Ale*_* K. 57

双倍逃脱;

SELECT *
  FROM Header
 WHERE userID LIKE '%''%'
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cod*_*ger 12

SELECT     *
FROM Header WHERE (userID LIKE '%''%')
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  • 这里没有正义.最早评论0票 (5认同)

Cat*_*key 7

SELECT *   FROM Header  WHERE userID LIKE '%' + CHAR(39) + '%' 
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