Interaction PHP and Mysql form does not work

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Fellas, I’m doing a pro bono job for a public service, and I’ve been trying for days to figure out what’s wrong with my code. If anyone can help me, I really appreciate it.

This is the form:

<html>

<style>

::-webkit-input-placeholder  { color:#CDCDCD; }
input:-moz-placeholder { color:#CDCDCD; }
textarea:-moz-placeholder { color:#CDCDCD; }​

</style>


<form name="saque" action="https://equadsaude.000webhostapp.com/bancodados_atualizar.php" method="POST">

<table>

<tr>
<td>Processo</td>         </tr>

<tr>
<td><input name="n1" placeholder="somente algarismos"></td>
</tr>

<tr>
<td>Valor total sacado</td>   </tr>

<tr>
<td><input name="n4" placeholder="00000.00"></td>
</tr>

<tr>
<td>Observações e Data </td> </tr>

<tr>
<td><input type="text" name="n3" ></td>
</tr>

<tr>
<td col span="3"><input type="submit" name="submit" value="Atualizar"></td>
</tr>
</table>
</form>
</html>

And the file . php (which is on the local server) that he should call is this:

<?php
if (isset($_POST["submit"])) 
{

$conectar = new mysqli("localhost","id1019345_dados_sobras","equadsaude", "id1019345_sobras");
$processo = $_POST[ 'n1' ] ;
$valor_sacado = $_POST[ 'n4' ] ;
$observacoes = $_POST[ 'n3' ] ;

//variavel de teste do POST no banco de dados
$teste = mysqli_query($conectar, "SELECT 'Processo' FROM 'Tab_Index' WHERE 'Processo' = '$processo' ");
//variavel para cálculo do Valor da Sobra no banco de dados
$sql_seleciona = mysqli_query($conectar, "SELECT 'Valor_sobra' FROM 'Tab_Index' WHERE 'Processo' = '$processo' ");
//variavel para selecao das Observacoes no banco de dados
$sql_seleciona2 = mysqli_query ($conectar, "SELECT 'Observacoes' FROM 'Tab_Index' WHERE 'Processo' = '$processo' ");

if(mysqli_num_rows($conectar, $teste) == 0)
{
echo "<p>Não existe o registro informado. Verifique novamente no Banco de Dados.</p>";  exit(mysqli_error());
}

else
{
$query1 = mysqli_query ($conectar, $sql_seleciona);
$resultado1 = $query1 - $valor_sacado;
$query2 = mysqli_query ($conectar, $sql_seleciona2);
$resultado2 = $query2."/". $observacoes;
                  
}

//Update do banco de dados
$sql_alterar = mysqli_query($conectar, "UPDATE 'Tab_Index' SET 'Valor_sobra' =  '$resultado1', 'Observacoes' =  '$resultado2' WHERE 'Processo' = '$processo' ");

if  ( isset ($sql_alterar) )
{
print "<h3> Valor da sobra atualizado com sucesso </h4>\n" ;
}
else 
{ 
print "<h3> Erro ao atualizar </h4>\n" ;
}

}
  ?>

php mysql form

  • If you are sending the data to another file, why are you opening a condition if (isset($_POST["submit"]))? It is preventing you from allocating variables.

1 answer

0

Dear Danilo, I have tried to remove this condition, and it does not work either. The error that appears is the same:

The equadsaude.000webhostapp.com page is not working

equadsaude.000webhostapp.com is unable to fulfill this request at the moment. HTTP ERROR 500

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