jQuery随机数组编号与上一个不同

Question

I'm applying a random background colour (from a palette of 3) to different sections of a page. However, I want to make sure that the same colour doesn't appear twice in a row.

I thought a little do while loop would work, but it's not quite there by the looks of it.

var colours = new Array('#FF5A5A', '#FFBE0D', '#00DDB8');    

var divs = $('.row');
var last;
var next;

// for each section
divs.each(function(i){

    // get a random colour
    do {

        next = Math.floor(Math.random()*3);

        // if it's the same as the last one, try again!
        } while( next === last ) {

            next = Math.floor(Math.random()*3);

        }

        // when it's different to the last one, set it
        $(this).css('background-color', colours[next] );

        // tell it this is the last one now
        next = last;

});

Any ideas?

Answer 1

That's kind of a syntax error - you couldn't decide whether you want a do-while-loop or a normal while-loop? What you put there is going to be interpreted as a simple block :

do {
    next = Math.floor(Math.random()*3);
} while( next === last ) // end of the do-while-loop!
// Block here - the braces could be omitted as well:
{
    next = Math.floor(Math.random()*3);
}
$(this).css('background-color', colours[next] );
…

This will correctly compute a different number than the last, but then it will overwrite it with a new (unrestricted) random number. Also, the assignment next = last; does the opposite of what you want.

So change your script to

do {
    next = Math.floor(Math.random()*3);
} while( next === last ) // if it's the same as the last one, try again!

// tell it this is the last one now
last = next;

// now that we've made sure it's different from the last one, set it
$(this).css('background-color', colours[next] );

Answer 2

Revised - (because I felt up to the challenge!) http://jsfiddle.net/6vXZH/2/

var last, colours = ['#ff5a5a', '#ffbe0d', '#00ddb8'];    

$('.row').each(function() {
  var color = colours.splice(~~(Math.random()*colours.length), 1)[0];
  $(this).css('background-color', color);   
  last && colours.push(last), last = color;
});

---

Hope this helps! I'd be happy to give you the play-by-play if you'd like it.

Using a little array magic, no inner loops necessary ( http://jsfiddle.net/6vXZH/1/ ) -

var colours = ['#ff5a5a', '#ffbe0d', '#00ddb8'];    
var used = [];

// for each section
$('.row').each(function(i){

  var color = colours.splice(~~(Math.random()*colours.length), 1)[0];
  $(this).css('background-color', color);

  used.push(color);

  if(used.length > 1) {
    colours.push(used.shift());
  }
});

Answer 3

Defining next and last before the function will be better.

var next=last=Math.floor(Math.random()*3);

$divs.each(function(i){        
    do {
        next = Math.floor(Math.random()*3);
    } while( next === last );  
    $(this).css('background-color', colours[next] );
    next = last;
});

Answer 4

{
var colours     = ['#FF5A5A', '#FFBE0D', '#00DDB8'],
    divs        = $('.row'),
    coloursSize = colours.length,
    last,
    next;


divs.each(function(i){

// get a random colour
    do {
            next = Math.floor( Math.random() * coloursSize );
            // if it's the same as the last one, try again!
        } while( next === last ) 
        // when it's different to the last one, set it
        $(this).css('background-color', colours[next] );
        // tell it this is the last one now
        last = next;

});

}