get value from MySQL database with PHP

$from = $_POST['from'];
$to = $_POST['to'];
$message = $_POST['message'];

$query  = "SELECT * FROM Users WHERE `user_name` = '$from' LIMIT 1";
$result = mysql_query($query);
while($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
    $fromID = $row['user_id'];
} 

I'm trying to have $formID be the user_id for a user in my database. Each row in the Users table is like:

user_id | user_name | user_type
   1    |  Hristo   |   Agent

So I want $from = 1 but the above开发者_运维百科 code isn't working. Any ideas why?

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Try this:

$from = mysql_real_escape_string($_POST['from']);
$to = mysql_real_escape_string($_POST['to']);
$message = mysql_real_escape_string($_POST['message']);

$query  = "SELECT * FROM Users WHERE user_name = '$from' LIMIT 1";
$result = mysql_query($query) or die(mysql_error());

while($row = mysql_fetch_assoc($result)) {
    $fromID = $row['user_id'];
}

Also, make sure that:

• You have connected to the database
• You do get data from the post, try var_dump with your vars eg var_dump($from)

---

Use mysql_fetch_assoc instead

---

while($row =mysql_fetch_assoc($result)){ $fromID = $row['user_id']; }

---

though it should. try this code

$from    = mysql_real_escape_string($_POST['from']);
$to      = mysql_real_escape_string($_POST['to']);
$message = mysql_real_escape_string($_POST['message']);

$query  = "SELECT * FROM Users WHERE `user_name` = '$from' LIMIT 1";
$result = mysql_query($query) or trigger_error(mysql_error().$query);
$row = mysql_fetch_array($result, MYSQL_ASSOC);
$fromID = $row['user_id'];
echo $fromID;

if it will throw no errors but still print no id, add this line

var_dump($row);

and post here it's output

not that you shouldn't use a user name but user id to address particular user.