Javascript indexOf没有按预期工作

Question

I am trying to test whether a value already exists in an array.

In this case, the value of "node1" does not change. both arrays are identical.

However, the same node is added to the array twice, despite the indexOf(node1) test. http://jsfiddle.net/v9yxj5hm/2/

var tree_rows = []; 

var node1 = ['Workplace','Revenue Overall',0];  
if (tree_rows.indexOf(node1) == -1){ tree_rows.push(node1); } 

node1 = ['Workplace','Revenue Overall',0]; 
if (tree_rows.indexOf(node1) == -1){ tree_rows.push(node1); } 

alert(tree_rows)

Answer 1

The .indexOf() method compares as if it were using the === strict equality operator. No two distinct arrays are equal to each other under those conditions.

When you use array (or, for that matter, object) literal expressions to create objects, you get distinct individual objects. The fact that one literal expression looks exactly like another doesn't mean they share identity.

Answer 2

The second use of declaration

node1 = ['Workplace','Revenue Overall',0];

when node1 already existed created a new entry in the lexical environment for node1. As a result, the check for indexOf will not find a previous value for that entry. This ends with node1 being placed in the array a second time.

Answer 3

You have to compare whether each item in node1 is in tree_rows . Maybe you want this?

var tree_rows = []; 

var node1 = ['Workplace','Revenue Overall',0];  
for (i in node1) { if (tree_rows.indexOf(node1[i]) == -1){ tree_rows.push(node1[i]); }} 

node1 = ['Workplace','Revenue Overall',0]; 
for (i in node1) { if (tree_rows.indexOf(node1[i]) == -1){ tree_rows.push(node1[i]); }}

alert(tree_rows)