SQL - Return the result from a percentage

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I need to increment in the command below, where it is possible to return a result bringing only a.2PRECO_CUST different (<>) from a1.PRECOUNIT when the difference is above 30% between these two columns.

From now on I thank you all.

select a3.NUMNOTA,a2.CODPROD,a2.DESCRICAO,a2.BARRA,a2.PRECO_CUST as Custo_Cadastro,
a1.PRECOUNIT as Custo_Nota from ITMENTRADANF as a1 
inner join PRODUTOS as a2 on a1.CODPROD = a2.CODPROD
inner join ENTRADANF as a3 on a3.CODENTRADANF=a1.CODENTRADANF
where a1.PRECOUNIT <> a2.PRECO_CUST
  • Replaces your WHERE for: WHERE a1.PRECOUNIT NOT BETWEEN 0.7 * a2.PRECO_CUST AND 1.3 * a2.PRECO_CUST

  • you can use a case when.

  • @Sorack, sorry perhaps my ignorance, but this Where, the values you defined '0.7' and '1.3', could answer me which is alogical that will be done when executing, I did, I executed but did not understand the return of the result within this Where you sent me. Thanks in advance.

  • 0.7 = 70% and 1.3 = 130%, two values which have a greater difference than 30%

  • So I get it. Thanks for helping my friend.

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