[英]Nested ternary operator for react components
I want to come up with a nested if ternary operator with react components but am being challenged, this is my logic:我想提出一个带有反应组件的嵌套 if 三元运算符,但受到挑战,这是我的逻辑:
if (value==='bank' && is_valid_iban && is_completed) {
return <Checked/>
}
else if (is_completed) {
return <Checked/>
}
else if (value==='businessplan' && is_required) {
return <NotChecked/>
}
This was my change:这是我的改变:
{
(!value=== 'bank' || is_valid_iban) &&
is_completed ? (<Checked/>) : (value ==='businessplan' && is_required && (<NotChecked/>))
}
What could be the best way of coming up with a ternary operator for the logic above.为上述逻辑提出三元运算符的最佳方法可能是什么。
Thanks谢谢
This solution seeks to simplify the below given set of conditions:该解决方案旨在简化以下给定的一组条件:
if (value==='bank' && is_valid_iban && is_completed) {
return <Checked/>
} else if (is_completed) {
return <Checked/>
} else if (value==='businessplan' && is_required) {
return <NotChecked/>
}
It is observed from the above that is_completed
is part of the first & second conditions.从上面观察到,
is_completed
是第一个和第二个条件的一部分。 Something like this: ((A && B && C) || C)
which can be represented as (A && B) || C
像这样:
((A && B && C) || C)
可以表示为(A && B) || C
(A && B) || C
. (A && B) || C
。
Using a standard if..else
structure:使用标准的
if..else
结构:
if ((value === 'bank' && is_valid_iban) || is_completed) return <Checked />
else if (value === 'businessplan' && is_required) return <NotChecked />
else return null;
Using ternary ?:
使用三元
?:
return (
((value === 'bank' && is_valid_iban) || is_completed)
? <Checked />
: (value === 'businessplan' && is_required)
? <NotChecked />
: null
);
When using ?:
please always indent the code so it is readable.使用
?:
时,请始终缩进代码以使其易于阅读。 Same applies for if else
as well;同样适用于
if else
; however, the if else
structure is a lot more readable than ?:
.但是,
if else
结构比?:
可读性强得多。
Make it Simple:让它变得简单:
{(value === "bank" && is_valid_iban && is_completed) || is_completed ? (
<Checked />
) : (value === "businessplan" && is_required) ? (
<NotChecked />
) : (
<NotChecked />
)}
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.