通过正则表达式从 Oracle 中的字符串中删除简单的 HTML 标签，需要解释
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Remove simple HTML-Tags from String in Oracle via RegExp, Explanation needed
提问by Basti
I do not understand, why my columns reg1 and reg2 remove "bbb" from my string, and only reg3 works as expected.
我不明白，为什么我的 reg1 和 reg2 列从我的字符串中删除了“bbb”，而只有 reg3 按预期工作。
WITH t AS (SELECT 'aaa <b>bbb</b> ccc' AS teststring FROM dual)
SELECT
teststring,
regexp_replace(teststring, '<.+>') AS reg1,
regexp_replace(teststring, '<.*>') AS reg2,
regexp_replace(teststring, '<.*?>') AS reg3
FROM t
TESTSTRING REG1 REG2 REG3
aaa <b>bbb</b> ccc aaa ccc aaa ccc aaa bbb ccc
Thanks a lot!
非常感谢！
回答by Olivier Jacot-Descombes
Because regex is greedy by default. I.e. the expressions .*or .+try to take as many characters as possible. Therefore <.+>will span from the first <to the last >. Make it lazy by using the lazy operator ?:
因为默认情况下正则表达式是贪婪的。即表达式.*或.+尝试采用尽可能多的字符。因此<.+>将从第一个跨越<到最后一个>。使用惰性运算符使其惰性?：
regexp_replace(teststring, '<.+?>')
or
或者
regexp_replace(teststring, '<.*?>')
Now, the search for >will stop at the first >encountered.
现在，搜索>将在>遇到的第一个停止。
Note that .includes >as well, therefore the greedy variant (without ?) swallows all the >but the last.
请注意，.包括>也包括在内，因此贪婪的变体（没有?）吞下>除了最后一个之外的所有。
回答by DevilPinky
Because the first one and the second one are finding this match:
<b>bbb</b>- in this case b>bbb</bmatches both .*and .+
因为第一个和第二个正在找到这个匹配：
<b>bbb</b>- 在这种情况下b>bbb</b匹配.*和.+
The third one also won't do what you need. You are looking for something like this: <[^>]*>. But you also need to replace all matches with ""
第三个也不会做你需要的。你正在寻找这样的东西：<[^>]*>。但是您还需要将所有匹配项替换为“”
回答by A_Developer_in_Austin_TX
If you are merely trying to display the string without all the HTML tags, you can use the function: utl_i18n.unescape_reference(column_name)
如果你只是想显示没有所有 HTML 标签的字符串，你可以使用函数：utl_i18n.unescape_reference(column_name)
