I have this query
$CeanLibre = "SELECT min('3') + 1 as prox_ean_libre FROM (SELECT 0 AS '3' union all SELECT '3' FROM 'Direcciones&Codigos' WHERE '3' BETWEEN
'".$CCodEan[2].$CCodEan[3].$CCodEan[4]."' AND
'".$CCodEan[2].$CCodEan[3].$CCodEan[5]."' ) 't1' WHERE not exists (select null
FROM 'Direcciones&Codigos' t2 WHERE 't2.3' = 't1.3' + 1 AND 't2.3' BETWEEN
'".$CCodEan[2].$CCodEan[3].$CCodEan[4]."' AND
".$CCodEan[2].$CCodEan[3].$CCodEan[5]."' )"; //miramos de nuevo el ultimo codigo por si aca
$REanLibre = mysqli_query($Conectar ,$CeanLibre); //pongo un resultado de la consulta
$DEanLibre=mysqli_fetch_array($REanLibre);
$EanLibre = trim($DEanLibre["prox_ean_libre"]);
But I get an error.
Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' )' at line 1 <b>Warning</b>: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in
I think it's for some '
and I do not know if the syntax would be that. "3"
is the name of a mysql column
I also have this one that is similar and does not fail:
$CIdmaxFactS = "SELECT MAX(CAST('3' AS UNSIGNED)) AS ultimoean FROM 'Direcciones&Codigos' WHERE '3' BETWEEN '".$CCodEan[2].$CCodEan[3].$CCodEan[4]."' AND '".$CCodEan[2].$CCodEan[3].$CCodEan[5]."' ";
$RIdmaxFactS = mysqli_query($Conectar ,$CIdmaxFactS);
$DIdmaxFactS=mysqli_fetch_array($RIdmaxFactS);
$idmaxFactS = trim($DIdmaxFactS["ultimoean"]);
The only thing that in this second looks for the last number and I need you to look at the first one that is free.
Can someone help me out? to see if you see something that escapes me.