0
148
VistasIncorrect value returns
I use this function:
function search (){
v = [[2.0], [3.0], [4.0], [5.0]]
if(v.indexOf(2.0) != -1){
Logger.log ('if')
}else{
Logger.log('else')
}
}
editing
My real need is in the values of words like this I was answered about numbers so I did not change and only added
function search (){
v = [["One"], ["two"], ["three"], ["four"]]
if(v.indexOf("One") != -1){
Logger.log ('if')
}else{
Logger.log('else')
}
}
And instead of "if" returning to me, "else" returns to me. I apologize for the broken English
3 Respuestas
Responde la pregunta0
You input array is a multi/two dimensional array. You are searching for 2.0 which is actually embedded inside another smaller array. Try changing the function to below.
function search() {
v = [
["One"],
["two"],
["three"],
["four"]
]
if (v.findIndex(e => e[0] === 'One') != -1) {
console.log('if')
} else {
console.log('else')
}
}
search();0
Did you mean to write
v = [2.0, 3.0, 4.0, 5.0];?
Because that will print out 'if'.
Reason is that in your example, each element of v array is an array of it's own.
If you really want it to be an array of arrays, then to make it work you can do this in the if statement:
if ( v.flat().includes(2.0) )
flatconverts it from an array of array that contain a number, to an array of numbers.includesis a nicer method to check if the array contains a value - no need to check for index :)
0
array.indexOf() is not applicable for nested array. If your array is primitive types then it's working, but if array is non-primitive types then through indexOf you can't match the value.
Use this instated of your function:
v.findIndex(ele => [...new Set(ele)][0] == 2.0) >=0 ? 'if' : 'else'
