Question

我有这样的字符串。

$50 From here i need only 50
$60.59 From here i need only 60, Need to remove $ and .59
€360 From here i need only 360.
€36.99 From here i need only 36 need to remove € and .99.
£900 From here i need only 900.
£90.99 From here i need only 90.

换句话说，我需要从字符串的开头删除所有货币标签，我只需要第一笔金额。希望我的问题很明确。感谢

Answer 1

此RegEx应该执行

~~(\$|€|£)\d+~~

这更好（感谢Jan）

[$€£]\d+

将它与PHP Preg Match

一起使用

preg_match - 执行正则表达式匹配

Answer 2

我建议不要使用正则表达式，因为它对这种情况来说太过分了。

$str = (int)ltrim($str, '$£€');

这就是你所需要的一切。

Performance vs Regex

我通过脚本运行上述测试，看看我的答案与使用RegEx之间的时差是多少，平均而言RegEx解决方案的速度慢了约20％。

<?php
function funcA($a) {
    echo (int)ltrim($a, '$£€');
};
function funcB($a) {
    echo preg_replace('/^.*?([0-9]+).*$/', '$1', $a);
};
//setup (only run once):
function changeDataA() {}
function changeDataB() {}

$loops = 50000;
$timeA = 0.0;
$timeB = 0.0;
$prefix =  str_split('€$€');

ob_start();
for($i=0; $i<$loops; ++$i) {
    $a = $prefix[rand(0,2)] . rand(1,999) . '.' . rand(10,99);

    $start = microtime(1);
    funcA($a);
    $timeA += microtime(1) - $start;

    $start = microtime(1);
    funcB($a);
    $timeB += microtime(1) - $start;
}
ob_end_clean();

$timeA = round(1000000 * ($timeA / $loops), 3);
$timeB = round(1000000 * ($timeB / $loops), 3);

echo "
TimeA averaged $timeA microseconds
TimeB averaged $timeB microseconds
";

时间因系统负载而异，因此应仅考虑相对于彼此的时间，而不是在执行之间进行比较。此外，这不是一个完美的性能基准测试脚本，有外部影响可以影响这些结果，但这提供了一个大致的想法。

TimeA averaged 5.976 microseconds
TimeB averaged 6.831 microseconds

Answer 3

你可以去：

<?php

$string = <<<DATA
$50 From here i need only 50
$60.59 From here i need only 60, Need to remove $ and .59
€360 From here i need only 360.
€36.99 From here i need only 36 need to remove € and .99.
£900 From here i need only 900.
£90.99 From here i need only 90.
DATA;

# look for one of $,€,£ followed by digits
$regex = '~[$€£]\K\d+~';

preg_match_all($regex, $string, $amounts);
print_r($amounts);
/*
Array
(
    [0] => Array
        (
            [0] => 50
            [1] => 60
            [2] => 360
            [3] => 36
            [4] => 900
            [5] => 90
        )

)
*/

?>

请参阅a demo on ideone.com。

Answer 4

使用正则表达式。例如：

$toStr = preg_replace('/^.*?([0-9]+).*$/', '$1', $fromStr);

请参阅preg_replace文档。

Answer 5

以下方式使用

    $str = '$50 From here i need only 50
    $60.59 From here i need only 60, Need to remove $ and .59
    €360 From here i need only 360.
    €36.99 From here i need only 36 need to remove € and .99.
    £900 From here i need only 900.
    £90.99 From here i need only 90.';

    $arr_ = array('$','€','£');

    echo str_replace($arr_,'',$str);

Answer 6

$newString=$string;
$currencyArray = array("$","€","£"); //just add the new item if you want that to add more
foreach($currencyArray  as $value) 
  $newString= str_replace($value,"",$newString);

$newString有你需要的东西。

从字符串php中删除一些字符

6 个答案:

Performance vs Regex