如何使用 if [on hold] 检测数字是否大于 20 或大于 50

Question

I try to detect when a number is greater than 10,20,30,40, etc. But this code does not work.

var number = 35;
var text;
if(number >= 10){
text = "> 10"
}
else if(number >= 20){
text = "> 20"
}
else if(number >= 30){
text = "> 30"
}

It only takes me the first if

Answer 1

35 is ≥ 10. It is also ≥20 and ≥30, but you're checking for ≥10 first, so the statement within the first if is executed. Since the rest of the conditionals are else if rather than if , none of the others are tested.

Ideally, check for ≥30 first. Then ≥20. Then ≥10.

var number = 35;
var text;

if(number >= 30){
    text = "> 30"
} else if(number >= 20) {
    text = "> 20"
} else if(number >= 10) {
    text = "> 10"
}

Answer 2

Reverse the order of the if 's. The program will leave the if/else statement as soon as it encounters an if that is true, so you need to put the ones that will be false before the first that will be true.

if(number >= 30){
    text = "> 30"
}
else if(number >= 20){
    text = "> 20"
}
else if(number >= 10){
    text = "> 10"
}

This will give you the highest one it's greater than.

if you want to know all of the numbers it's greater than, you can do something like

text = "";

if(number >= 10){
    text += "> 10\n"
}
if(number >= 20){
    text += "> 20\n"
}
if(number >= 30){
    text += "> 30\n"
}

Since this isn't scalable (any number greater than 30 will just return 30), you can do something like this which will give you the multiple of 10 it's greater than

text = "> " + (number/10) + "0";

Answer 3

You should do the same but in a different order, beginning from bigger to little, a sort will be a good idea if it were an array.

var number = 35;
var text;
if(number >= 40){
text = ">= 40"
}
else if(number >= 30){
text = ">= 30"
}
else if(number >= 20){
text = ">= 20"
}
else if(number >= 10){
text = ">= 10"
}