Material UI 5 class name styles

Question

I migrated from Mui 4 to 5 and wonder how to use class names. If I want to apply certain styles to just one component there is the SX property. However, I'm struggling with using the same class for multiple components. In v4 my code looked like this:

export const useStyles = makeStyles((theme: Theme) =>
  createStyles({
    root: {
      padding: theme.spacing(1),
      margin: 'auto',
    },
  })
)

I could import this useStyles hook in any component and use it like this:

const classes = useStyles()
...
<div className={classes.root}>...</div>

This docs say, that I can 'override styles with class names', but they don't tell how to do it: https://mui.com/customization/how-to-customize/#overriding-styles-with-class-names

Do I have to put these styles in an external CSS file?

.Button {
  color: black;
}

I would rather define the styles in my ts file.

I also found this migration guide: https://next.material-ui.com/guides/migration-v4/#migrate-makestyles-to-emotion

I don't like approach one, because using this Root wrapper, it is inconvenient to apply a class conditionally. (Especially for typescript there is some overhead) Approach two comes with an external dependency and some boilerplate code.

Ideally I would use styles like this, perhaps with one rapper function around the styles object:

export const root = {
  padding: theme.spacing(1),
  margin: 'auto',
}

<div className={root}>...</div>

Of course, the last approach doesn't work, because className wants a string as input. Does anybody know an alternative with little boilerplate code?

Answer 1

I suggest you take a look at emotion's documentations for details. The sx prop is actually passed to emotion.

You can do something like this:

const sx = {
  "& .MuiDrawer-paper": {
    width: drawerWidth
  }
};
<Drawer sx={sx}/>

Equivalent to MUI v4

const useStyles = makeStyles({
  drawerPaper: {
    width: drawerWidth,
  }
});

const classes = useStyles();

<Drawer
  classes={{
    paper: classes.drawerPaper,
  }}
/>

Answer 2

Answering your exact question, there are use cases (I think yours is not one of them and you should use styled components) however for those like me who stumble upon it and want a "exact answer to this question" and not a "do this instead", this is how you achieve to retrieve the class names.

This is so far undocumented.

For functional components, using emotion, here an use case where you have a 3rd party component that expects, not one, but many class names, or where the className property is not where you are meant to pass the property.

import { css, Theme, useTheme } from "@mui/material/styles";
import { css as emotionCss } from "@emotion/css";

const myStyles = {
  basicClass: {
    marginLeft: "1rem",
    marginRight: "1rem",
    paddingLeft: "1rem",
    paddingRight: "1rem",
  },
  optionClass: (theme: Theme) => ({
    [theme.breakpoints.down(theme.breakpoints.values.md)]: {
      display: "none",
    }
  })
}

function MyComponent() {
  cons theme = useTheme();

  // first we need to convert to something emotion can understand
  const basicClass = css(myStyles.basicClass);
  const optionClass = css(myStyles.optionClass(theme));

  // now we can pass to emotion
  const basicClassName = emotionCss(basicClass.styles);
  const optionClassName = emotionCss(optionClass.styles);

  return (
    <ThirdPartyComponent basicClassName={basicClassName} optionClassName={optionClassName} />
  )
}

When you have a Class Component, you need to use the also undocumented withTheme from @mui/material/styles and wrap your class, if you use the theme.

WHEN IT IS NOT AN USE CASE

1. When your component uses a single className property just use styled components.

import { styled } from "@mui/material/styles";

const ThrirdPartyStyled = styled(ThirdPartyComponent)(({theme}) => ({
  color: theme.palette.success.contrastText
}))

2. Even if you have dynamic styles

import { styled } from "@mui/material/styles";

interface IThrirdPartyStyledExtraProps {
  fullWidth?: boolean;
}

const ThrirdPartyStyled = styled(ThirdPartyComponent, {
  shouldForwardProp: (prop) => prop !== "fullWidth"
})<IThrirdPartyStyledExtraProps>(({theme, fullWidth}) => ({
  color: theme.palette.success.contrastText,
  width: fullWidth ? "100%" : "auto",
}))

Even if each one has some form of custom color, you just would use "sx" on your new ThrirdPartyStyled.

