- Vue.js 2.x by Example
- Mike Street
- 315字
- 2021-07-02 20:00:28
Filtering our filters
Now we have our filtering in place, we need to only show the radio buttons when the isActive option is selected in our dropdown. Using the knowledge we've learned, this should be relatively straightforward.
Create a new method that checks the select box value and returns true when Active User is selected in our dropdown:
isActiveFilterSelected() {
return (this.filterField === 'isActive');
}
We can now use v-show for both the input and radio buttons, reversing the effect when on the query box:
<label for="filterQuery" v-show="!isActiveFilterSelected()">
Query:
<input type="text" id="filterQuery" v-model="filterQuery">
</label>
<span v-show="isActiveFilterSelected()">
Active:
<label for="userStateActive">
Yes:
<input type="radio" v-bind:value="true" id="userStateActive"
v-model="filterUserState">
</label>
<label for="userStateInactive">
No:
<input type="radio" v-bind:value="false" id="userStateInactive" v-
model="filterUserState">
</label>
</span>
Take note of the exclamation point before the method call on the input field. This means not, and effectively reverses the result of the function, for example not true is the same as false and vice versa.
To improve user experience, we can also check that the filtering is active at all before showing either of the inputs. This can be added by including a secondary check in our v-show attribute:
<label for="filterQuery" v-show="this.filterField &&
!isActiveFilterSelected()">
Query:
<input type="text" id="filterQuery" v-model="filterQuery">
</label>
This now checks that filterField has a value and that the select box is not set to isActive. Make sure you add this to the radio buttons too.
A further user experience enhancement would be to ensure all the users don't disappear when the isActive option is chosen. This currently happens because the default is set to a string, which does not match with either the true or false values of the field. Before filtering in this field, we should check that the filterUserState variable is either true or false, that is a Boolean. We can do this by using typeof once more:
if(this.filterField === 'isActive') {
result = (typeof this.filterUserState === 'boolean') ?
(this.filterUserState === person.isActive) : true;
}
We are using a ternary operator to check that the result to filter on is boolean. If it is, then filter as we were; if it is not then simply show the row.
- VMware View Security Essentials
- 從零構建知識圖譜:技術、方法與案例
- 零基礎學Scratch少兒編程:小學課本中的Scratch創意編程
- 機器人Python青少年編程開發實例
- 深度學習:算法入門與Keras編程實踐
- Python高效開發實戰:Django、Tornado、Flask、Twisted(第3版)
- 零基礎學Java程序設計
- PHP+MySQL網站開發項目式教程
- Java 9模塊化開發:核心原則與實踐
- Haxe Game Development Essentials
- ExtJS高級程序設計
- 深入剖析Java虛擬機:源碼剖析與實例詳解(基礎卷)
- Visualforce Developer’s guide
- Python:Deeper Insights into Machine Learning
- Solutions Architect's Handbook