javascript - 在本机反应中隐藏动态生成的按钮
问题描述
我想在本机反应中隐藏动态生成的按钮下面是我的代码。我使用了溢出:隐藏但它不起作用。
renderSeats() {
var seats = [];
var columnCount = 2;
for(var index = 0; index < columnCount; index++) {
var title = (index + 1).toString();
var isDisabled = false, isHidden = false;
if(title == 1) {
isDisabled = true;
isHidden = false;
}
if(isHidden)
seats.push(
<View hide={this.state.isHidden}>
<Button title={title} style={{overflow: "hidden"}} disabled={isDisabled}></Button>
</View>
);
else
seats.push(
<View style={{overflow: "hidden"}}>
<Button title={title} style={{overflow: "hidden"}} disabled={isDisabled}></Button>
</View>
);
}
return seats;
}
解决方案
overflow: hidden
用于溢出的内容。
你可以试试display: none
或visibility: hidden
。但是,这也可能不适用于 react-native。
相反,你不能渲染它:
if ( !isHidden ) {
seats.push( <withButton> )
}
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