Sam*_*iel 12 php mysqli prepared-statement
嘿伙计们,当我尝试运行此代码时,我收到了上述警告:
$mysqli=new mysqli("localhost", "***", "***","***") or die(mysql_error());
function checklogin($username, $password){
global $mysqli;
$result = $mysqli->prepare("SELECT * FROM users WHERE username = ?");
$result->bind_param("s", $username);
$result->execute();
if($result != false){
$dbArray=mysql_fetch_array($result);
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Cha*_*ndu 43
您正在代码中混合mysql和mysqli调用.使用mysqli_fetch_array而不是mysql_fetch_array.