mysqli_fetch_assocが動作するのはなぜですか？

Question

私は自分のデータベースにURLを入れた場合、URLに行き、どのHTTPコードが取得されているかを確認するコードを取得しますが、現在はHTTP-ライトの代わりにコード0。

$username = "root"; 
$password = "luca170385"; 
$hostname = "localhost"; 

$dbhandle = mysqli_connect($hostname, $username, $password) 
or file_put_contents($filename, date("Y-m-d H:i:s")."Unable to connect to database", FILE_APPEND); 
echo "Connect to MySQL<br>"; 

$selected = mysqli_select_db($dbhandle , "accounts") 
or file_put_contents($filename, date("Y-m-d H:i:s")."Could not select database<br>", FILE_APPEND); 
echo "Selected Database<br>"; 

$sql = "SELECT `address` FROM `url` WHERE `code` IS NULL"; 

if ($result = mysqli_query($dbhandle, $sql)){ 

    while ($row = mysqli_fetch_assoc($result));{ 
    printf ($row['address']); 
    $output = "{$row['address']}"; 
    echo "{$row['address']}"; 
    $handle = curl_init($sql); 
    curl_setopt($handle, CURLOPT_RETURNTRANSFER, TRUE); 

    $response = curl_exec($handle); 

    $httpCode = curl_getinfo($handle, CURLINFO_HTTP_CODE); 
    $update = "UPDATE url SET `code`= $httpCode WHERE `address` = $sql"; 
    echo "$update.<br>"; 
    echo "$handle.<br>"; 
    echo "$httpCode.<br>"; 

    mysqli_query($dbhandle, $update); 
    echo "{$row['address']}"; 

    curl_close($handle); 
    } 
}else { 
    echo 'Not working'; 
} 

Answer 1

のコメントをご確認ください： -

<?php 
error_reporting(E_ALL);// check all type of error 
ini_set('display_errors',1);// display those errors 
$username = "root"; 
$password = "luca170385"; 
$hostname = "localhost"; 

$dbhandle = mysqli_connect($hostname, $username, $password, "accounts"); 
/*or file_put_contents($filename, date("Y-m-d H:i:s")."Unable to connect to database", FILE_APPEND); 
echo "Connect to MySQL<br>";*/// from where this $filename come from? 

/*$selected = mysqli_select_db($dbhandle ,) 
or file_put_contents($filename, date("Y-m-d H:i:s")."Could not select database<br>", FILE_APPEND); 
echo "Selected Database<br>";*/ //not needed 

$sql = "SELECT `address` FROM `url` WHERE `code` IS NULL"; 

if ($result = mysqli_query($dbhandle, $sql)){ 

    while ($row = mysqli_fetch_assoc($result)){ // remove ; 
     printf ($row['address']); 
     $output = $row['address']; // remove {} and "" 
     echo $row['address']; // remove {} and "" 
     $handle = curl_init($sql); 
     curl_setopt($handle, CURLOPT_RETURNTRANSFER, TRUE); 

     $response = curl_exec($handle); 

     $httpCode = curl_getinfo($handle, CURLINFO_HTTP_CODE); 
     $update = "UPDATE url SET `code`= $httpCode WHERE `address` = $output"; // not $sql 
     echo "$update.<br>"; 
     echo "$handle.<br>"; 
     echo "$httpCode.<br>"; 
     mysqli_query($dbhandle, $update); 
     echo $row['address'];// remove {} and "" 
     curl_close($handle); 
    } 
}else { 
    echo 'Not working'; 
} 

Answer 2

;は終了するため、空です。

;を削除するだけで済みます。