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Jan 20th, 2017
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  1. UPDATE table
  2. SET stringcol = REPLACE(stringcol, 'tester1', 'tester2')
  3. WHERE stringcol LIKE '%tester1%'
  4.  
  5. UPDATE table
  6. SET stringcol = REPLACE(stringcol, 'tester1', 'tester2')
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