3. When you are just trying to reuse a style around (your use case)

const myReusableStyle = {
  color: "red",
}

// better
const MyStyledDiv = styled("div")(myReusableStyle);
// questionable
const MySpanWithoutStyles = styled("span")();

// better
const MyDrawerStyled = styled(Drawer)(myReusableStyle);

function MyComponent() {
  return (
    <MyStyledDiv>
      questionable usage because it is less clean:
      <MySpanWithoutStyles sx={myReusableStyle}>hello</MySpanWithoutStyles>
      <MySpanWithoutStyles sx={myReusableStyle}>world</MySpanWithoutStyles>

      these two are equivalent:
      <MyDrawerStyled />
      <Drawer sx={myReusableStyle} />
    </MyStyledDiv>
  )
}

Now what is "presumably" cool about this is that your style, is just an object now, and you can just import it and use it everywhere without makeStyles or withStyles, supposedly an advantage, even when to be honest, I have never used that of exporting/importing around; the code seems a bit cleaner nevertheless.

You seem to want to use it so all you do is.

export const myStyles {
  // your styles here
}

because this object is equivalent in memory, and it is always the same object, something that is easier to mess up with styles, it should be as effective or even more than your hook, theoretically (if it re-renders often even when setup may be longer), which stores the same function in memory but returns a new object every time.

Now you can use those myStyles everywhere you deem reasonable, either with styled components or by assigning to sx.

You can further optimize, say if it's always a div that you use that is styled the same way, then the styled component MyStyledDiv should be faster, because it is the same and done each time. How much faster is this? According to some sources 55% faster, to me, it is taking 4 weeks of refactor and the JSS compatibility with emotion is bad, all mixed with SSR is making everything unusable and slow and broken, so let's see until then when the whole is refactored.

Answer 3

I'm in the same boat, about six months behind, i.e., starting to make the transition to v5 from v4 now... Just when I thought I had a handle on it all!

Having read this post and trying a few things out, I was able to replicate the ability to re-use a chunk of css. I'm a big fan of what used to be the overrides prop; that feature hasn't gone away, it's just under a different prop (loosely speaking). Regardless, I mention it because it provides access to what I like a lot about css: selectors.

To hit all MUI-Drawers my pref is for whatever the new overrides is. For targeted reuse of css I like the following:

import { reuseThisCss } from 'sharedCss`;

export default styled(Drawer)(({ theme, ownerState }) => {
  ...
  return {
    '& .MuiDrawer-paper': {
      boxShadow: xxl,
      border: 'none',

      '& .MuiListItemText-root': reuseThisCss,
    },
  };

export default ThisSpecificDrawerVariant;

Note: The focus is not on using styled (It's not my goto approach).

The css in the return value is the equivalent to the following css: .MuiDrawer-paper .MuiListItemText-root {...}.

This says, "select all .MuiListItemText-root under the .MuiDrawer-paper parent. If I want to optimize the render, while increasing the dependency on a specific hierarchy, I'll specify/expand on the selector that much more with whatever lies between the .MuiDrawer-paper and MuiListItemText-root. For instance, in my case:

...
return {
    '& .MuiDrawer-paper': {
      boxShadow: xxl,
      border: 'none',
      '& > a > li > div > .MuiListItemText-root': reuseThisCss,
    },
  };

Finally, per a question in the comments, generally this will not prevent a nested application of the style. In my experience, marking each level with a className is useful. I only "mark" the element that signals the start of a new level. So, if it were Drawer in the above example, I would start the css selector with .MUI-Drawer.level-3. The rest of css remains the same.

I still have not figured out if whether setting the className dynamically remains a performant and sufficiently flexible goto... TBD.

Answer 4

Here is a pattern that I've found useful in MUI 5. It allows you to keep style definitions in the same file but isolated, & avoids repeated function calls for every CSS property where you need to access your theme (e.g. width: ({ spacing }) => spacing(12))). It also feels similar to MUI's native CSS API.

Create a function that takes your theme as an argument & returns an object of named style groups. Then reference those groups directly in your sx props. This also allows for the use of classNames in a way similar to Material-UI 4.

import { useTheme } from '@mui/material';
import clsx from 'clsx';

export const NavItem = (props) => {
  // Bring in style groups
  const sx = styles(useTheme());

  // Define classNames
  const classNames = clsx({
    isActive: props.isActive
  });

  return (
    {/* Use classNames and style groups */}
    <ListItemButton className={classNames} sx={sx.button}>
      <ListItemAvatar sx={sx.avatar}>{props.icon}</ListItemAvatar>
      <ListItemText>{props.label}</ListItemText>
    </ListItemButton>
  );
}

// Define style groups
function styles(theme) => {
  return {
    button: {
      paddingX: 6,

      '&.isActive': {
        backgroundColor: theme.palette.secondary.light 
      }
    },

    avatar: {
      '.isActive &': {
        border: '2px solid green'
      }
    }
  };
}

Answer 5

If you are using makeStyles or withStyles to provide CSS class, you can follow the instruction below.

CSS overrides created by makeStyles