如何使indexof（）区分数组中的两个相似数字

Question

I have an array say

x = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16]

I am using javascript indexof() to check if an element exist or not in the array and if it doesn't exits to do something....

my problem is that when an element like, 1,2,3,4,5,6 doesn't exist in the array, it still judge elements like 10 for 1, 12 for 2, 13 for 3, 14 for 4, 15 for 5, 16 for 6. and so it will still return that the element exists eventhough it is not there

eg:

if x = [2,3,4,5,6,7,8,9,10,11,12,13,14,15,16]

this will still not work as it will see 10 as 1

 if(data1.indexOf(1)===-1){ document.getElementById("s1").style.backgroundColor= "red"; $("input[name='seat1']").attr("disabled", "disabled"); } 

Answer 1

您可以parse字符串并将indexOf用作数组方法。

JSON.parse(data1).indexOf(1) ...

Answer 2

Array.indexOf works in any browser > IE8: MDN Browser compatibility

Polyfill It if you need to support anything.

 // POLYFILL TO SUPPORT ANY BROWSER // https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/indexOf#Polyfill // Production steps of ECMA-262, Edition 5, 15.4.4.14 // Reference: http://es5.github.io/#x15.4.4.14 if (!Array.prototype.indexOf) { Array.prototype.indexOf = function(searchElement, fromIndex) { var k; // 1. Let o be the result of calling ToObject passing // the this value as the argument. if (this == null) { throw new TypeError('"this" is null or not defined'); } var o = Object(this); // 2. Let lenValue be the result of calling the Get // internal method of o with the argument "length". // 3. Let len be ToUint32(lenValue). var len = o.length >>> 0; // 4. If len is 0, return -1. if (len === 0) { return -1; } // 5. If argument fromIndex was passed let n be // ToInteger(fromIndex); else let n be 0. var n = +fromIndex || 0; if (Math.abs(n) === Infinity) { n = 0; } // 6. If n >= len, return -1. if (n >= len) { return -1; } // 7. If n >= 0, then Let k be n. // 8. Else, n<0, Let k be len - abs(n). // If k is less than 0, then let k be 0. k = Math.max(n >= 0 ? n : len - Math.abs(n), 0); // 9. Repeat, while k < len while (k < len) { // a. Let Pk be ToString(k). // This is implicit for LHS operands of the in operator // b. Let kPresent be the result of calling the // HasProperty internal method of o with argument Pk. // This step can be combined with c // c. If kPresent is true, then // i. Let elementK be the result of calling the Get // internal method of o with the argument ToString(k). // ii. Let same be the result of applying the // Strict Equality Comparison Algorithm to // searchElement and elementK. // iii. If same is true, return k. if (k in o && o[k] === searchElement) { return k; } k++; } return -1; }; } // TEST ARRAY var arr = [0,1,2,3,4,6,7,8,9,10,12,13,14,16,17,18,20] // TESTS var offset = 0; for(var i = 0; i <= 20; i++) { if(arr.indexOf(i) === -1) { console.log(i, "not found"); } else { console.log(i, "found") } } 

Answer 3

I would convert them to strings like this.

 function existsInArray(numberArrays, number){ const numAsStrings = numberArrays.join(',').split(',') return numAsStrings.indexOf(String(number)) > -1 } console.log(existsInArray([1,2,3,4,5,6,33], 33)) //true console.log(existsInArray([1,2,3,4,5,6,33], 34)) //false 

Hope this helps.

Answer 4

Your example snippet is not helpful, as it does not show the data ("data1") you are working on.

The behaviour you describe is not the one which one gets with your example array:

 x = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16]; alert("indexOf(1):" + x.indexOf(1)); alert("indexOf(10):" + x.indexOf(10)); alert("indexOf(0):" + x.indexOf(0)); 

It works as intended.

Your real data proably contains strings and not integers.