JS array learning to determine whether array elements meet given conditions

In the previous article "JS Array Learning: Traverse the Array and Double the Element Value", we introduced the method of traversing the array and processing each element in the array. This time we will continue to talk about array traversal and introduce several methods to detect whether all array elements meet the specified conditions. Friends in need can learn about it~

So how to determine whether the elements in the array meet the specified conditions? For example, traverse an array and check whether all elements in the array are even numbers. How to do this?

Let's introduce 3 methods below, starting with the familiar for loop, and then introducing 2 built-in functions - see how to use them to detect.

Method 1: Use for loop

Implementation idea:

• ##Use for statement to traverse the array

for(var i=0;i<a.length;i++){
}

• Check whether the elements in the array are all even numbers, and conversely: as long as one is not an even number, then you can get "not all even numbers".

So the implementation code can be written like this:

var a = [2,4,5,6,8],b=0;
for(var i=0;i<a.length;i++){
	if (a[i] % 2 != 0) {
		b=0
		break;
	}else{
		b++;
	}
}
if (b) {
	console.log("都是偶数");
} else {
	console.log("不全为偶数");
}

Analysis code:

• We use an intermediate variable

  bAssign a value of 0. If the array element is an even number, b will increase by 1; when an array element is not an even number, the value of b will become 0 again, and use "break;" to exit the loop.

• At this time, we can know whether the elements in the array are even numbers by judging the value of variable b: if

  b>0, then they are all even numbers; if b=0 is not all even numbers.
  

Because element 5 in the array is not an even number, the output result is:

Method 2: Use the some() method

some() method to detect whether there are elements that meet the conditions in the array. Thinking about it from another angle, it can also be used to detect whether all elements in the array do not meet the specified conditions. If they do not meet the specified conditions, it will return false. If one or more elements match, it will return true. (The same idea as the above code)

array.some(function callbackfn(Value,index,array),thisValue)

function callbackfn(value,index,array): A callback function, which cannot be omitted, and can accept up to three parameters:

• value: The value of the current array element, cannot be omitted.

• index: The numeric index of the current array element.

• array: The array object to which the current element belongs.

Let’s take a look at the implementation code:

function f(value, index, ar) {
    if (value % 2 != 0) {
        return true;
    }
}
var a = [2,4,6,8,10,12];
var b = a.some(f);
if (b) {
	console.log("不全为偶数");
} else {
	console.log("都是偶数");
}

It can be seen that because the elements in the array are all even numbers, the output result is:

We change the array element 12 to 11, then the output result is:

Method 3: Use every() method

Every() method can detect whether all array elements meet the specified conditions. The syntax is as follows. For parameter values, please refer to the some() method above

array.every(function callbackfn(Value,index,array),thisValue)

Let’s take a look at the implementation code:

function f(value, index, ar) {
    if (value % 2 == 0) {
        return true;
    }else {
        return false;
    }
}
var a = [2,4,6,8,10,11];
var b = a.every(f);
if (b) {
	console.log("都是偶数");
} else {
	console.log("不全为偶数");
}

The output result is:

不全为偶数

The difference between every() and some() methods:

Every() returns true only when all elements in the array meet the conditions; as long as one element in the array meets the conditions, some( ) will return true.

Okay, that’s all. If you need it, you can watch it:

javascript video tutorial

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