php - set int from sql query -

i trying set int value sql query. in ios app can assign int value photo id, store , retrieve fine. problem comes if want overwrite photo new jpg still using existing idphoto , therefore same filename, e.g. 1.jpg. first check whether user exists. if update photo (this need set idphoto) otherwise create photo new id (works fine).

function uploaddetails($name, $location, $photodata, $idphoto) {      $uploads = query("select name, idphoto users name = '%s'  limit 1",$name);       if (count($uploads['result'])>0) {          $result = query("update users set name='$name', location='$location', idphoto='$idphoto' name = '%s'", $name);      //need define idphoto users table          if (move_uploaded_file($photodata['tmp_name'], "icons/".$idphoto.".jpg")) {              thumb("icons/".$idphoto.".jpg", 180);                      //i can print out confirmation iphone app                     print json_encode(array('$idphoto'=>$idphoto));          } else {                     //print out error message iphone app                     errorjson('upload on server problem');         };      }     else {         if ($photodata['error']==0) {           $result = query("insert users(name, location) values('%s','%s')", $name, $location);               if (!$result['error']) {                  // fetch active connection database (it's initialized automatically in lib.php)                 global $link;                  // last automatically generated id in table                 $idphoto = mysqli_insert_id($link);                  if (move_uploaded_file($photodata['tmp_name'], "icons/".$idphoto.".jpg")) {                      thumb("icons/".$idphoto.".jpg", 180);                      print json_encode(array('successful'=>1));                 } else {                      errorjson('upload on server problem');                 };              } else {                 errorjson('upload database problem.'.$result['error']);             }           }     } } 

so problem lies in first part of code need update photo still use same idphoto

---

edit

the piece of code needed follows:

$getid = mysqli_fetch_assoc(mysqli_query($link, "select idphoto users name = '$name'")); $idphoto = $getid['idphoto']; 

although got answer myself, appreciate feedback of how phrase questions better in future. , writing out full code helped me @ bigger picture , see going wrong.

your question extremely vague i'm going take following assumptions:

1. your query ever return single record
2. you using mysqli

once have run query , stored results $result can use following code idphoto:

//store row results variable $row = $result->fetch_assoc();  //store idphoto variable $idphoto = $row['idphoto']; 

note: selecting name database when have name since you're using fetch record