How to change button icon when button is clicked in react.js - css

I have a react function which returns a button.
<div className="col-6 btn-group btn-group w-100">
<AuditMenuButtons buttonValue='Pending' buttonName='Inbox' changeFilterForButton={this.props.changeFilterForButton} icon={icon_inbox}/>
<AuditMenuButtons buttonValue='Rejected' buttonName='Rejected' changeFilterForButton={this.props.changeFilterForButton} icon={icon_rejected}/>
<AuditMenuButtons buttonValue='Accepted' buttonName='Accepted' changeFilterForButton={this.props.changeFilterForButton} icon={icon_accepted}/>
</div>
Function is added below
function AuditMenuButtons(props) {
return(
<button className="w-25 btn menu-btn p-lg-3" name={props.buttonName} value={props.buttonValue} onClick={props.changeFilterForButton}><img src={props.icon} className="pr-3 menu-btn-icons">
</img>{props.buttonName}</button>
);
}
You will see 3 buttons in above code. I want to change the button icon when one button is clicked. actually button icon color should be green when button is clicked. Images are .png file (with green and silver border). I tried button:active in css it didn't work for me. Image should remain until I clicked another button or page was refreshed

In this case, the icon part is a UI state, it has to maintained in your state and passed down to AuditMenuButtons has props.
use these props in AuditMenuButtons to do the desired check.
import React,{Component} from 'react';
class demoComponent extends from Component{
this.state={
isClicked:false,
buttonIcons:{
pending:{active_Icon:"../IconURL",Icon:"../IconURL"},
rejected:{active_Icon:"../IconURL",Icon:"../IconURL"},
accepted:{active_Icon:"../IconURL",Icon:"../IconURL"}
}
}
clickHandler = (event) =>{
this.setState(
{
isClicked:!this.state.isClicked // this is gonna toggle everytime you click //
}
);
}
render(){
return <div className="col-6 btn-group btn-group w-100">
<AuditMenuButtons clickhandler={this.clickHandler} buttonValue='Pending' buttonName='Inbox' isClicked={this.state.isClicked} buttonIcons={this.state.buttonIcons} changeFilterForButton={this.props.changeFilterForButton} icon={icon_inbox}/>
<AuditMenuButtons clickhandler={this.clickHandler} buttonValue='Rejected' buttonName='Rejected' isClicked={this.state.isClicked} buttonIcons={this.state.buttonIcons} changeFilterForButton={this.props.changeFilterForButton} icon={icon_rejected}/>
<AuditMenuButtons clickhandler={this.clickHandler} buttonValue='Accepted' buttonName='Accepted' isClicked={this.state.isClicked} buttonIcons={this.state.buttonIcons} changeFilterForButton={this.props.changeFilterForButton} icon={icon_accepted}/>
</div>
}
}
export default demoComponent;

You can try something like this in your .css file.
.button:focus{background: url('your new green image');

You can manage the image path in react state and call a method attach
it to onClick, where you use setState() and update the state.
Refer
https://reactjs.org/docs/handling-events.html
https://reactjs.org/docs/react-component.html#setstate
this.state = {
image_path: 'your image url here'
}
changeUrl = () => {
this.setState({image_path:'new path'});
}
<AuditMenuButtons onClick={this.changeUrl} src={this.state.image_path}/>

You can try with this:
changeFilterForButton: function (props) {
props.currentTarget.style.backgroundColor = '#ccc';
}
function AuditMenuButtons(props) {
return(
<button className="w-25 btn menu-btn p-lg-3" name={this.props.buttonName}
value={this.props.buttonValue} onClick={this.props.changeFilterForButton}><img src={this.props.icon} className="pr-3 menu-btn-icons">
</img>{this.props.buttonName}</button>
);
}
or if you want to use react methodology then you can use construtor like this
constructor(props) {
super(props);
this.state = {isColor: false};
// This binding is necessary to make `this` work in the callback
this.changeFilterForButton= this.changeFilterForButton.bind(this);
}
changeFilterForButton() {
this.setState(state => ({
isColor: !state.isColor
}));
}
function AuditMenuButtons(props) {
return (
<button className="w-25 btn menu-btn p-lg-3" name={props.buttonName} value={props.buttonValue} onClick={props.changeFilterForButton} style="background-color: {this.state.isColor? '#CCC' : ''} "><img src={props.icon} className="pr-3 menu-btn-icons">
</img>{props.buttonName}</button>
);
}

