PHP使用else来回显一个数字，如果且大于或小于 [英] PHP Echo a number using else if and greater than or less than

问题描述

我有一个类似于whois数据库的脚本。此函数返回站点视图，我想在值之间进行回显。

I have a script which is like a whois database. This function returns site views and I want to echo between value.

如何回显并返回一个结果？如果数字是4000，它应该只返回 1k-10k

How can I echo and return one result? If the number is say 4000, it should return only 1k-10k

行像

1550330

1550330

1000000

代码：

$siteTotalViews=1000000;
if($siteTotalViews <= 100){
    echo '0-100';
}
if($siteTotalViews <= 1000){
    echo '100-1k';
}
if($siteTotalViews <= 10000){
    echo '1k-10k';
}
if($siteTotalViews <= 100000){
    echo '10k-100k';
}
if($siteTotalViews <= 1000000){
    echo '100k-1 mil';
}
if($siteTotalViews <= 2000000){
    echo '1 mil-2 mil';
}
if($siteTotalViews <= 5000000){
    echo '2 mil-5 mil';
}
if($siteTotalViews <= 10000000){
    echo '5 mil-10 mil';
}
if($siteTotalViews >= 10000000){
    echo '10 mil +';
}

推荐答案

快速修复：

$siteTotalViews=1000000;
if($siteTotalViews <= 100){
    echo '0-100';
}
//next else is new
else if($siteTotalViews <= 1000){
    echo '100-1k';
}
//next else is new
else if($siteTotalViews <= 10000){
    echo '1k-10k';
}
//next else is new
else if($siteTotalViews <= 100000){
    echo '10k-100k';
}

更好的修复：

$names=array(
  100 => '0-100',
  1000 => '100-1k',
  10000 => '1k-10k',
  ...
}

foreach ($names as $count=>$name)
  if ($siteTotalViews<$count) break;

echo $name;

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