小编use*_*919的帖子

我是PHP的新手,我复制了这段代码;

$sql = "SELECT userID, username, password, first_name FROM users WHERE email='$e' AND activated='1' LIMIT 1";
        $query = mysqli_query($connection, $sql);
        $row = mysqli_fetch_row($query);
        $db_id = $row[0];
        $db_username = $row[1];
        $db_pass_str = $row[2];
        $db_firstname = $row[3];

但它只是不起作用......这就是我的桌子的样子;

UserID -- username -- firstname -- lastname --  email  --------- password  

1   --    CJ_65 --    Craig -----  Johnson  --  CJ@email.com  -- passwordd 

有人可以帮帮我吗？

我正在尝试使用 php 发送多封电子邮件。但是每次我尝试发送电子邮件时，我都会收到“errorerrorerror”——每封电子邮件都有一个“错误”——就在表中。这是代码

$emailsql = "SELECT Username FROM Companyuserinfo WHERE Company_ID = '$cid'";
$emailquery = mysqli_query($connection, $emailsql);


while($emailrow = mysqli_fetch_array ($emailquery)){  
 $Usernamesend = $emailrow['Username'];

$sendsql = "SELECT * FROM users WHERE username = '$Usernamesend'";
$sendquery = mysqli_query($connection, $sendsql);

$sendrow = mysqli_fetch_array ($sendquery);
    $emailtosend = $sendrow['email'];

    $to="$emailtosend";
    $from = "info@site.org";
    $subject="TEST!";
    $message="HEY MY BROTHER!! I AM TESTING TdfdHIS BABY! WOOHOO!";
    $headers = "From: $from\n";
        $headers .= "MIME-Version: 1.0\n";
        $headers .= "Content-type: text/html; charset=iso-8859-1\n";
     mail($to, $subject, $message, $headers);

     if (!mail($to, $subject, $message, …

我试图尽快摆脱一个错误.我正在使用mysql_num_rows但它总是返回0.而我不知道是不是因为我的语法错误或者什么......你们可以帮帮我吗？这是代码;

<?php
include_once("checklogin.php");
$u = "";
if(isset($_GET["u"])){
    $u = preg_replace('#[^a-z0-9]#i', '', $_GET['u']);
} else {
    header("location: http://www.myswesite.com/login.php");
    exit();    
}
$sql = "SELECT * FROM users WHERE username='$u' AND activated='1' LIMIT 1";
$user_query = mysqli_query($connection, $sql);
$numrows = mysqli_num_rows($user_query);
if($numrows < 1){
    echo "User does not exist or is not yet activated, press back";
    exit(); 
}
?>