如何处理包含负号的字符串

发布于 2025-02-01 09:59:09 字数 491 浏览 4 评论 0原文

我使用的是简单的正则[^0-9。]，该与给定的字符串中的0到9和一个段之间的任何数字不匹配并与其余的匹配，我将其替换为空字符串然后，以字符串格式将我的有效值（整数或双）转换为整数或double。

例如：

'123A.478' => '123.478'
'123.48' => '123.48'
'123AX' => '123'

我也想处理负有价值的字符串，即，如果字符串具有“ -123”，我想保留它。因此，我需要将字符串转换为如下：

'--123.46' => '-123.46',
'123A-.46' => '123.46',
'-123--.46' => '-123.46', 
'A-123-.46' => '-123.46'

我尝试使用量词，但我无法与现有的量发架构建正确的正则拨号。

有什么办法可以使用Regex实现这一目标？

原文

I am using a simple regex [^0-9.] that doesn't match any numbers between 0 to 9 and a period in my given string and matches the rest, and I replace that with empty string and then convert my valid value (integer or double) in string format into an integer or double.

For example:

'123A.478' => '123.478'
'123.48' => '123.48'
'123AX' => '123'

I want to also handle negative valued strings as well i.e. if a string has '-123', I want to retain it. So, I need to convert the strings as follows:

'--123.46' => '-123.46',
'123A-.46' => '123.46',
'-123--.46' => '-123.46', 
'A-123-.46' => '-123.46'

I tried using a quantifier but I was unable to build a correct regex with my existing one.

Is there any way I can achieve this using regex?

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笙痞 2025-02-08 09:59:09

您可能会使用交替删除示例字符串中的零件，然后在替换中使用一个空字符串。

^[^\d\s]+(?=-)|[^\d\s.]+(?!\d)

说明

^ String的启动
[^\ d \ s]+匹配数字以外的1+ chars或Whitespace Char
（？= - ）断言-直接在右
|或
[^\ d \ s。]+匹配1+除了数字，Whitespace char或dot
（？！\ d）以外的其他字符以外，直接向右边的数字声明

a REGEX DEMO 和a php demo 。

示例代码

$strings = [
    "123A.478",
    "123.48",
    "123AX",
    "--123.46",
    "123A-.46",
    "-123--.46",
    "A-123-.46"
];

$pattern = '/^[^\d\s]+(?=-)|[^\d\s.]+(?!\d)/';

foreach ($strings as $str) {
    echo preg_replace($pattern, "", $str) . PHP_EOL;
}

输出

You might use an alternation to remove the parts from the example strings, and in the replacement use an empty string.

^[^\d\s]+(?=-)|[^\d\s.]+(?!\d)

Explanation

^ Start of string
[^\d\s]+ Match 1+ chars other than a digit or a whitespace char
(?=-) Assert a - directly to the right
| Or
[^\d\s.]+ Match 1+ chars other than a digit, whitespace char or dot
(?!\d) Assert not a digit directly to the right

See a regex demo and a PHP demo.

Example code

$strings = [
    "123A.478",
    "123.48",
    "123AX",
    "--123.46",
    "123A-.46",
    "-123--.46",
    "A-123-.46"
];

$pattern = '/^[^\d\s]+(?=-)|[^\d\s.]+(?!\d)/';

foreach ($strings as $str) {
    echo preg_replace($pattern, "", $str) . PHP_EOL;
}

Output

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幽梦紫曦～ 2025-02-08 09:59:09

您可以使用

preg_replace('~^(-)|-+~', '$1', preg_replace('~[^\d.-]+~', '', $s))

php演示：

<?php

$strings = ['--123.46', '123A-.46', '-123--.46', 'A-123-.46'];
foreach ($strings as $s) {
    echo $s . " => " . preg_replace('~^(-)|-+~', '$1', preg_replace('~[^\d.-]+~', '', $s)) . PHP_EOL;
}

输出：

--123.46 => -123.46
123A-.46 => 123.46
-123--.46 => -123.46
A-123-.46 => -123.46

有两个步骤：

所有出现[^\ d .-]+模式（除了数字，点或连字符以外的任何其他字符）均已从输入字符串，然后
^（ - ）| - +模式出现（字符串中的第一个字符将连字符放置在捕获组1中，所有其他连字符都刚刚匹配） $ 1，第1组值（因此，所有连字符均被删除，但在字符串中是第一个）。

You can use

preg_replace('~^(-)|-+~', '$1', preg_replace('~[^\d.-]+~', '', $s))

See the PHP demo:

<?php

$strings = ['--123.46', '123A-.46', '-123--.46', 'A-123-.46'];
foreach ($strings as $s) {
    echo $s . " => " . preg_replace('~^(-)|-+~', '$1', preg_replace('~[^\d.-]+~', '', $s)) . PHP_EOL;
}

Output:

--123.46 => -123.46
123A-.46 => 123.46
-123--.46 => -123.46
A-123-.46 => -123.46

There are two steps:

All occurrences of the [^\d.-]+ pattern (any char other than a digit, dot or hyphen) are removed from the input string and then
The ^(-)|-+ pattern occurrences (the first char in the string that is a hyphen is placed into Capturing group 1 and all other hyphens are just matched) are replaced with $1, Group 1 value (so, all hyphens are removed but the first one in the string).

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