update

[magcr23]

Hi, i'm having a problem with update data:

 

I have a page to create a category (category = categoria ,in my code) and it's working, but now i need another page to update the category if it's need, and that's where the error begings...

 

First i create this form: 

<form action='altcategoria.php' method='POST' enctype='multipart/form-data'>
<select name="updateCategoria">
<?php
$consulta = "SELECT nome FROM categoria ORDER BY id";
$resultado = mysqli_query($con, $consulta);
while ($row = mysqli_fetch_array($resultado)){
echo "<option value=''>", $row{"nome"}, "</option>";
}
?>
</select>
<?php 	
echo"<br />";
?>
			
			
<input type="text" name="novoCat" placeholder="Novo nome">
<input type='file' name='myfile2' id='myfile2'>
<input type="submit" name="alterarCat" value="Alterar">
</form>

And then i created the page altcategoria.php with this code:

<?php
$con=mysqli_connect("localhost","root","6794","website");
if (mysqli_connect_errno()){
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$novoCat = $_POST["novoCat"];
$selectCat = $S_POST["updateCategoria"];

	
	
//INSERIR IMAGEM

$name = $_FILES["myfile2"]["name"];
$type = $_FILES["myfile2"]["type"];
$size = $_FILES["myfile2"]["size"];
$temp = $_FILES["myfile2"]["tmp_name"];
$error = $_FILES["myfile2"]["error"];
	
$i = 1;
$actual_name = pathinfo($name,PATHINFO_FILENAME);
$original_name = $actual_name;
$extension = pathinfo($name, PATHINFO_EXTENSION);

$nome = $actual_name.".".$extension;
$caminho = $name;
	
if($error > 0){
die("Erro");
}
else{
if($type == "image/jpg" || $type == "image/png" || $type == "image/jpeg"){
	
if(file_exists('imgcategoria/' . isset($_FILES[$name]['name']))){
				
while(file_exists('imgcategoria/'.$actual_name.".".$extension)){           
$actual_name = (string)$original_name.$i;
$nome = $actual_name.".".$extension;
$i++;
}
			
echo "O upload do ficheiro " . $nome .  " deu certo.";
				
//mover ficheiro para pasta imgcategoria
move_uploaded_file($temp,"imgcategoria/".$nome);
				
//inserir caminho do ficheiro na base dados
$caminho2 = $nome;
				
$update2 = "INSERT INTO imagem(id, caminho) VALUES (DEFAULT, '$caminho2')";
mysqli_query($con, $update2);
				
$updateCat= "UPDATE categoria SET id = DEFAULT, nome = '$novoCat', caminhoImagem = '$caminho2' WHERE nome = '$selectCat' ";

mysqli_query($con, $updateCat);			
}
else{

//mover ficheiro para pasta imgcategoria
move_uploaded_file($temp,"imgcategoria/".$name);
			
//inserir caminho do ficheiro na base dados
$caminho = $name;
	
$insereIMG = "INSERT INTO imagem(id, caminho) VALUES (DEFAULT, '$caminho')";
mysqli_query($con, $insereIMG);				
				
$updateCat2= "UPDATE categoria SET id = DEFAULT, nome = '$novoCat', caminhoImagem = '$caminho' WHERE nome = '$selectCat' ";

mysqli_query($con, $updateCat2);
			
}
}
else{
die("Tipo/tamanho nao suportado");
}		
}	
	
header('Location: inicio.php');

mysqli_close($con);
?>

With that i can upload the file to the table "imagem" but the update of data in the table "categoria" it's not working.

Anyone can help me?

 

Btw: i'm sending the code in attachment to a better read.

altcategoria.php

[magcr23]

Hi again, i uses an 

echo $updateCat;

and that's what i got:

 

O upload do ficheiro bianca5.PNG deu certo. 

UPDATE categoria SET id = DEFAULT, nome = 'tas', caminhoImagem = 'bianca5.PNG' WHERE nome = ''

 

I know where the error is, but not how fix it...

[Ch0cu3r]

$S_POST should be $_POST on this line

$selectCat = $S_POST["updateCategoria"];

[magcr23]

 

$S_POST should be $_POST on this line

$selectCat = $S_POST["updateCategoria"];

yeah i found that, but keeps the same error:

 

UPDATE categoria SET id = DEFAULT, nome = 'tas', caminhoImagem = 'bianca5.PNG' WHERE nome = ''

[Barand]

 

while ($row = mysqli_fetch_array($resultado)){

echo "<option value=''>", $row{"nome"}, "</option>";

}

All your options have an empty string as their value. Either

• omit the value=''so it defaults to to the "nome" value.
• Explicitly specify the $row['nome'] as the value
• Better still, specify the id as the value and use the id instead of name in related records

Edited June 18, 2015 by Barand

[magcr23]

 

All your options have an empty string as their value. Either

• omit the value=''so it defaults to to the "nome" value.
• Explicitly specify the $row['nome'] as the value
• Better still, specify the id as the value and use the id instead of name in related records

 

 thx for the help, i made it 

echo "<option value='", $row{"nome"}, "'>", $row{"nome"}, "</option>";

Edited June 18, 2015 by magcr23

[magcr23]

I have another problem, when i do 

$updateCat2= "UPDATE categoria SET id = DEFAULT, nome = '$novoCat', caminhoImagem = '$caminho' WHERE nome = '$selectCat' ";

all the updates categorys will get id=0. how can i make it continue the auto increment? 

I mean, if i had id = 1,2,3,4,5 and i update the id=3 it will keep id=3

[Barand]

1. Autoincrement applies to insert queries.

Don't set a value for the id in the update or inserts, let the autoincrement look after the values.

Once an id has been allocated on INSERT, it should never be changed