Form, PHP and MYSQL - mysqli_num_rows always equal to zero

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Form, PHP and MYSQL - mysqli_num_rows always equal to zero

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#1 by (1 votes)

1

Friends, I'm doing pro bono work for a public service, and I've been trying to figure out what's wrong with my code for days. If anyone can help me, thank you very much.

O formulário é esse:

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

     <?php

$conectar = new mysqli("localhost","id1019345_dados_zzzz","xxxx", "id1019345_sobras") or die(mysqli_error());

    $processo = $_POST[ 'n1' ] ;
    $valor_sacado = $_POST[ 'n4' ] ;
    $observacoes = $_POST[ 'n3' ] ;

    //variavel de teste do POST no banco de dados
    $teste = mysqli_query($conectar, "SELECT 'id' FROM 'Tab_Index' WHERE 'Processo' = '$processo' ");
    while (mysqli_num_rows($conectar, $teste) == 0)
    {
    echo "<p>Não existe o registro informado. Verifique novamente no Banco de Dados.</p>";  exit(mysqli_error());
    }


    //variavel para cálculo do Valor da Sobra no banco de dados
    $sql_seleciona = mysqli_query($conectar, "SELECT 'Valor_sobra' FROM 'Tab_Index' WHERE 'Processo' = '$processo' ");
    while ($query_row = mysqli_fetch_assoc($conectar, $sql_seleciona))
        {
        foreach($query_row as $key => $value)
                           {
                           $resultado = $value-$valor_sacado;
                           }
        }

    //variavel para selecao das Observacoes no banco de dados
    $sql_seleciona2 = mysqli_query ($conectar, "SELECT 'Observacoes' FROM 'Tab_Index' WHERE 'Processo' = '$processo' ");
    while ($query_row2 = mysqli_fetch_assoc($conectar, $sql_seleciona2))
        {
        foreach($query_row2 as $key => $value)
                           {
                           $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" ;
    }


      ?>

The problem is that the message "There is no record entered, please check the database again", which I warned you if you did not find anything in the database. But even when I put a Process number that exists in the database, it keeps giving that information. And if I remove this conference from the script, the final message "Replace value successfully updated" appears as if the database update had occurred, but when I check the database, nothing has changed.

Impression I have is that the interaction with the database is not occurring, for some reason I do not know what it is.

The table in the DB has 4 columns: id, Process (BIGINT), Value_sobra (DECIMAL 7,2), Remarks (VARCHAR). HOST: localhost USERNAME: id1019345_dados_zzzz PASSWORD: xxxx DB: id1019345_sobras

php mysql

asked by anonymous 14.03.2017 / 18:12

1 answer

How to insert data coming from a form in the database [PDOstatment - CRUD] Safe way to work with cookies

score 1 · Answer 1

Dude, first, never ever ever pass a form string straight to the query this way. The correct way, using mysqli is as follows:

$stmt = $dbConnection->prepare('SELECT * FROM employees WHERE name = ?');
$stmt->bind_param('s', $name);

$stmt->execute();

$result = $stmt->get_result();
while ($row = $result->fetch_assoc()) {
    // do something with $row
}

reference: link

Now let's get down to the problem:

At the time of picking up the number of rows that came back, you only have to pass the query result, you do not need the connection. That is:

$resultado = mysqli_query($conectar, "SELECT * FROM teste");
$linhas = mysqli_num_rows($resultado);
echo $linhas; // 256

reference: link

This is because the object returned to the result variable will be used in the count, the connection to the database is no longer necessary, only if you are going to do another query.

The other problem

But even the check is wrong after its UPDATE , why are you checking to see if the variable was set at some point ( isset() ). Even if it can not perform the operation, the variable has been defined yes, but for a very effective value: FALSE ! But since your question was "Is this variable set?", It will always return TRUE and follow the block within the condition.

Then the correct form can be two:

if ($sql_alterar) {
  (...)
}

I do not like it personally, because it's not very clear what you're checking, but it's sure to follow else if you can not UPDATE . I prefer this:

if ($sql_alterar !== false) {
   (...)
}

Why I'm sure, at the time I hit the eye, that I'm worried if this variable is false.