2017-06-03 44 views
1
利率

請任何人都可以幫我解決問題,是新的PHP開發領域如何存儲日常基礎到數據庫

正在開發的投資興趣門戶網站,我越來越問題存放利率每日以進入數據庫,實際的問題是存儲興趣與用戶的詳細信息,我使用cron作業運行功能,每24小時它的工作完美,但當我登錄爲用戶它的存儲該用戶的利率值只,其他用戶不存儲,我想我正在使用會話用戶登錄的詳細信息可能是這個問題,請任何人都可以幫助我存儲每個用戶的興趣值與用戶的詳細信息

session_start(); 
    $_SESSION['sess_user']; 
    $conn = mysqli_connect("localhost", "root", "", "bitmining"); 

    $investedusername = $_SESSION['sess_user']; 
    $sql3="SELECT sum(inv_amount) as total FROM buyinvest WHERE invested_username='$investedusername'";   

    $result = mysqli_query($conn, $sql3);   
    while($row = mysqli_fetch_assoc($result)){ 

     $investment = $row['total']; 
     $days = 1; 
     $interest_rate = (350/100/365) * $days ; 
     $finalvalue = $interest_rate * $days; 

    } 

    $date = "SELECT invest_timestamp FROM buyinvest ORDER BY invest_timestamp DESC LIMIT 1"; 
    $result = mysqli_query($conn, $date); 
    $d_test = mysqli_fetch_assoc($result); 

    $d_test2 = $d_test["invest_timestamp"]; 

    $timestamp = time(); 

    $investeduser = $_SESSION['sess_user']; 
    $interestamount = $finalvalue; 
    $date_interest = date("y-m-d h:i:s"); 

    if ($d_test2 + 86400 == $timestamp){ 
     if ($conn){ 
       $sql3 = "INSERT INTO user_investments(invested_user, interest_amount, date_interest) VALUES('$investeduser','$interestamount','$date_interest')"; 
       if (mysqli_query($conn, $sql3)) 
       { 
        echo "Updated!"; 
       } 
       else 
       { 
        echo "There is an error."; 
       } 
      } 
     } 
     else { 
     echo "Not yet!"; 
     } 

請人幫我解決上面的代碼

+0

只有當您希望所有做你只是存儲爲單個用戶數據選擇的所有用戶,並運行上面的代碼在循環中 –

+0

請給我示例代碼 – maffidesigners

+0

只需獲取所有用戶並在循環中運行查詢 –

回答

0

像這樣的事情

$sql = "SELECT * from table_name"; 
$result = $conn->query($sql); 
if($result->num_rows > 0){ 

while ($row = $result->fetch_assoc()) { 

$sql3="SELECT sum(inv_amount) as total FROM buyinvest WHERE invested_username='".$row['username']"'"; 
// your code 
} 
}