Related

React ignores one of two style classes

Two buttons are rendered in my button component. Depending on the label, the button receives the class addButton or clearButton.
Button Component:
import './Button.css'
interface ButtonProps {
lable: string,
disabled: boolean,
onClick: MouseEventHandler<HTMLButtonElement>
}
export const Button: FunctionComponent<ButtonProps> = ({ lable, disabled, onClick}): ReactElement => {
let style: string = lable == 'ADD' ? 'addButton' : 'clearButton';
console.log(lable);
console.log(style);
return(
<div>
<button className= {`button ${style}`} type='button' disabled= { disabled } onClick= { onClick }>
{lable}
</button>
</div>
);
}
In the stylesheet Button.css are the two classes addButton and clearButton. Strangely, only the style class that is in second place is loaded. In this situation, the addButton receives its colour and the clearButton remains colourless. If I place the style class of the clearButton in second place in the stylesheet, it receives its colour and the addButton remains colourless.
Button.css
.clearButton {
background-color: #8c00ff;
}
.addButton {
background-color: #7ce0ff;
}
If I outsource both style classes to separate stylesheets, it works but this is not an elegant solution.
I would also like to understand the problem
I've execute your code, and it works, I think the problem is where you call the button you are not passing props to
change import type of button style
import styles from './Button.css';
return(
<div>
<button className={`button ${label === 'ADD' ? styles.addButton : styles.clearButton}`}>
{lable}
</button>
</div>
);
and reolace background-color to background
.clearButton {
background: #8c00ff;
}
.addButton {
background: #7ce0ff;
}
you can test code in this codesandbox project

Add color on every button Click React js

Hello guys I currently have a buttons like category. I want that when I click a button it will have a color, and when I click it again it will turn to it's original color which is white. When I click 2 button both will have dark color, then click again to remove single color.
this is my div when I'm adding a the category id
<div className={classes.scrollMenu}>
{categories.map((category) => {
return (
<>
<Button
key={category._id}
className={classes.button}
onClick={(e) => {
let values = {
price: [],
category: [category._id],
}
}}
>
{category.name}
</Button>
</>
)
})}
</div>
This is the image that when I click single button it will color one button.
Thank you
code Solution: https://codesandbox.io/s/stoic-meadow-y5cei?file=/src/App.js
App.js
import "./styles.css";
import React, { useState } from "react";
export default function App() {
let categories = ["one", "two", "three"];
const [activeFilter, setActiveFilter] = useState(["one"]);
const categoryOnClick = (category) => {
activeFilter.includes(category)
? removeCategory(category)
: setCategory(category);
};
const setCategory = (category) => {
setActiveFilter([...activeFilter, category]);
};
const removeCategory = (category) => {
const index = activeFilter.findIndex((cat) => cat === category);
activeFilter.splice(index, 1);
setActiveFilter([...activeFilter]);
};
return (
<div className="chip-list my-3">
{categories.map((category, index) => {
return (
<button
key={index}
className={`${activeFilter.includes(category) ? "active" : ""}`}
onClick={() => categoryOnClick(category)}
>
<span>{category}</span>
</button>
);
})}
</div>
);
}
css
.active {
background-color: black;
color: white;
}
check if this solution works for you
used useState hook to hold the state of buttons which you will select
.active class will apply to the button which is selected
On click of that button we will check if the button is already selected or not if selected removeCategory() function run
or if button is not selected then setCategory() function will run and it will update the state
if you need clarification please let me know thanks
Few tips to start with:
Fragment is unnecessary when wrapping single DOM element
Inline function initialisation inside a render is a bad thing. On each new re-render, it allocates extra client memory to newly initialised function. That means, for every map object you will have that many functions, that gets newly created and referenced on each reload
You can easily go with single line return statement of arrow function here. () => <hi> instead of () => { return <hi> }
As for solutions, there are quite a few ways to change button colour during execution. I will suggest the most simple (in my opinion) way to do it. Just have classname variable, then add subclass that will style button accordingly.
Example:
By default it has class name of .button, after click you simply add styling and it ends up having .button .button--red, all is left to do, declaration in css.
.button {
style button here
. . .
add additional stylings here
. . .
&.button--red { color: red }
}
As for how handler should look like, if that is what you asking. Button could be used in your new component let's say, named StyledButton or ColourfulButton that will have internal state to handle what kind of colour is represented.

css or react Keep border focus when clicking my eye button

I have a login page where I set border colors on focus and I want to keep the focus even when I'm clicking my eye password toggleI have attached a gif that shows how I loose my focus state when toggling my eye
One way to do it is to store a ref to the password input field, and call focus() on it in the eye's onClick handler:
import React, { useRef } from "react";
const Eye = ({ onClick }) => <span onClick={onClick}>eye</span>;
export default function App() {
const inputRef = useRef();
const handleEyeClick = () => {
if (inputRef.current) {
inputRef.current.focus();
}
};
return (
<div>
<input ref={inputRef} type="text" />
<Eye onClick={handleEyeClick} />
</div>
);
}

