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Home/ Questions/Q 6097997
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Editorial Team
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Editorial Team
Asked: May 23, 20262026-05-23T13:06:24+00:00 2026-05-23T13:06:24+00:00

SELECT * FROM Header WHERE (userID LIKE [%’%])

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SELECT *
  FROM Header
 WHERE (userID LIKE [%'%])
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  1. Editorial Team
    Editorial Team
    2026-05-23T13:06:24+00:00Added an answer on May 23, 2026 at 1:06 pm

    Double them to escape;

    SELECT *
      FROM Header
     WHERE userID LIKE '%''%'
    
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