'none',
      overflow: 'hidden'
    }
  },
}))
    
export const LoadingBackdrop = (props: any) => {
  const classes = useStyles()
  const backdropProps = {
    disableEscapeKeyDown: true,
    disableAutoFocus: true,
    disableEnforceFocus: true,
    disableScrollLock: true,
    disablePortal: true,
  }
  return (
    <Dialog
      {...backdropProps}
      fullWidth
      className={classes.dialog.root}
      classes={{
        paperScrollPaper: classes.dialog.paperScrollPaper,
        paper: classes.dialog.paper,
        paperFullWidth: classes.dialog.paper
      }}
      BackdropProps={{
        classes: { root: classes.dialog.backdrop }
      }}
      style={{ position: 'absolute' }}
      {...props}
    >
      <Loading />
    </Dialog>
  )
}

在LoadingBackDrop成分,我收到以下错误ts eslint:

酒店'root'上不存在类型'string'.ts(2339)

这同样适用于其他用途的课程。对话,仅仅更改名称的财产...

我怎么可以注释或类型的makestyles功能(或者挂钩的话)以解决这个错误?

1

最好的答案

1

你不能巢的风格你的方式这样做-他们需要的是一个级别的深度。 例如:

const useStyles = makeStyles(() => ({
  root: {
    position: 'absolute'
  },
  backdrop: {
    position: 'absolute'
  },
  paperScrollPaper: {
    overflow: 'visible'
  },
  paper: {
    background: 'none',
    boxShadow: 'none',
    overflow: 'hidden'
  }
}))

并且:

<Dialog
  {...backdropProps}
  fullWidth
  className={classes.root}
  ...

它们 风格的文件 具有进一步的例子。

2021-11-23 21:12:21

这就是它! 谢谢你
Roger Peixoto

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