将单行传输到数组并将其回显到各自的文本框中

Question

I'm doing a search function where the user need to input ID number in order to search a person's info. I did search the web on how to do this, but I got stuck when transferring the data back. In my case, I want to transfer every data into their respective textbox. For example :

fullname => fullname textbox
gender => gender textbox

So how do I convert the array into individual php variable and echo it in the textbox? Or there might be any better way? Here is the function on what I've done so far :

function userRetrieve ($cardid) {
$cardid = mysql_real_escape_string($cardid);
$sql = mysql_query("SELECT * FROM `users` WHERE `card_id` = '$cardid'");
$res = mysql_query($sql) or trigger_error(mysql_error().$sql);
$row = mysql_fetch_assoc($res);
    if ($row){
        //What should I do here? 
    }
}

Any help would be appreciated. Thank you.

Answer 1

Since you are querying only one user do

if ($row){
   $user = new user($row['id'], $row['name'], $row['birthdate'], ...);//if you have a class user,which is adivsable
   //if you do not just save the row or use it like
   echo "User name = " . $row['name'];
}

When you query multiple users you can do

$users = array();
while($row = mysql_fetch_assoc($sql)){//do not call mysql_query twice
   $users[] = new user($row['id'], $row['name'], $row['birthdate'], ...); //if you have a class user,which is adivsable
   //if you do not, just save the whole row
   $users[] = $row;
}

If you use the second option you can access the users like so:

foreach($users as $user){
   echo $user['name'];
}

the values ['here'] should match your column names in the database.

To get the values from the function do

return $row;

---

Your function should look like:

function userRetrieve ($cardid) {
  $cardid = mysql_real_escape_string($cardid);
  $sql = mysql_query("SELECT * FROM `users` WHERE `card_id` = '$cardid'") or trigger_error(mysql_error().$sql);
  $row = mysql_fetch_assoc($sql);
  if ($row){
     return $row;
  }
  return false;
}

---

Now you can call your function like so:

$current_user = userRetrieve(14); //id example, should be a variable
echo $current_user['name'];