How to differentiate two different CSS props to be conditionally styled, that have the same name in React Native?

Question

I have this style: <TouchableOpacity style={[styles.container, { backgroundColor }]}> and this switch case:

    let backgroundColor = "white";
    switch (this._getName()) {
        case "bot":
            backgroundColor = "#F6F9FC";
            break;
    }

That correctly change the background color of the TouchableOpacity component when the name is "bot".

Inside that TouchableOpacity, I have this: that correctly changes the color of a circle based on a certain state when mixed with these switch cases:

    const { report } = this.props;
    let backgroundColor = "gray";
    switch (report.status) {
        case "active":
            backgroundColor = "green";
            break;
        case "inQueueForMaintenance":
            backgroundColor = "yellow";
            break;
        case "inMaintenance":
            backgroundColor = "yellow";
            break;
        case "defeated":
            backgroundColor = "red";
            break;
    }

However, if I mix both switch cases, it'll conflict because both props are backgroundColor. How to avoid this?

score 0 · Accepted Answer · answered Aug 06 '18 at 17:15

Why don't you try the classic way of creating objects. Use two different names.

let firstBg = "white";
switch (this._getName()) {
    case "bot":
        firstBg = "#F6F9FC";
        break;
}

{ backgroundColor: firstBg }

and then,

const { report } = this.props;
let secondBg = "gray";
switch (report.status) {
    case "active":
        secondBg = "green";
        break;
    case "inQueueForMaintenance":
        secondBg = "yellow";
        break;
    case "inMaintenance":
        secondBg = "yellow";
        break;
    case "defeated":
        secondBg = "red";
        break;
}

{ backgroundColor: secondBg}

How to differentiate two different CSS props to be conditionally styled, that have the same name in React Native?

1 Answers1