Question

我想删除以WITH data_table AS ( SELECT 'A' user_code, 'Root ID' description, 'B' parent_code FROM dual UNION SELECT 'B' user_code, 'Level1' description, 'C' parent_code FROM dual UNION SELECT 'C' user_code, 'Level2' description, 'D' parent_code FROM dual UNION SELECT 'D' user_code, 'Level3' description, 'E' parent_code FROM dual UNION SELECT 'E' user_code, 'Level4' description, 'F' parent_code FROM dual ) SELECT MAX(SYS_CONNECT_BY_PATH(user_code,'->')) FROM data_table CONNECT BY prior parent_code = user_code START WITH user_code = 'A'和space符号开头的文字末尾的所有字词。不应删除句子中的URLS或主题标签。

示例文字：

我尝试了这个但它删除了所有的标签：

hello world #dontremoveme foobar http://example.com/#dontremoveme #remove #removeme #removeüäüö

我的想法是检查文本末尾的每个单词，检查前面有$tweet = "hello world #dontremoveme foobar http://example.com/#dontremoveme #remove #removeme #removeüäüö"; preg_match_all("/(#\w+)/", $tweet, $matches); var_dump( $matches );的前导#，直到不再是这种情况。如何将其翻译成正则表达式？

Answer 1

您可以使用类似的内容：( #[^# ]+?)+$并将其替换为空字符串。

有一个例子here。由于您有非ASCII字符，.运算符（与任何字符匹配）应该可以帮助您处理任何字符。

Answer 2

以下正则表达式匹配行尾的所有以[Space]#开头的字词。

/( #\S+)*$/g

https://regex101.com/r/eH4bJ2/1

Answer 3

这将完成这项工作：

$tweet = "hello world #dontremoveme foobar http://example.com/#dontremoveme #remove #removeme #removeüäüö";
$res = preg_replace("/ #\p{L}+\b(?!\s+\p{L})/u", '', $tweet);
echo $res,"\n";

<强>输出：

hello world #dontremoveme foobar http://example.com/#dontremoveme

从句子末尾删除主题标签

3 个答案: