Question

我不明白为什么第二个mysql外观不起作用。从我所知道的$sqlb = "SELECT * FROM emp_list WHERE password = '$pw' AND username = '$un' LIMIT 1 ";在第二行的位置，$ pw或$ un中没有值，实际上我无法回应任何

$un = $_POST['username'];
$pw = $_POST['password'];
$store = $_POST['store'];   
$pw = md5($pw);
$ipaddress = $ipaddress;

从一开始或$ nip从第一个循环我做错了什么？

     <?php     

    $ipaddress = '';
        if (isset($_SERVER['HTTP_CLIENT_IP']))
            $ipaddress = $_SERVER['HTTP_CLIENT_IP'];
        else if(isset($_SERVER['HTTP_X_FORWARDED_FOR']))
            $ipaddress = $_SERVER['HTTP_X_FORWARDED_FOR'];
        else if(isset($_SERVER['HTTP_X_FORWARDED']))
            $ipaddress = $_SERVER['HTTP_X_FORWARDED'];
        else if(isset($_SERVER['HTTP_FORWARDED_FOR']))
            $ipaddress = $_SERVER['HTTP_FORWARDED_FOR'];
        else if(isset($_SERVER['HTTP_FORWARDED']))
            $ipaddress = $_SERVER['HTTP_FORWARDED'];
        else if(isset($_SERVER['REMOTE_ADDR']))
            $ipaddress = $_SERVER['REMOTE_ADDR'];
        else {
            $ipaddress = 'UNKNOWN';

 }     
    $servername = "localhost";
    $username = "******";
    $password = "**********";
    $dbname = "************";


// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
$connb = new mysqli($servername, $username, $password, $dbname, true);


// Check connection
if ($conn->connect_error) 
{
    die("Connection failed: " . $conn->connect_error);
} 


$un = $_POST['username'];
$pw = $_POST['password'];
$store = $_POST['store'];   
$pw = md5($pw);
$ipaddress = $ipaddress;

$sql = "SELECT * FROM acclst  WHERE s = '$ipaddress' LIMIT 1 ";
$result = $conn->query($sql);

if ($result->num_rows > 0) 
{
   // There is an ipaddress and it is approved
    while($row = $result->fetch_assoc()) 
{

        echo 'welcome ';
        echo $nip = $row['s'];   


    if ($conn->connect_error) 
{
    die("Connection failed: " . $conn->connect_error);
}   

// this does not echo       
echo $store;

$sqlb = "SELECT * FROM emp_list  WHERE `username` = '$un' AND password = '$pw' LIMIT 1 ";
$resultb = $connb->query($sqlb);

if ($result->num_rows > 0) {
   // There is an ipaddress and it is approved
    while($rowb = $result->fetch_assoc()) {
// none of this echo's

        echo $pw;
        echo 'welcome ';
        echo $rowb['namel'];        


         }
      } 
    }
} 


else 
{
    // There is no ip address and you do not have access.
    echo "0 results";
}

?>

Answer 1

这是我发现的。现在它工作正常。

// Create connection
$conn = new mysqli($servername, $username, $dbpassword, $dbname);


// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 


$un = $_POST['username'];
$pw = $_POST['password'];
$store = $_POST['store'];   
//  $pw = password_hash('donavons', PASSWORD_DEFAULT);


$ipaddress = $ipaddress;

$sql = "SELECT * FROM acclst  WHERE s = '$ipaddress' LIMIT 1 ";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
   // There is an ipaddress and it is approved
    while($row = $result->fetch_assoc()) {

        echo '<pre>IP Address Verified.' .  $row['s'] . '</pre>';   


$sqlb = "SELECT * FROM emp_list WHERE username = '$un' LIMIT 1 ";
$result = $conn->query($sqlb);

if ($result->num_rows > 0) {
   // There is an ipaddress and it is approved
    while($rowb = $result->fetch_assoc()) {

$hash = $rowb['password'];

if (password_verify($pw, $hash))
{

Answer 2

不确定我是否理解你的问题，但尝试这样的事情：

$sqlb = "SELECT * FROM emp_list  WHERE password = " . $pw . " AND username = " . $un . " LIMIT 1 ";

$ var无法正常运行php mysql脚本

2 个答案: