How to do conditional rendering in react native with more than 1 condition?
Following is some portion of my code
Index
.then(response => response.json())
.then((responseData) => {
this.setState({
progressData:responseData,
});
.....
......
render() {
const { progressData }= this.state;
return(
<View style={{flex:1}}>
<HeaderExample />
<View>
{progressData == "1"}
(<View>
<Text style={{fontSize:28,color:"#8470ff",fontWeight: 'bold',paddingTop:20,alignSelf:'center'}}>Pending</Text>
</View>)}
{ progressData == "2" &&
(<View>
<CardSection>
<Text style={{fontSize:28,color:"#8470ff",fontWeight: 'bold',paddingTop:20,alignSelf:'center'}}>InProgress </Text>
<View style={styles.buttonContainer}>
<Button
title="Report"
color="#8470ff"
onPress={() =>onPressReport()}
/>
</View>)}
But here it is for a single case means if responseData
contains only one field. But now the reponseData contains 2 arrays. Each with 3 objects. So how do I check conditional rendering here?My responseData looks like this. I want to populate some UI on each condition. That means if status = 1 && work_type ="plumber"
then render some UI.
Also if status = 2 && work_type="electrical" && assigend_to="worker_45"
then render some ui. So how do I do this?
Please help
You can move your render in a new variable, or function. to keep clear the render
function
render() {
const { progressData }= this.state;
return(
<View style={{flex:1}}>
<HeaderExample />
<View>
{renderProgressData(progressData)}
... //rest of your code
)
}
and in your renderProgressData
function you can create a switch
renderProgressData = (progress) => {
switch(progress) {
case 1:
return (<View>1</View>)
case 2:
return (<View>1</View>)
// ... and so on
default:
return (<View>Default View</View>)
}
}
It is a little cleaner in this way for me.