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hide RoomSubList if has 0 tiles and either searchFilter or emptyContent

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Signed-off-by: Michael Telatynski <7t3chguy@gmail.com>
This commit is contained in:
Michael Telatynski
2018-06-25 09:41:28 +01:00
parent 5b027c6410
commit a174555452
1 changed files with 10 additions and 3 deletions
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11
src/components/structures/RoomSubList.js
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@@ -1,6 +1,7 @@
/* /*
Copyright 2017 Vector Creations Ltd
Copyright 2015, 2016 OpenMarket Ltd Copyright 2015, 2016 OpenMarket Ltd
Copyright 2017 Vector Creations Ltd
Copyright 2018 New Vector Ltd
Licensed under the Apache License, Version 2.0 (the "License"); Licensed under the Apache License, Version 2.0 (the "License");
you may not use this file except in compliance with the License. you may not use this file except in compliance with the License.
@@ -349,8 +350,14 @@ var RoomSubList = React.createClass({
var label = this.props.collapsed ? null : this.props.label; var label = this.props.collapsed ? null : this.props.label;
let content; let content;
if (this.state.sortedList.length === 0 && !this.props.searchFilter && this.props.extraTiles.length === 0) { if (this.state.sortedList.length === 0 && this.props.extraTiles.length === 0) {
// if no search filter is applied and there is a placeholder defined then show it, otherwise show nothing
if (!this.props.searchFilter && this.props.emptyContent) {
content = this.props.emptyContent; content = this.props.emptyContent;
} else {
// don't show an empty sublist
return null;
}
} else { } else {
content = this.makeRoomTiles(); content = this.makeRoomTiles();
content.push(...this.props.extraTiles); content.push(...this.props.extraTiles);