我有一個更大的字符串和一個字符串列表。我想更改較大的字符串，以便 - 對于列表中較大字符串中出現的任何字符串，刪除后綴部分直到下一個空格。更大的字符串WITH dataTab0 AS (SELECT TO_CHAR(to_date(tab_0_0.times),'YYYYMMDD') AS TIME_ID_CATEGORYe93bc60a0041,tab_0_0.request_id AS PAGE_IMPRESSIONf6beefc4b44e4b FROM full_contents_2列表TIME_ID_CATEGORY
PAGE_IMPRESSION
...我需要刪除像 e93bc60a0041 和 f6beefc4b44e4b 這樣的后綴，它們在 TIME_ID_CATEGORY 和 PAGE_IMPRESSION 之后出現我期待以下結果。我需要在 java 中基于正則表達式/有效的解決方案來實現相同的目的。WITH dataTab0 AS (SELECT TO_CHAR(to_date(tab_0_0.times),'YYYYMMDD') AS TIME_ID_CATEGORY,tab_0_0.request_id AS PAGE_IMPRESSION FROM full_contents_2
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TA貢獻1828條經驗 獲得超6個贊
像這樣的事情怎么樣？本質上將 TIME_ID_CATEGORY 或 PAGE_IMPRESSION 匹配到組 1，并將其后的任何內容（即后綴）匹配為組 2。
(TIME_ID_CATEGORY|PAGE_IMPRESSION)(\w+)
然后只需將第 2 組的內容替換為空字符串即可?；蛘咧皇翘鎿Q為 Group 1，這也將消除后綴（請參見下面的代碼片段）。
示例代碼片段：
public static void main(String args[]) throws Exception {
String line = "WITH dataTab0 AS (SELECT TO_CHAR(to_date(tab_0_0.times),'YYYYMMDD') AS TIME_ID_CATEGORYe93bc60a0041,tab_0_0.request_id AS PAGE_IMPRESSIONf6beefc154b44e4b FROM full_contents_2";
Pattern p = Pattern.compile("(TIME_ID_CATEGORY|PAGE_IMPRESSION)(\\w+)");
Matcher m = p.matcher(line);
if (m.find()) {
String output = m.replaceAll("$1");
System.out.println(output);
//WITH dataTab0 AS (SELECT TO_CHAR(to_date(tab_0_0.times),'YYYYMMDD') AS TIME_ID_CATEGORY,tab_0_0.request_id AS PAGE_IMPRESSION FROM full_contents_2
}
}
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