Olá ta dando esse erro no meu codigo e nao to conseguindo enxergar o erro.
Erro: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 'cod_ose like 79' at line 1
public Ose buscaporOse(Integer nome_ose)throws Exception{
OpenDatabase();
String SQL= "select * "
+ "from ose,cliente,servico "
+ "where ose.cod_cliente=cliente.cod_cliente "
+ "AND ose.cod_servico=servico.cod_servico"
+ "AND cod_ose like ?";
pstmt=con.prepareStatement(SQL);
pstmt.setInt(1,'%'+ nome_ose +'%');
rs=pstmt.executeQuery();
Ose o= null;
while(rs.next()){
o=new Ose();
o.setCod_ose(rs.getInt("cod_ose"));
o.setNome(rs.getString("nome"));
o.setNomeServico(rs.getString("nomeServico"));
o.setSituacao(rs.getString("situacao"));
o.setObservacao(rs.getString("observacao"));
}
return o;
}
else if(action.equalsIgnoreCase("buscaporOse")){
try{
Integer nome_ose=Integer.parseInt(request.getParameter("nome_ose"));
OseBD ob = new OseBD();
Ose o = ob.buscaporOse(nome_ose);
if(o.getCod_ose()>0){
request.getRequestDispatcher("osesConsulta.jsp").forward(request, response);
} else {
request.setAttribute("msg", "Ose não encontrada");
request.getRequestDispatcher("oses.jsp").forward(request, response);
}
<form action="controleOse.jsp?action=buscaporOse" method="post">
<label for="nome_ose">Digite o codigo da ose:</label><br />
<input type="search" name="nome_ose" id="nome_ose" value="" />
<input type="submit" value="Buscar" /><br /><br />
</form>