Testing Style on React component

I have the following class with a function, that opens a modal (open_modal(...)) in a separate file to a component as I have a large number of modals that use this functionality.
import open from "open";
import $ from "jquery";
class ReactHelpers {
static open_webpage(page_url) {
open(page_url);
}
static open_modal(overlay_id, modal_id) {
$(overlay_id).css("display", "block");
$(modal_id).css("display", "block");
}
static close_modal(overlay_id, modal_id) {
$(overlay_id).css("display", "none");
$(modal_id).css("display", "none");
}
}
export default ReactHelpers;
I am trying to assert that the open_modal function has added css to the divs in question as below:
it('should close the modal', function () {
const wrapper = shallow(
<div id="overlay_id">
<div id="modal_id">
<p>modal</p>
</div>
</div>
)
const overlay = wrapper.find('#overlay_id')
const modal = wrapper.find('#modal_id')
ReactHelpers.open_modal(overlay, modal);
console.log('OVERLAY ', overlay);
expect(overlay.prop('style')).toHaveProperty('display', 'block');
expect(modal_style).toHaveProperty('display', 'block');
});
Further, I'm sure to how the open_webpage function would be tested as this is a library function. In my other tests in my other components, I'm mocking this so it's never actually been tested.
Any help is greatly appreciated.
Thanks
To test style of dom elements:
You should mount the component (using mount), instead of just creating it (using shallow).
Since you're changing the style of dom element directly, You should test the style of the dom element (component.getDOMNode().style.display), instead of testing the react style property (component.prop.style).
example:
import $ from "jquery";
it("should create a div and changes its color to red", () => {
const wrap = mount(
<div id="red_el"></div>
);
const el = wrap.find("#red_el").getDOMNode()
$(el).css("color", "red");
expect(el.style.color).toEqual("red");
});
In your case:
it("should open modal", () => {
const wrapper = mount(
<div>
<div id="overlay" style={{ display: "none" }}>
<div id="modal" style={{ display: "none" }}>
overlay
</div>
</div>
</div>
);
const overlay = wrapper.find("#overlay").getDOMNode();
const modal = wrapper.find("#modal").getDOMNode();
ReactHelpers.open_modal(overlay, modal);
expect(overlay.style.display).toEqual("block");
expect(modal.style.display).toEqual("block");
});
See it live on codesandbox (switch to the tests tab to run the tests .)

ReactJS transition for progress bar not working

I am just diving into ReactJS so I am quite a newbie in this Reactjs world. I have read the FB documentations and some tutorials on the internet and started my test project.
In my test project I am trying to include a progress bar for users to see their progress of filling down some forms across 3 pages. This part works all great till I wanted to add some transition magic to the process bar.
I've written the code below and I thought it would be the right way to archive my goal by pushing a prop from the parent to this child progressBar component for determining the percentage of the progress bar.
In my constructor I set the default width at 0 to update it by componentDidMount to a percentage which comes from the parent. I've managed to receive and set te style but the transition isn't working at all. I try to archive a fancy progress bar which runs from 0% width to the given width in percentage via the props.
My code look likes as follow:
ProgressBar component
import './style.scss';
import React from 'react';
import classnames from 'classnames';
class ProgressBar extends React.Component {
constructor(props) {
super(props);
this.state = { progressionStyle : { } }
}
componentDidMount() {
this.setState({
progressionStyle : {
width : this.props.progression,
transition : 'all 1500ms ease'
},
scene1 : (this.props.scene1 == 'active') ? 'active' : (this.props.scene1 == 'done') ? 'done' : '',
scene2 : (this.props.scene2 == 'active') ? 'active' : (this.props.scene2 == 'done') ? 'done' : '',
scene3 : (this.props.scene3 == 'active') ? 'active' : (this.props.scene3 == 'done') ? 'done' : '',
});
}
/**
*
* Render
* #return {JSX}
*/
render() {
return (
<div className="progress-bar">
<div className="progress-bar__inner">
<div className="progress-bar__progress">
<div className={classnames('progress-bar__progress-fill', this.props.active)} style={this.state.progressionStyle}></div>
</div>
<div id="scene1" className="progress-bar__element">
<i className={classnames('progress-bar__icon', this.state.scene1)}></i>
<span className="progress-bar__label">Scene 1</span>
</div>
<div id="scene2" className="progress-bar__element">
<i className={classnames('progress-bar__icon', this.state.scene2)}></i>
<span className="progress-bar__label">Scene 2</span>
</div>
<div id="scene3" className="progress-bar__element">
<i className={classnames('progress-bar__icon', this.state.scene3)}></i>
<span className="progress-bar__label">Scene 3</span>
</div>
</div>
</div>
);
}
}
export default ProgressBar;
You cannot set the nested state directly, you should be doing it like
componentDidMount() {
var style = {...this.state.style}
style.width = this.props.progression
style.transition = 'all 500ms ease-in'
this.setState({style});
}
Also, you need to update your state in the componentWillReceiveProps function as you are updating state based on the props.
componentWillReceiveProps(nextProps) {
var style = {...this.state.style}
style.width = nextProps.progression
style.transition = 'all 500ms ease-in'
this.setState({style});
}
To make this effect work, I've found out the solution was to wrap the style into a function and call via request animation frame via as fol
componentDidMount() {
requestAnimationFrame(()=> {
this.showProgress();
});
}
showProgress() {
var style = { };
style.width = this.props.progression;
style.transition = 'all 1500ms ease-in';
this.setState({style});
}
https://stackoverflow.com/a/43779273/968898

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