Returning the NULL value

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Someone can help me, I’m making a code to insert data into a table in the database and update too, but unfortunately it’s not working, all variables are giving NULL and the very code that selects the table also gives NULL when I do the var_dum().

Yet where is the error? I have checked several times, the connection to the database is okay because it shows other tables k also has connection only when I do POST to insert k not accepted. Only shows the variable value when I make a purposeful error in table selection. Someone has an idea?

<?php if(isset($_POST['button'])) {

$code_aluno=$_POST['code_aluno'];
$disciplina=$_POST['disciplina'];
$id_trabalho= $_POST['id_trabalho'];
$nota = $_POST['nota'];
$id = $_GET['id'];


$sql_trb_bimestra = "SELECT * FROM trabalhos WHERE id ='$id_trabalho'";

$resulta_trb_bimestre = mysqli_query($conexao, $sql_trb_bimestra);

while ($res_trb_bimestre = mysqli_fetch_assoc($resulta_trb_bimestre)) {

$bimestre = $res_trb_bimestre['bimestre'];

$sql_post = "UPDATE envio_trabalho SET status = 'Accepted', nota = '$nota' WHERE id = '$id_trabalho' AND discipline ='$disciplina'";
mysqli_query($conexao, $sql_post);

$sql_insert = "INSERT INTO nota_trabalho (code, bimestre, discipline, nota) VALUES ('$code_aluno', '$bimestre', '$nota' )";

mysqli_query($conexao, $sql_insert);

echo "<script language='javascript'>window.location='';</script>";
}
}var_dump($disciplina)

?>
  • Just below if(isset($_POST['button'])) { puts a print_r($_POST); and shows us what is being sent in the post.

1 answer

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In this passage:

INSERT INTO nota_trabalho (code, bimestre, discipline, nota)
VALUES ('$code_aluno', '$bimestre', '$nota' )

The value corresponding to discipline. Behold:

INSERT INTO nota_trabalho (code, bimestre, discipline, nota)
VALUES ('$code_aluno', '$bimestre', '$disciplina' , '$nota' )

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