Don't show 'unpin' for persistent widgets
This will mostly fix https://github.com/vector-im/element-web/issues/15139 although it could still break in a race condition.pull/21833/head
parent
017765df24
commit
eeb4c5696f
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@ -26,8 +26,8 @@ export default class WidgetContextMenu extends React.Component {
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// Callback for when the revoke button is clicked. Required.
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onRevokeClicked: PropTypes.func.isRequired,
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// Callback for when the unpin button is clicked. Required.
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onUnpinClicked: PropTypes.func.isRequired,
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// Callback for when the unpin button is clicked. If absent, unpin will be hidden.
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onUnpinClicked: PropTypes.func,
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// Callback for when the snapshot button is clicked. Button not shown
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// without a callback.
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@ -86,11 +86,13 @@ export default class WidgetContextMenu extends React.Component {
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);
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}
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options.push(
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<MenuItem className="mx_WidgetContextMenu_option" onClick={this.onUnpinClicked} key="unpin">
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{_t("Unpin")}
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</MenuItem>,
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);
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if (this.props.onUnpinClicked) {
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options.push(
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<MenuItem className="mx_WidgetContextMenu_option" onClick={this.onUnpinClicked} key="unpin">
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{_t("Unpin")}
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</MenuItem>,
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);
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}
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if (this.props.onReloadClicked) {
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options.push(
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@ -841,7 +841,7 @@ export default class AppTile extends React.Component {
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contextMenu = (
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<ContextMenu {...aboveLeftOf(elementRect, null)} onFinished={this._closeContextMenu}>
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<WidgetContextMenu
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onUnpinClicked={this._onUnpinClicked}
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onUnpinClicked={ActiveWidgetStore.getWidgetPersistence(this.props.app.id) ? null : this._onUnpinClicked}
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onRevokeClicked={this._onRevokeClicked}
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onEditClicked={showEditButton ? this._onEditClick : undefined}
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onDeleteClicked={showDeleteButton ? this._onDeleteClick : undefined}
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