從SQL 中的全名解析名字、中間名和姓氏
問題:
我們如何從SQL 中的單一全名字段中提取名字、中間名和姓氏?這對於匹配全名不完全匹配的名稱非常有用。
解決方案:
讓我們考慮以下提供 90% 準確率的實用解決方案:
WITH FIRST_NAME AS ( SELECT TITLE.ORIGINAL_INPUT_DATA, TITLE.TITLE, CASE WHEN 0 = CHARINDEX(' ', TITLE.REST_OF_NAME) THEN TITLE.REST_OF_NAME ELSE SUBSTRING(TITLE.REST_OF_NAME, 1, CHARINDEX(' ', TITLE.REST_OF_NAME) - 1) END AS FIRST_NAME, CASE WHEN 0 = CHARINDEX(' ', TITLE.REST_OF_NAME) THEN NULL -- no more spaces? assume rest is the last name ELSE SUBSTRING(TITLE.REST_OF_NAME, 1 + CHARINDEX(' ', TITLE.REST_OF_NAME), LEN(TITLE.REST_OF_NAME)) END AS REST_OF_NAME FROM ( SELECT CASE WHEN SUBSTRING(TEST_DATA.FULL_NAME, 1, 3) IN ('MR ', 'MS ', 'DR ', 'MRS') THEN LTRIM(RTRIM(SUBSTRING(TEST_DATA.FULL_NAME, 1, 3))) ELSE NULL END AS TITLE, CASE WHEN SUBSTRING(TEST_DATA.FULL_NAME, 1, 3) IN ('MR ', 'MS ', 'DR ', 'MRS') THEN LTRIM(RTRIM(SUBSTRING(TEST_DATA.FULL_NAME, 4, LEN(TEST_DATA.FULL_NAME)))) ELSE LTRIM(RTRIM(TEST_DATA.FULL_NAME)) END AS REST_OF_NAME, TEST_DATA.ORIGINAL_INPUT_DATA FROM ( SELECT REPLACE(REPLACE(LTRIM(RTRIM(FULL_NAME)), ' ', ' '), ' ', ' ') AS FULL_NAME, FULL_NAME AS ORIGINAL_INPUT_DATA FROM ( SELECT 'GEORGE W BUSH' AS FULL_NAME UNION SELECT 'SUSAN B ANTHONY' AS FULL_NAME UNION SELECT 'ALEXANDER HAMILTON' AS FULL_NAME UNION SELECT 'OSAMA BIN LADEN JR' AS FULL_NAME UNION SELECT 'MARTIN J VAN BUREN SENIOR III' AS FULL_NAME UNION SELECT 'TOMMY' AS FULL_NAME UNION SELECT 'BILLY' AS FULL_NAME UNION SELECT NULL AS FULL_NAME UNION SELECT ' ' AS FULL_NAME UNION SELECT ' JOHN JACOB SMITH' AS FULL_NAME UNION SELECT ' DR SANJAY GUPTA' AS FULL_NAME UNION SELECT 'DR JOHN S HOPKINS' AS FULL_NAME UNION SELECT ' MRS SUSAN ADAMS' AS FULL_NAME UNION SELECT ' MS AUGUSTA ADA KING ' AS FULL_NAME ) RAW_DATA ) TEST_DATA ) TITLE ) SELECT FIRST_NAME.ORIGINAL_INPUT_DATA, FIRST_NAME.TITLE, FIRST_NAME.FIRST_NAME, CASE WHEN 0 = CHARINDEX(' ', FIRST_NAME.REST_OF_NAME) THEN NULL -- no more spaces? assume rest is the last name ELSE SUBSTRING(FIRST_NAME.REST_OF_NAME, 1, CHARINDEX(' ', FIRST_NAME.REST_OF_NAME) - 1) END AS MIDDLE_NAME, SUBSTRING( FIRST_NAME.REST_OF_NAME, 1 + CHARINDEX(' ', FIRST_NAME.REST_OF_NAME), LEN(FIRST_NAME.REST_OF_NAME) ) AS LAST_NAME FROM FIRST_NAME;
此解決方案處理以下特殊情況情況:
- NULL、前導/尾隨空格、連續空格
- 沒有中間名
- 原始全名作為單獨的列
- 具體列表前綴作為單獨的「標題」列
以上是如何從 SQL 中的全名字段中高效解析名字、中間名和姓氏?的詳細內容。更多資訊請關注PHP中文網其他相關文章!

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