Re: Upload a file question

On 20 Jun, 11:22, Captain Paralytic <paul_laut...@xxxxxxxxx> wrote:
On Jun 20, 10:54 am, Pépê <josemariabar...@xxxxxxxxx> wrote:



Hi all.

Im a newbie in PHP and im trying touploada file to the server.

I use a form touploada pdf file and some text information about it.

The client uploads the file and the system renames that file and puts
all the information in the database.

The problem is when the client goes again to edit the information, i
always have to choose a file touploador else it will put blank the
pdf column and he cant find the old one!

i do a $_POST['file'] to the UPDATE statement but i think i need to do
a if clause(and dont know what im going to put )...but where? i tried
it in the UPDATE statement and i cant..

Build your update statement dynamically. This is the sort of thing,
but you should sanitise the $_POST input.

if($_POST['file'])
  $fileup = ",file = '{$_POST['file']}'";
else
  $fileup = '';

$qry = "
INSERT INTO fred SET
  id = {$id},
  info1 = '{$info1}',
  info2 = '{$info2}
  {$fileup}
ON DUPLICATE KEY UPDATE
  info1 = '{$info1}',
  info2 = '{$info2}'
  {$fileup}
";

Hi Captain,

I tried what you ve done but with the update statment:

if($_POST['relatorio_pdf']){
$fileup = ",relatorio_pdf = '{$_POST['relatorio_pdf']}'";
}else{
$fileup = '';

if (empty($error) ) {

$sql = "UPDATE relatorio SET
relatorio_nome = '{$_POST['relatorio_nome']}',
relatorio_ano = '{$_POST['relatorio_ano']}',
relatorio_pdf = '$fileup',
relatorio_activo = '{$_POST['relatorio_activo']}'
WHERE relatorio_id = {$_GET['relatorio_id']}";


}
But it didnt worked..

And i didnt quite understand this line: $fileup = ",relatorio_pdf =
'{$_POST['relatorio_pdf']}'"; (why the comma, and then a variable name?
.

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