从字符串中删除特殊字符“|”如果它在开始之间包含“,”|“结束”|则忽略

Question

字符串：

(STR,0:30) + |"Outdoor"| + |"Chair Mat","Commercial Floor Mat","Door Mat"| + [Post Type] + (STR,0:30)

结果：

(STR,0:30) + "Out door" + |"Chair Mat","Commercial Floor Mat","Door Mat"| + [Post Type] + (STR,0:30)

我试过了

preg_replace('/\|\"[a-z A-Z]\"\|/i', '"', $string);

我找不到任何解决方案，我知道的唯一条件是字符串将以 |" 开头并以 "| 结尾，如果它不包含 ","我们可以删除或替换 |" 和 "| 为 "

Answer 1

我想你可以像这样使用正则表达式：

\|("[^,]+")\|

Regex demo

$re = '/\|("[^,]+")\|/m';
$str = '(STR,0:30) + |"Outdoor"| + |"Chair Mat","Commercial Floor Mat","Door Mat"| + [Post Type] + (STR,0:30)';
$subst = '';

$result = preg_replace($re, $subst, $str);

echo "The result of the substitution is ".$result;
// (STR,0:30) + "Outdoor" + |"Chair Mat","Commercial Floor Mat","Door Mat"| + [Post Type] + (STR,0:30)

Answer 2

虽然问题中没有明确解释，但此任务的目标是询问管道封闭的子串。 如果子字符串中有多个双引号子字符串，则管道必须保留。如果管道内只有一个双引号子字符串，则应删除这些管道。问题是关于检查逗号，但如果其中一个双引号子字符串包含一个逗号。如果双引号子字符串中存在转义双引号，则可能会出现同样的猴子扭曲场景。

不明白我在说什么？我能理解，所以我有一个演示和一个新的正则表达式模式，它利用了 the wisdom of this post.

代码：(Demo)

$strings = [
    '(STR,0:30) + |"Outdoor"| + |"Chair Mat","Commercial Floor Mat","Door Mat"| + [Post Type] + (STR,0:30)',
    '(STR,0:30) + |"Outdoor"| + |"Red, Yellow, and Green Jamaican Mat"| + [Post Type] + (STR,0:30)',
    '(STR,0:30) + |"Outdoor"| + |"USA \"MAGA\" Mat"| + [Post Type] + (STR,0:30)',
];

print_r(
    preg_replace('~\|("[^"\\]*(?:\\"[^"\\]*)*")\|~', '', $strings)
);

输出：

Array
(
    [0] => (STR,0:30) + "Outdoor" + |"Chair Mat","Commercial Floor Mat","Door Mat"| + [Post Type] + (STR,0:30)
    [1] => (STR,0:30) + "Outdoor" + "Red, Yellow, and Green Jamaican Mat" + [Post Type] + (STR,0:30)
    [2] => (STR,0:30) + "Outdoor" + "USA \"MAGA\" Mat" + [Post Type] + (STR,0:30)
)

注意如何将 "Chair Mat","Commercial Floor Mat","Door Mat" 正确地视为多个条目。相反 "Red, Yellow, and Green Jamaican Mat" 和 "USA \"MAGA\" Mat" 是单个条目，因此它们适当地删除了管道。

从字符串中删除特殊字符“|”如果它在开始之间包含“,”|“结束”|则忽略

Remove special character " | " from a string ignore if it contains "," between starts |" End "|

php

regex

preg-replace