更好的写评级PHP功能的方式

Question

我已经在下面写了这个评级功能,并且想知道它是否可以做得更小/更简单或更好的编码,它使用显示基于被调用函数内的百分比数字的图像

function rating( $number ) {

if ($number >= .0 && $number <= .099999) { 
  $star = '<img src="images/0.gif" alt="' . $number . '" />';
}

if ($number >= .10 && $number <= .199999){
  $star = '<img src="images/10.gif" alt="' . $number . '" />';
}

if ($number >= .20 && $number <= .299999){
  $star = '<img src="images/20.gif" alt="' . $number . '" />';
}

if ($number >= .30 && $number <= .399999){
  $star = '<img src="images/30.gif" alt="' . $number . '" />';
}

if ($number >= .40 && $number <= .499999){
  $star = '<img src="images/40.gif" alt="' . $number . '" />';
}

if ($number >= .50 && $number <= .599999){
  $star = '<img src="images/50.gif" alt="' . $number . '" />';
}

if ($number >= .60 && $number <= .699999) { 
  $star = '<img src="images/60.gif" alt="' . $number . '" />';
}

if ($number >= .70 && $number <= .799999){
  $star = '<img src="images/70.gif" alt="' . $number . '" />';
}

if ($number >= .80 && $number <= .899999){
  $star = '<img src="images/80.gif" alt="' . $number . '" />';
}

if ($number >= .90 && $number <= .999999){
  $star = '<img src="images/90.gif" alt="' . $number . '" />';
}

if ($number > .999999){
  $star = '<img src="images/100.gif" alt="' . $number . '" />';
}

return $star;
}

继承人如何使用:

rating(0.51087)

Answer 1

当然,这需要一点点数学.

function rating($number) {
    return '<img src="images/' . floor($number * 10) * 10 . '.gif" alt="' . $number . '" />';
}