If else condition not working properly React

This is mine If else condition No matter what the className is its rendering the same form i am getting this Problem i want to solve this Problem

const square = "nodesquare";
var form;
if (square) {
  form = <form>
    <input className="process" type="text" name="process" id="process" />
    <Button type="primary" className="button center" onClick={this.clearAll}>Add</Button>
  </form>
  console.log('Form')
} else {
  form = <p>Nothing to Watch</p>
}

return (
  <div
    key={index}
    className={'node ' + node.className}
    id={node.id}
    ref={nodes => this.refs.nodes[index] = nodes}
    style={node.style}
    onClick={this.activeElem}
    square={square}>
    {form}
    <div className="delete-btn" onClick={event => this.deleteNode(event, node)}>X</div>
  </div>
)
            })}

      </div >
    );

I'm not sure to understand what is your problem. As your square const is always equal to "nodesquare", your code will always trigger the if condition and never the else.

If else condition not working properly React, I'm not sure to understand what is your problem. As your square const is always equal to "nodesquare" , your code will always trigger the if� But the ifelse() condition is working opposite i.e it is giving red color where C1 has greater value than C2 and vice versa as shown from the plot below, also this can be seen from the legend: Also I want to change the legend title to "Suposed title" and labels as "Column1" for data points from C1 and "Column2" for data points from C2.

You say:

const square = "nodesquare";
var form;
if (square) { ....

It looks like squareis allways truthy. Check the concepts of truthy and falsy in javascript.

Conditional Rendering – React, Conditional rendering in React works the same way conditions work in JavaScript. a variable and using an if statement is a fine way to conditionally render a component, If the value of the prop is false , then the component does not render:. if else condition is not properly working; if else condition is not properly working. Answered Active Solved. Ask Question. Last Reply 19 hours ago By dharmendr.

The Main Issue i was getting here is that i was in the Div binding Square thats why i am getting this error look here

             key={index}
             className={'node ' + node.className}
             id={node.id}
             ref={nodes => this.refs.nodes[index] = nodes}
             style={node.style}
             onClick={this.activeElem}
             square={square}
      >

thats why it was always true thanks for your Precious Time

4 React conditional rendering methods with props and state, You can use if-else statements, inline if statements, ternary operators, and evern return If you're not familiar with React state, props or their hooks please hop on over to the actually need to have one evaluation on the left side and the output on the right side. And inside the hook function I'm setting yellAtPerson to true . Teams. Q&A for Work. Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information.

5 Ways to Implement Conditional Rendering in React, So, try spending the right amount of time optimizing your code while always pushing to reuse The if-else statements can cause wasted re-renders in React. It might The problem here comes from React's diffing algorithm. if-else statements don't work inside JSX. This is because JSX is just syntactic sugar for function calls and object construction. Take this basic example:

ifelse, The if statement executes a statement if a specified condition is truthy. If the function testNum(a) {. 2 To see how this works, this is how it would look if the nesting were properly indented: Do not confuse the primitive Boolean values true and false with truthiness or falsiness of the Boolean object. IF-Else is a type of conditional statement which executes a part of code based on condition. The condition can be anything with different type of operators. So in this tutorial we would going to create a react native application with IF-Else & Nested IF-Else conditional statement for both android and iOS applications.

Conditional rendering in React Native may crash your app , Conditional rendering done wrong leads to errors in React Native that When using it widely in your App without proper attention sooner or In function, you can call return inside if statement and it will not execute further on. An if/else block is the easiest way to solve the problem, but I’m sure you know this is not a good implementation. It works great for simple use cases, and every programmer knows how it works. But there’s a lot of repetition, and the render method looks crowded.