特定用戶會話的SQL UPDATE

Question

我可以基於會話ID向數據庫添加特定數據或從數據庫刪除特定數據，但是，當我嘗試為該特定用戶更新數據時，出現錯誤。 該重復的密鑰聲明無法正常工作，因為密鑰已設置為記錄ID與股票的代碼。 任何解決方案的想法都可以解決，因為現在當一個人嘗試兩次添加相同的股票時，他們被允許這樣做，因此這會在以后引發問題。 不知道我現在是否會使用該Update語句來進行正確的處理，而不是更新部分，因此，感謝您的反饋/幫助。 謝謝！

INSERT＆UPDATE的代碼：

// When the Buy button is pressed, specific action will be triggered according to the input given.
    if(isset($_POST['Buy']))
    { 
        // Checking whether first line is completely filled.
        if(empty($_POST['sym1']) or empty($_POST['pri1']) or empty($_POST['q1']))
        {
            ?><h2><center>To add values, please fill out at least the first row completely.</center></h2><?php
        // die();
    }
    // Loop through the form to allow for an appropriate db update.
    for($x=1;$x<=4;$x++)
    {
        $sym = [];
        $pri = [];
        $q = [];
        // If input provided is correct then update the db.
        if (!empty($_POST['sym'.$x]) and !empty($_POST['pri'.$x]) and !empty($_POST['q'.$x])) 
        {
            $sym[$x] = $_POST['sym'.$x];
            $pri[$x] = $_POST['pri'.$x];
            $q[$x] = $_POST['q'.$x];
            $memberid = $_SESSION['memberID'];
            $sql = "INSERT INTO portfolio2 
                (stocks_symbol, price, quantity, memberID)
                VALUES ('$sym[$x]', $pri[$x], $q[$x], $memberid)
                ON DUPLICATE KEY UPDATE
                price=$pri[$x], quantity=$q[$x]";

            // Check if values are added successfully and if so, then display a message to the user.
            if(mysqli_query($conn, $sql))
            {
                ?><h2><center><?php
                echo "Stocks added successfully!";
                ?></h2><center><?php
            }
            else
            {
                ?><h2><center><?php
                echo "Error- Stocks weren't added!". "<br>". $sql.
                "<br>". $conn->error;
                ?></h2><center><?php
            }
        }
    }
    mysqli_close($conn);
}
// UPDATE 
    elseif(isset($_POST['Update']))
{
    // Check to see whether the stock symbol has been provided
    if(empty($_POST['sym1']))
    {
        ?><h2><center>To update values, please enter the symbol of the stock to be updated.</center></h2><?php
        // die();
    }

    // Loop through the form to allow for an appropriate db update.
    for($x=1;$x<=4;$x++)
    {
        $sym = [];
        $pri = [];
        $q = [];

        // When all three values to be updated are given and are correct, update the db accordingly.
        if (!empty($_POST['sym'.$x]) and !empty($_POST['pri'.$x]) and !empty($_POST['q'.$x])) 
        {
            $sym[$x] = $_POST['sym'.$x];
            $pri[$x] = $_POST['pri'.$x];
            $q[$x] = $_POST['q'.$x];
            $memberid = $_SESSION['memberID'];
            $sql = "UPDATE portfolio2 SET price=$pri[$x] and quantity=$q[$x] WHERE stocks_symbol='$sym[$x]' and memberid=$memberid";

            // Check to see whether the values are updated successfully and if so, then display a message to the user.
            if(mysqli_query($conn, $sql))
            {
                ?><h2><center><?php
                echo "Stocks updated successfully!";
                ?></h2><center><?php
            }
            else
            {
                ?><h2><center><?php
                echo "Error- Couldn't update stocks from the table". "<br>". $sql.
                "<br>". $conn->error;
                ?></h2><center><?php
            }
        }   
    }
    mysqli_close($conn);
}

表結構：Portfolio2

CREATE TABLE `portfolio2` (
 `stockID` int(11) NOT NULL AUTO_INCREMENT,
 `stocks_symbol` varchar(30) NOT NULL,
 `price` decimal(30,2) DEFAULT NULL,
 `quantity` int(30) DEFAULT NULL,
 `memberid` int(11) NOT NULL,
 PRIMARY KEY (`stockID`)
) ENGINE=InnoDB AUTO_INCREMENT=10 DEFAULT CHARSET=latin1

Answer 1

如果要防止用戶兩次添加相同的庫存，可以通過創建UNIQUE索引來做到這一點：

ALTER TABLE `portfolio2` ADD UNIQUE `unique_idx`(`memberid`, `stocks_symbol`);