Send data to Database according to <select> tag via PHP

Question

I’m making a form to send some data to the bank, to expedite a process, my question is the following, I have 5 systems that I need to update, so I made a select html.

I need to send each option of select for a different bank table, for example, if I select the "System 1" he goes to the table "System 1" on the bench, if it is "System 2" on the table "System 2" and so on, in each table would have the same fields only would change the name, then I list it on another screen of the site.

What would be the simplest way for me to do this in PHP ?

<div class="form-group">
        <form action="envia.php" method="post" enctype="multipart/form-data">
            <div class="form-group">
                <label class="col-sm-3 control-label"><b>Sistema</b></label>    
                <div class="col-md-10">
                    <select name="sistema" class="form-control" required/>
                        <option>Escolha o sistema que deseja importar</option>
                        <option value="sis1">Sistema 1</option>
                        <option value="sis2">Sistema 2</option>
                        <option value="sis3">Sistema 3</option>
                        <option value="sis4">Sistema 4</option>
                        <option value="sis5">Sistema 5</option>
                        <option value="sis6">Sistema 6</option>
                        <option value="sis7">Sistema 7</option>
                    </select>
                </div>
            </div>

Answer 1

Taking into account that you are performing a POST and that the <select> has the following syntax:

<select name="sistema">
    <option selected>Selecione uma opção* </option>
    <option value="sistema1" > Sistema 1 </option>
    <option value="sistema2" > Sistema 2 </option>
    <option value="sistema3" > Sistema 3 </option>
    <option value="sistema4" > Sistema 4 </option>
    <option value="sistema5" > Sistema 5 </option>
</select>

The drawback is to enter the name of the tables in the value of the options and capture it in the PHP.

In his PHP would look like this:

$table = addslashes($_POST['sistema']); //nome da tabela

And when it’s time to ride your query, would look something like:

$query = "INSERT INTO $table (campo1, campo2, campo3) VALUES (valor1, valor2, valor3)";

If you do not want to pass the table name directly to us options, pass generic names and treat on PHP.

Example:

<select name="sistema">
        <option selected>Selecione uma opção* </option>
        <option value="table1" > Sistema 1 </option>
        <option value="table2" > Sistema 2 </option>
        <option value="table3" > Sistema 3 </option>
        <option value="table4" > Sistema 4 </option>
        <option value="table5" > Sistema 5 </option>
    </select>

In the PHP:

$table = addslashes($_POST['sistema']);

switch($table) {
    case 'table1':
        $table = 'sistema1'; //definindo o valor da table como sistema1 (nome da tabela)
    break;
    case 'table2':
        $table = 'sistema2'; //definindo o valor da table como sistema2 (nome da tabela)
    break;
    case 'table3':
        $table = 'sistema3'; //definindo o valor da table como sistema3 (nome da tabela)
    break;
    case 'table4':
        $table = 'sistema4'; //definindo o valor da table como sistema4 (nome da tabela)
    break;
    case 'table5':
        $table = 'sistema5'; //definindo o valor da table como sistema5 (nome da tabela)
    break;
}

$query = "INSERT INTO $table (campo1, campo2, campo3) VALUES (valor1, valor2, valor3)";

Since you didn’t inform your structure, I made a more generic example possible.

Answer 2

<?php

$select = $_POST['seu_select'];

switch ($select) {
    case 0:
    $sql = "INSERT INTO tabela1(campo1,campo2,campo3) VALUES ('valor1', 'valor2', 'valor3')";
    mysqli_query($conexao,$sql);

    break;

    case 1:
    $sql = "INSERT INTO tabela2(campo1,campo2,campo3) VALUES ('valor1', 'valor2', 'valor3')";
    mysqli_query($conexao,$sql);
    break;

    case 2:
    $sql = "INSERT INTO tabela3(campo1,campo2,campo3) VALUES ('valor1', 'valor2', 'valor3')";
    mysqli_query($conexao,$sql);
    break;
}

?>

Answer 3

if the name of your <select> for table do so:

$sql="INSERT into `".$_POST[tabela]."` set campo1='".$_POST[var1]."'";
mysqli_query($con,$sql);

Obs: this if you are using the mysqli if it is mysql only

$sql="INSERT into `".$_POST[tabela]."` set campo1='".$_POST[var1]."'";
mysql_query($sql);

where campo1 is an ex you can enter the names of the fields in your table and the names as variable, and the method of your form has to be as post