How do I fix this unwanted button behavior in ReactJS

So I’m trying to conditionally switch between two html buttons in ReactJS base on wether the form is shown or not. Basically if the form is shown, show a submit button and if not shown, show a regular button to show the form.

This is the code that I have. (I’m using Antd’s button component but I also tried using the regular html button and had the same problem.

<div>
        {showForm ? (
          <Button
            loading={isUploading}
            htmlType="submit"
            form="videoForm"
            className={Styles.button}
          >
            {isUploading ? 'Uploading' : 'Upload Video'}
          </Button>
        ) : (
          <Button
            htmlType="button"
            onClick={() => setShowForm(true)}
            className={Styles.button}
          >
            {successMsg ? 'Upload Another Video' : 'Show Form'}
          </Button>
        )}
      </div>

The issue I’m having is that when the form is not shown and I click the Show Form button, the form shows correctly and the buttons switch, but the form submit event is triggering which is not what I expected or want.

Any ideas why? And how I can fix this issue? I also tried doing the following but got the same results.

<div>
        <Button
          loading={isUploading}
          htmlType={showForm ? 'submit' : 'button'}
          form="videoForm"
          className={Styles.button}
        >
          {showForm && (isUploading ? 'Uploading' : 'Upload Video')}
          {!showForm && (successMsg ? 'Upload Another Video' : 'Show Form')}
        </Button>
      </div>

Any help would be greatly appreciated.