Make choice via combobox from DB table to insert data

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Make choice via combobox from DB table to insert data

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#1 by (1 votes)

-3

Exemplifying: Choose in the Options in which of the Database table you want to insert your information.

<!---
EXE: Banco de Dados
CREATE TABLE tabela_acao, tabela_comedia, tabela_ficcao(
  id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
   nome varchar(20) not null,
   genero varchar(20) not null,
   clasificacao varchar(20) not null,
);
--->

EXE: Tabela que vai inserir conteudo no BD
<form action="" method="">
   <label for="id1">Nome do Filme</label>
   	<input type="text" name="nome" id="id1" required>
   <label for="id2">Genero do Filme</label>
   	<input type="text" name="genero" id="id2" required>
   <label for="id3">Classificação do Filme</label>
  	<input type="text" name="clasificacao" id="id3" required>
   <input type="submit" value="Adicionar">
</form>

EXE: combobox que vai determinar em gual tabela os filmes ira ser colocados
<select>
    <option>tabela_acao</option>
    <option>tabela_comedia</option>
    <option>tabela_ficcao</option>
</select>

php mysql

asked by anonymous 10.10.2016 / 20:25

1 answer

Search filter for some reason you are not looking for the product. Queries sending data to another table [closed]

score 1 · Answer 1

your insertion code in mysql would be this

cod html

<form action="?acao=grava" method="post">
   <label for="id1">Nome do Filme</label>
    <input type="text" name="nome" id="id1" required>
   <label for="id2">Genero do Filme</label>
    <input type="text" name="genero" id="id2" required>
   <label for="id3">Classificação do Filme</label>
    <input type="text" name="clasificacao" id="id3" required>
   <input type="submit" value="Adicionar">
</form>

EXE: combobox que vai determinar em gual tabela os filmes ira ser colocados
<select name='tabelas'>
    <option value='0'>Ação</option>
    <option value='1'>Comédia</option>
    <option value='2'>Ficção</option>
</select>

php that will stay before or after the form

if($_GET['acao']=="grava"){
   $categoria=array('tabela_acao', 'tabela_comedia', 'tabela_ficcao');
   $sql="INSERT INTO ".$categoria[$tabelas]." set nome='".$_POST['nome'].", genero='".$_POST['genero'].", classificacao='".$_POST['classificacao']."";
   mysqli_query($con,sql);
   echo "dados adicionados";
}

This does exactly what you asked for.