检查数组中是否存在值

Question

I would like to check if values in Array1 exist in Array2. If it exist, then add the index for that value in a new array as a value.

My solution below is doing almost that but it dosent alert me every same number, duplicates. For exampel the value 4355 in array1, its 3 of that value, but it only alert me 1 time. I would like to alert me every time.

var array1 = [];
array1.push(3545);
array1.push(4355);     
array1.push(5435);
array1.push(3457); 
array1.push(4355);
array1.push(4355);
var array2 = [];
array2.push(1022);
array2.push(4355);
array2.push(2333);
array2.push(5656);
array2.push(1234);
array2.push(5555);
array2.push(7777);
array2.push(3455);

for(var i=0; i<array2.length; i++){
 checkIfExist(array2[i], array1);
}

function checkIfExist(searchValue, searchArray){
  if ($.inArray(searchValue, searchArray) < 0) {
      alert(searchValue + " not found");
    }
    else {
        alert(searchValue + " found with index(" + $.inArray(searchValue, array2) + ") in array");
    }
}

Answer 1

To find all the indices of a particular value in an array you could use jQuery.each() :

var indices = [];
$.each(searchArray, function(index, value) {
    if (value == searchValue) {
        indices.push(index);
    }
});

Now, indices is an array with the indices at which the value was found, or empty if there were no matches.

Answer 2

I'm not sure to well understand what you want... hope it will help you!

 var array1 = []; array1.push(3545); array1.push(4355); array1.push(5435); array1.push(3457); array1.push(4355); array1.push(4355); var array2 = []; array2.push(1022); array2.push(4355); array2.push(2333); array2.push(5656); array2.push(1234); array2.push(5555); array2.push(7777); array2.push(3455); var array3={}; for(var i=0; i<array1.length; i++){ if( (index= array2.indexOf(array1[i])) > 0 ){ if(!array3[array1[i]]) array3[array1[i]]= [] array3[array1[i]].push({index1:i,index2:index}); } } document.write(JSON.stringify(array3, undefined, 2)); console.log(array3); 

Answer 3

Maybe try looping through the array and every time you find a match sending out the alert

alert(searchValue + " found with index(" + $.inArray(searchValue, array2) + ") in array");

(Sorry I couldn't put this as a comment, I don't have 50 reputation yet.

Answer 4

Use the regular javascript way to get an index from an Array instead of these angular things:

array.indexOf( value )

This will return the index of the value, or -1 if it is not found.

function checkIfExist(value, array) {
  var index = array.indexOf(value);

  if (index === -1) {
    alert(searchValue + " not found");
  } else {
    alert(value + " found with index(" + index + ") in array");
  }
}

Answer 5

What about something like this? See the jsfiddle :

var arr = [3545, 4355, 5435, 3457, 4355, 4355],
arr2 = [1022, 4355, 2333, 5656, 1234, 5555, 7777, 3455],
occurences = {};

arr2.forEach(function(el, i, array) {
    occurences[el] = $.map(arr, function(element, index) {
        return element == el ? index : null;
    });
});

console.log(occurences);

$('body').html('Indexes: '+JSON.stringify(occurences));

// ==> Indexes: {"1022":[],"1234":[],"2333":[],"3455":[],"4355":[1,4,5],"5555":[],"5656":[],"7777":[]}

Basically, this loops through arr2 to match each of those values to a value in arr . Each value from arr2 is made a key to a new object, called occurences , and there is an array of the indexes of arr where the values match the ones that are in arr2 .

As you can see, the only value from arr2 that appears in arr is 4355 , and it occurs at the indexes of 1, 4, and 5 . Since none of the other values from arr2 match those in arr , the arrays of indexes for those values are simply empty.

Obviously you can pretty the code up a bit and adjust it as it fits your needs.

Answer 6

$.inArray gives the index only for the first occurrence in array starting from 0 else -1 .

function checkIfExist(searchValue, searchArray){
  var index = [];//trace all matching index
  for(var i = 0, l = searchArray.length; l > i; i ++) {
    if(searchValue === searchArray[i]) {//matched value
      index.push(i);//add index to array
    }
  }
  return index;
}

var arr = checkIfExist(0, [1,0,3,9,0]);

if(arr.length) {//match found
    alert('Matched at index: ' + arr.join(','))
} else {
  alert('No match found');
}