問題修飾

Question

這裏我的代碼 -

if($pin == '105') { 
$sqltree = "INSERT INTO tbltree (`userId`, `level`, `superId`, `rootId`, `childcount`) 
          VALUES ('$child1', '1', '$newid', '$myroot', '0');"; 
    mysql_query($sqltree); 
    update_level($newid); 
} 
function update_level() 
    { 
     //for 1st level 
      $newid = $_SESSION['newid']; 
      //getting senior's level 1 and to increase by 1 
      $sqlgetlevel = "SELECT superId,level1 FROM tbltree WHERE userID='$newid'"; 
      echo "<br>test:".$sqlgetlevel; 
      $result = mysql_query($sqlgetlevel,$link)or die(mysql_error()); //line 340 
      $row = mysql_fetch_array($result, MYSQL_ASSOC); 
      $level1 = $row["level1"]; 
      $level1 = $level1 + 1; 

      //update increased level 
      $sqlupdate = "UPDATE tbltree SET level1='$level1' WHERE userId='$newid';"; 
      mysql_query($sqlupdate,$link)or die(mysql_error()); 

      //change superId for new level 
      $superid = $row["superId"]; 
} 

錯誤 -test:SELECT superId,level1 FROM tbltree WHERE userID='29277640'
Warning: mysql_query() expects parameter 2 to be resource, null given in C:\xampp\htdocs\303\levelupdate.php on line 340

Answer 1

$鏈接是你的函數裏面未知的，使其成爲一個全局變量或只是不使用它。

$結果= mysql_query（$ sqlgetlevel）或死亡（mysql_error（））; // line 340