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select * from a where columna like '%/%%' escape '/' UP SELECT * from a WHERE columnA LIKE '%\%%' ESCAPE'\'; ÆäÖУ¬ESCAPE'\'±íʾ£º\Ϊ»»Âë×Ö·û£¬Æ¥Åä´®ÖиúÔÚ\ºóÃæµÄ×Ö·û%²»ÔÙ¾ßÓÐͨÅä·ûµÄº¬Ò壻 תÒåΪÆÕͨµÄ¡°%¡±×Ö·û¡£ÒýÓà select * from a where columna like '%/%%' escape '/' UP Â¥Éϵķ½·¨¹ÜÓ㬵«ÊÇżÏëÎÊÒ»¸öÈç¹û×Ö·û´®Àï°üº¬ÓÐ'£¨µ¥ÒýºÅ£©µÄÒªÔõôÓÃLike²éѯ°¡£¿£¿ select * from t1 where instr(f1,'%') != 0 ; ÓÃÕýÔò±í´ïʽ°É£º select * from dual where 'tian%dd'like '%/%%' escape '/';ÒýÓà ¥Éϵķ½·¨¹ÜÓ㬵«ÊÇżÏëÎÊÒ»¸öÈç¹û×Ö·û´®Àï°üº¬ÓÐ'£¨µ¥ÒýºÅ£©µÄÒªÔõôÓÃLike²éѯ°¡£¿£¿ select * from t1 where instr(f1,'''') != 0;ÒýÓà ¥Éϵķ½·¨¹ÜÓ㬵«ÊÇżÏëÎÊÒ»¸öÈç¹û×Ö·û´®Àï°üº¬ÓÐ'£¨µ¥ÒýºÅ£©µÄÒªÔõôÓÃLike²éѯ°¡£¿£¿ where col1 like '%''%' »ò where col1 like '%'||chr(39)||'%' »ò where instr(col1,chr(39))>0 Â¥Ö÷ÓÃOracle ÕýÔòÊÔÊÔ SQL code: with tt as ( select '33%44''' COL from dual ) select * from
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type Exception report message description The server encountered an internal error () that prevented it from fulfilling this request. exception org.apache.jas ......
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