CODEWITHSUNDEEP
javascriptreactjsreact-reduxredux-formreact-redux-form
max li
max li

Reputation: 2457

React redux form update pristine after change

I need to make dispatch and change the form data after page loaded:

dispatch(change("project", field, value));

however this will change "pristine" to false and "dirty" to true, anyone has a good solution for such case?

Upvotes: 3

Views: 6100

Answers (3)

viniciusalvess
viniciusalvess

Reputation: 814

I don't know if it is the correct way, but it is working for me.

this.props.dispatch({type: '@@redux-form/BLUR', payload: yourPayload, meta: {form: "formName", field: "formField", touch: true}});

I dispatch the code above, at the onChange event of my field, forcing the redux-form to update the pristine status.

Upvotes: 0

lee
lee

Reputation: 246

You can set the silent flag to true, this will not update pristine and dirty flags. Example: actions.change(model, value, { silent: true }) You could also use actions.load(). Difference between load() and change()

This is different than calling actions.change(model, value, { silent: true }). Both are similar with one main difference: actions.load also updates the .loadedValue, whereas a silent change does not.

https://davidkpiano.github.io/react-redux-form/docs/api/actions.html#actions-load

Upvotes: -2

1ven
1ven

Reputation: 7016

In order to achieve this, you can provide initialValues to decorated with reduxForm component:

compose(
  connect((state) => ({
    initialValues: state.someData
  })),
  reduxForm({
    form: 'someForm'
  })
)(Component)

Or, alternatively, you can use initialize action creator from redux-form after data will be loaded from server:

dispatch(initialize('myForm', newInitialValues));

Dispatching initialize action will set state of your form to pristine

Upvotes: 4

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