Question

我的更新脚本有问题，我的脚本有问题吗？每当我运行它时，它只是显示回声“ERROR”。这是我的更新脚本：

<?PHP 
    $con = mysqli_connect("localhost","root","","testuser");
    $choosed2 = $_POST['choosed2'];
    $creditnew = $_POST['creditnew'];
    $debitnew = $_POST['debitnew'];

    $newid = $choosed2 - 1;
    $bal = "SELECT balance FROM `bal` WHERE id=$newid;";
    $chek = mysqli_query($con,$bal);
    $row = mysqli_fetch_assoc($chek);
    $old_balance = $row['balance'];
    if ($old_balance == NULL){
        $old_balance = 0;
    }
    $balancenew = $old_balance - $creditnew + $debitnew;
    $up = "UPDATE `bal` SET debit='$debitnew' credit='$creditnew' balance='$balancenew' WHERE id='$choosed2'";
    if(mysqli_query($con,$up))
        {
            echo "<div class='form' style='text-align:center; float:center;'><div style='font-weight:bold; font-size:20px; margin:10px'>UPDATE SUCCESS</div><br/><a href='index.php' style='margin:5px;' class='tombol'><< BACK</a><a href='Input.php' style='margin:5px;' class='tombol'>INPUT AGAIN</a></div>";
        }else{
            echo "<div class='form'><h3>ERROR</h3><br/><a href='balance.php'>Input again</a>";
        }

?>

info：choosed2是行ID。

任何人都可以帮助我吗？感谢。

Answer 1

您在SQL语句中错过了逗号（,）。您的SQL查询将是

$up = "UPDATE `bal` SET debit='$debitnew', credit='$creditnew', balance='$balancenew' WHERE id='$choosed2'";

PHP - 更新脚本不起作用

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