if声明与img.src

Question

The problem: whenever, if i have the single '=' the variable will show up and be fine.. but it completely ignores the if statement. if i only do the double '=='.. the variable doesnt show up, and it goes to the else state. lemme know if u see anything wrong.

var pic1 = document.getElementById('team1pic').src;
var win1 = document.getElementById('wins').innerHTML;
if (pic1 == 'pens.jpg') {
    document.getElementById('wins').innerHTML = PittWins;
} else {
    document.getElementById('wins').innerHTML = 'no';
}

here is some html to go along with it that i have

<a class="dock-item" id="pens" href="#" onclick="document.getElementById('team1pic').src='pens.jpg'"><span>Pittsburgh Penguins</span><img src="pens.jpg" alt="Pittsburgh Penguins" /></a>` 

and then

<th width="35%" ><img src="" / id='team1pic'></th>

Answer 1

Do an alert(pic1) before the if and you'll figure it out.

The only "problem" is that pic1 is not equal to 'pens.jpg' . If you replace with a single = , the assignment operator, you get an expression that evaluates to 'pens.jpg' , which is always true , so it always runs the first branch.

Answer 2

Better:

var s = document.getElementById('team1pic').src,
    w = document.getElementById('wins');

w.innerHTML = s == 'pens.jpg' ? PittWins : 'no';

I assume that PittWins is a variable.

Answer 3

I would be nice to see the HTML this is attached to.

The reason the single = works but == doesn't points to the conclusion that they are not equal. The reason it shows up for = is because the assignment evaluates to true. This will be the case for anything you assign to the variable that is "truthy", meaning the value would evaluate true in JS. In javascript everything but 0 , false , null & undefined evaluate to true.

You might want to do an alert(pic1) to see just what is in there, or if you are using Firefox or Chrome you can use the Javascript debugger tools.

Answer 4

use

var pic1 = document.getElementById('team1pic').getAttribute('src',2);

...to get the src like it's defined and not the complete path.