get data from database table to other database table php mysql

Question

I want to get user name from a database table from his id and put it in other data table

function upload_image($image_temp, $image_ext, $album_id, $image_n, $image_description) {
$album_id = (int)$album_id;
$image_n = mysql_real_escape_string(htmlentities($image_n));
$image_description = mysql_real_escape_string(htmlentities($image_description));
//$download_link = 'uploads/'. $album_id. '/'. $image['id']. '.'. $image_ext;
$mysql_date_now = date("Y-m-d (H:i:s)");
$user_name = mysql_query("SELECT `name` FROM `users` WHERE `user_id` = ".$_SESSION['user_id']);

mysql_query("INSERT INTO `images` VALUES ('', '".$_SESSION['user_id']."','$user_name', '$album_id', UNIX_TIMESTAMP(), '$image_ext', '$image_n', '$image_description', '','$mysql_date_now')");


$image_id = mysql_insert_id();
$download_link = 'uploads/'. $album_id. '/'. $image_id. '.'. $image_ext;

mysql_query("UPDATE `images` SET `download_link`='$download_link' WHERE `image_id`=$image_id ");
$selection = mysql_query("SELECT `user_two` FROM `follow` WHERE `user_one`='".$_SESSION['user_id']."'");

while ($row = mysql_fetch_array($selection)) {
mysql_query("INSERT INTO `notification` VALUES ('', '".$_SESSION['user_id']."', '".$row['user_two']."', '', UNIX_TIMESTAMP(), '$image_n', '$image_description', '$download_link')");
}

$image_file = $image_id.'.'.$image_ext;
move_uploaded_file($image_temp, 'uploads/'.$album_id.'/'.$image_file);

Thumbnail('uploads/'.$album_id.'/', $image_file, 'uploads/thumbs/'.$album_id.'/');

}

the problem is here

$user_name = mysql_query("SELECT `name` FROM `users` WHERE `user_id` = ".$_SESSION['user_id']);

mysql_query("INSERT INTO `images` VALUES ('', '".$_SESSION['user_id']."','$user_name', '$album_id', UNIX_TIMESTAMP(), '$image_ext', '$image_n', '$image_description', '','$mysql_date_now')");

i get this in database (Resource id #14)

"SELECT name FROM users WHERE user_id = ".$_SESSION['user_id']."") — Mihai
– Mihai, Commented Aug 21, 2013 at 22:46
mysql_query returns a resource, not a database field. You need to do $resource = mysql_query("SELECT `name`...) and then $record = mysql_fetch_array($resource) and then $user_name = $record['user_name']. You did it correctly with your $selection query. — SaganRitual
– SaganRitual, Commented Aug 21, 2013 at 22:50

user1864610 · Accepted Answer · 2013-08-21 22:54:08Z

2

Your query doesn't return data: it returns a resource. You then have to work with the resource to retrieve your data, so in this line:

$user_name = mysql_query("SELECT `name` FROM `users` WHERE `user_id` = ".$_SESSION['user_id']);

$user_name doesn't contain the information you want.

Try:

$result = mysql_query("SELECT `name` FROM `users` WHERE `user_id` = ".$_SESSION['user_id']);
list($user_name) = mysql_fetch_array($result);

Note: mysql is deprecated - use mysqli or PDO. The principle is the same.

answered Aug 21, 2013 at 22:54

user1864610

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Community · Accepted Answer · 2017-05-23 11:49:51Z

-1

I think you should look here: MySQL syntax for Join Update

Or here: MySQL Insert & Joins

You could do this in just one query with some joining of the tables.

edited May 23, 2017 at 11:49

CommunityBot

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answered Aug 21, 2013 at 22:47

denlau

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1 Comment

user1864610 Over a year ago

He might be able to improve the efficiency of the queries this way, but the problem is in PHP and the use of mysql.

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