冗余“if”语句较少

形式上的所有语句:

if(condition){
    return false
} else {
    return true
} 

可以简化为:

return !condition

因此,在您的情况下,它将导致:

return !(etReportRow1.text.toString() == "" || etReportRow2.text.toString() == "" || etReportRow3.text.toString() == "")

或:

return 
    etReportRow1.text.toString().isNotEmpty() && 
    etReportRow2.text.toString().isNotEmpty() && 
    etReportRow3.text.toString().isNotEmpty()

注: isNotEmpty() 是扩展方法:

public inline fun CharSequence.isNotEmpty(): Boolean = length > 0

为了避免重复的代码,还可以使用 Sequence :

public fun validateInput() = sequenceOf(etReportRow1, etReportRow2, etReportRow3)
    .map { it.text.toString() }
    .all { it.isNotEmpty() }

因为布尔表达式的计算结果是布尔值,所以只需返回表达式本身的结果,而不必显式返回 true 或 false .

您可以使用以下单表达式函数进一步简化操作:

 private fun validateInput() = etReportRow1.text.toString() != "" && 
                               etReportRow2.text.toString() != "" &&
                               etReportRow3.text.toString() != ""

尝试

return !(etReportRow1.text.toString() == "" || etReportRow2.text.toString() == "" || etReportRow3.text.toString() == "")

我认为这应该有效:

private fun validateInput() = !(etReportRow1.text.toString.isEmpty()  
                                || etReportRow2.text.toString().isEmpty()
                                || etReportRow3.text.toString().isEmpty() )

更简洁:

public fun validateInput() = setOf(
    etReportRow1, etReportRow2, etReportRow3
).none { 
    "${it.text}".isEmpty() 
}