PHP str_replace 和/或 number_format 失败

Question

      $BRL_PRICE = currency("USD", "BRL", $allmoney);
      echo "BRL1 = " . $BRL_PRICE."<br />";
      $BRL_PRICE = str_replace(" ", "", $BRL_PRICE);
      echo "BRL2 = " . $BRL_PRICE."<br />";  
      $BRL_PRICE = number_format($BRL_PRICE, 2);
      echo "BRL3 = " . $BRL_PRICE."<br />";

输出是..

BRL1 = 1 531.70922
BRL2 = 1 531.70922
BRL3 = 1.00

我真正想要的是将值四舍五入到最接近的美分..10位。我知道 number_format 失败，因为currency() google Money Converter 添加了一个空格而不是 comra 或空白。

我想

BRL3 = 1531.71

。

function currency($from_Currency,$to_Currency,$amount) {
    $amount = urlencode($amount);
    $from_Currency = urlencode($from_Currency);
    $to_Currency = urlencode($to_Currency);
    $url = "http://www.google.com/ig/calculator?hl=en&q=$amount$from_Currency=?$to_Currency";
    $ch = curl_init();
    $timeout = 0;
    curl_setopt ($ch, CURLOPT_URL, $url);
    curl_setopt ($ch, CURLOPT_RETURNTRANSFER, 1);
    curl_setopt($ch,  CURLOPT_USERAGENT , "Mozilla/4.0 (compatible; MSIE 8.0; Windows NT 6.1)");
    curl_setopt ($ch, CURLOPT_CONNECTTIMEOUT, $timeout);
    $rawdata = curl_exec($ch);
    curl_close($ch);
    $data = explode('"', $rawdata);
    $data = explode(' ', $data['3']);
    $var = $data['0'];
    return $var;
}

Answer 1

如果您懒得阅读文档，请不要责怪软件：

$BRL_PRICE = number_format($BRL_PRICE, 2, '.', '');

echo number_format(1531.70922, 2, '.', '');

Answer 2

为什么不使用 sprintf('%.2f',$var)

Answer 3

我不认为你的号码中有一个简单的空格。尝试 preg_replace 看看会发生什么：

$BRL_PRICE = preg_replace('/\s+/', '', $BRL_PRICE);

Answer 4

您从 google 获得的响应是​​无效的 json。你甚至不需要json_decode它，而只是为了日志。

str_replace 工作得很好（一如既往），您只需替换正确的字符串：

$BRL_PRICE = str_replace("\xc2\xa0", "", $BRL_PRICE);

即 Unicode 字符“NO-BREAK SPACE”(U+00A0)。