sqlite
database; ?? ??? ??????
英[la?k] 美[la?k]
vt. prep.(?? ??) like;(?? ??)...how(?? ??) like n. like; ??; ??, conj? ?????. like...; as )? ??? ????...; (????? ??, ?? ??, ?? ?? ?? ??? ? ???) ???
SQLite ?? ?? ???
??: SQLite? LIKE ???? ?????? ??? ??? ??? ?? ????? ? ?????. LIKE ???? ?? ???? ?? ???? ???? ?? true(1)? ?????. LIKE ???? ?? ???? ????? ??? ? ?????. ??? ??(%) ??(_) ??? ??(%)? 0?, 1? ??? ??? ??? ?????. ??(_)? ?? ?? ?? ??? ?????. ??? ??? ???? ??? ? ????.
??: ??% ? _? ?? ??? ??? ????.
SELECT FROM table_name
WHERE ? LIKE 'XXXX%'
??
SELECT FROM table_name
WHERE ? LIKE '%XXXX%'
or
SELECT F ROM table_name
WHERE ? LIKE 'XXXX_'
or
SELECT FROM table_name
WHERE ? LIKE '_XXXX'
or
SELECT FROM table_name
WHERE ? LIKE '_XXXX_'
AND ?? OR ???? ???? N?? ??? ??? ? ????. . ??? XXXX? ??? ?? ?? ??? ?? ? ????.
SQLite ?? ?? ?
COMPANY 表有以下記錄： ID NAME AGE ADDRESS SALARY ---------- ---------- ---------- ---------- ---------- 1 Paul 32 California 20000.0 2 Allen 25 Texas 15000.0 3 Teddy 23 Norway 20000.0 4 Mark 25 Rich-Mond 65000.0 5 David 27 Texas 85000.0 6 Kim 22 South-Hall 45000.0 7 James 24 Houston 10000.0 sqlite> SELECT * FROM COMPANY WHERE AGE LIKE '2%'; 這將產生以下結果： ID NAME AGE ADDRESS SALARY ---------- ---------- ---------- ---------- ---------- 2 Allen 25 Texas 15000.0 3 Teddy 23 Norway 20000.0 4 Mark 25 Rich-Mond 65000.0 5 David 27 Texas 85000.0 6 Kim 22 South-Hall 45000.0 7 James 24 Houston 10000.0 sqlite> SELECT * FROM COMPANY WHERE ADDRESS LIKE '%-%'; 這將產生以下結果： ID NAME AGE ADDRESS SALARY ---------- ---------- ---------- ---------- ---------- 4 Mark 25 Rich-Mond 65000.0 6 Kim 22 South-Hall 45000.0