PHP/SQL syntax: passing variable to SQL query

Question

I am getting tired trying to see what is wrong. I have two php. From the first I am sending a variable 'select1' (basically the id) to the second and than I want to update that record uploading a pdf file.

$id = "-1";
if (isset($_GET['select1'])) {
  $id = mysql_real_escape_string($_GET['select1']);
}



if(isset($_POST['Submit'])) {
    $my_upload->the_temp_file = $_FILES['upload']['tmp_name'];
    $my_upload->the_file = $_FILES['upload']['name'];
    $my_upload->http_error = $_FILES['upload']['error'];
    if ($my_upload->upload()) { // new name is an additional filename information, use this to rename the uploaded file
        mysql_query(sprintf("UPDATE sarcini1 SET file_name = '%s' WHERE id_sarcina = '%s'", $my_upload->file_copy, $id));
    }
}

If I put a line with a valid id, like:

$id = 14;

it is working. What I am doing wrong? Thank you!

what's the error that you got?

Ghassan Elias
– Ghassan Elias

2012-09-14 07:00:29 +00:00
Commented Sep 14, 2012 at 7:00 — Ghassan Elias
– Ghassan Elias, Commented Sep 14, 2012 at 7:00

dregad · Accepted Answer · 2012-09-14 07:19:12Z

1

If you need to accept both post & get, then you should try something like the code below to retrieve the variable.

$var = 'select1';
if( isset( $_POST[$var] ) ) {
    $id = $_POST[$var];
} else if( isset( $_GET[$var] ) ) {
    $id = $_GET[$var];
} else {
    $id = -1;
}

answered Sep 14, 2012 at 7:19

dregad

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7 Comments

Mugur Ungureanu Over a year ago

Still not working. And trying to get a feedback using <?php echo $id;?> i get the right value....

Mugur Ungureanu Over a year ago

The $id is still not recognized by the mysql_query(sprintf("UPDATE sarcini1 SET file_name = '%s' WHERE id_sarcina = '%s'", $my_upload->file_copy, $id));

dregad Over a year ago

I am not sure where you tried to echo $id; (assuming just before the call to mysql_query) so you're sure you actually get there. Also not sure if you get any error messages or any kind of output, if yes it may be useful to provide them.

dregad Over a year ago

As a side note, you probably should use mysqli extension instead of mysql (or ADOdb or PDO), and use bind variables instead of building the query with sprintf

Mugur Ungureanu Over a year ago

Thank you for your time. So, I inserted an echo $id after your ideea and before if(isset($_POST['Submit'])) {..... bla, bla and it shows the right id.I found a free php code here linkand I intend to update some of my records uploading a pdf file (the copy of the invoice) and saving the name as file_name in sarcini1 table.

|

Farhan Hafeez · Accepted Answer · 2012-09-14 07:12:16Z

1

You are using both GET and POST at the same time. As far as I can see, this condition is not returning True

if (isset($_GET['select1']))

Edit: If you don't find any answer in above; maybe some more information/code can help getting to a solution.

edited Sep 14, 2012 at 7:12

answered Sep 14, 2012 at 7:04

Farhan Hafeez

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