在2D Javascript数组中搜索值索引

Question

I am trying to write a jQuery that will find the index of a specific value within a 7x7 2D array.

So if the value I am looking for is 0 then I need the function to search the 2D array and once it finds 0 it stores the index of the two indexes.

This is what I have so far, but it returns "0 0" (the initial values set to the variable.

Here is a jsFiddle and the function I have so far:

http://jsfiddle.net/31pj8ydz/1/

$(document).ready( function() {

    var items = [[1,2,3,4,5,6,7], 
                 [1,2,3,4,5,6,7], 
                 [1,2,3,0,5,6,7], 
                 [1,2,3,4,5,6,7], 
                 [1,2,3,4,5,6,7], 
                 [1,2,3,4,5,6,7], 
                 [1,2,3,4,5,6,7]];

    var row = 0;
    var line = 0;

    for (i = 0; i < 7; ++i) {
        for (j = 0; i < 7; ++i) {
            if (items[i, j] == '0,') {
                row = i;
                line = j;
            }
        }
    }

    $('.text').text(row + ' ' + line);
});

HTML:

<p class="text"></p>

Answer 1

You access your array with the wrong way. Please just try this one:

items[i][j]

When we have a multidimensional array we access the an element of the array, using array[firstDimensionIndex][secondDimensionIndex]...[nthDimensionIndex] .

That being said, you should change the condition in your if statement:

 if( items[i][j] === 0 )

Please notice that I have removed the , you had after 0. It isn't needed. Also I have removed the '' . We don't need them also.

Answer 2

Your if statement is comparing

if (items[i, j] == '0,')

Accessing is wrong, you should use [i][j] .

And your array has values:

[1,2,3,4,5,6,7]
....

Your value '0,' is a string, which will not match numeric values inside the array, meaning that your row and line won't change.

Answer 3

First, you are accessing your array wrong. To access a 2D array, you use the format items[i][j] .

Second, your array doesn't contain the value '0' . It doesn't contain any strings. So the row and line variables are never changed.

You should change your if statement to look like this:

if(items[i][j] == 0) {

Notice it is searching for the number 0, not the string '0'.

Answer 4

There are following problems in the code

1) items[i,j] should be items[i][j] .

2) You are comparing it with '0,' it should be 0 or '0' , if you are not concerned about type.

3) In your inner for loop you should be incrementing j and testing j as exit condition.

Change your for loop like bellow and it will work

for (i = 0; i < 7; i++) {
        for (j = 0; j < 7; j++) {
            if (items[i][j] == '0') {
                row = i;
                line = j;
            }
        }
    }

DEMO

Note:-

1) Better to use === at the place of == , it checks for type also. As you see with 0=='0' gives true.

2) Better to say i < items.length and j<items[i].length instead of hard-coding it as 7.

Answer 5

var foo;
items.forEach(function(arr, i) {
    arr.forEach(function(val, j) {
        if (!val) { //0 coerces to false
            foo = [i, j];
        }
    }
}

Here foo will be the last instance of 0 in the 2D array.

Answer 6

You are doing loop wrong On place of

 for (i = 0; i < 7; ++i) {
        for (j = 0; i < 7; ++i) {
            if (items[i, j] == '0,') {
                row = i;
                line = j;
            }
        }
    }

use this

 for (i = 0; i < 7; i++) {
        for (j = 0; j < 7; j++) {
            if (items[i][j] == 0) {
                row = i;
                line = j;
            }
        }
    }

Here is the demo

Answer 7

looks like you are still learning how to program. But here is an algorithm I've made. Analyze it and compare to your code ;)

var itens = [[1,2,3,4,5,6,7], 
             [1,2,3,4,5,6,7], 
             [1,2,3,0,5,6,7], 
             [1,2,3,4,5,6,7], 
             [1,2,3,4,5,6,7], 
             [1,2,3,4,5,6,7], 
             [1,2,3,4,5,6,7]];

var row = null;
var collumn = null;

for (var i = 0; i < itens.length; i++) {
    for (var j = 0; j < itens[i].length; j++) {
        if (itens[i][j] == 0) {
            row = i;
            collumn = j;
        }
    }
}

console.log(row, collumn);