Question

我有两个表可以应用更改。但我需要回应基于某个标准所做的更改。现在对于第一个表，所做的任何更改都会被回显，但是如果对第二个表（else）进行更改，我不确定如何回显更改。

if (isset($_POST['submit']))
{
    if (isset($_POST['ID'])) {
        $sql = "SHOW COLUMNS FROM Employees";
        $result = mysqli_query($con,$sql);
        while($row = mysqli_fetch_array($result)){
            $tempname = $row['Field'];

            // Changes Function
            $sqlCheck = "SELECT * FROM Employees WHERE ID='".$_GET["id"]."' AND (".$row['Field']." NOT LIKE '".$_POST[$tempname]."')";
            $result3 = mysqli_query($con,$sqlCheck);
            if (mysqli_num_rows($result3) > 0) {
                // output data of each row
                while($row3 = mysqli_fetch_array($result3)) {
                    $sql3 = "INSERT INTO `Changes` (`Table`, `ID`, `Attribute`, `DateChanged`, `HRUser`, `OldValue`, `NewValue`) VALUES ('Employees', '".$_GET["id"]."', '".$row["Field"]."', '".date("d/m/Y h:i:sa")."', '$login_session', '$row3[$tempname]', '$_POST[$tempname]')";
                    if (mysqli_query($con,$sql3) === TRUE) {
                    } else {
                        echo "Error: " . $sql3 . "<br>" . mysqli_error($con);
                    }
                }
            }
            //End Changes Function


             $sql2 = "UPDATE Employees SET ".$row['Field']."= '$_POST[$tempname]' WHERE ID='".$_GET["id"]."'";
            $result2 = mysqli_query($con,$sql2);
            if (mysqli_query($con,$sql2) === TRUE) {
            } else {
                echo "Error: " . $sql2 . "<br>" . mysqli_error($con);
                echo '<script>swal("Error", "Something went wrong '.mysqli_error($con).'", "error");</script>';
            }

        }




        echo '<script>swal("Success", "Changes have been saved", "success");</script>';
    } 
    //End If POST Submit True
    else
    {

        $sql = "SHOW COLUMNS FROM Candidates";
        $result = mysqli_query($con,$sql);
        while($row = mysqli_fetch_array($result)){
            $tempname = $row['Field'];


            // Changes Function
            $sqlCheck = "SELECT * FROM Candidates WHERE ID='".$_GET["cid"]."' AND (".$row['Field']." NOT LIKE '%".$_POST[$tempname]."%')";
            $result4 = mysqli_query($con,$sqlCheck);
            if (mysqli_num_rows($result4) > 0) {
                // output data of each row
                while($row4 = mysqli_fetch_array($result4)) {
                    $sql4 = "INSERT INTO `Changes` (`Table`, `ID`, `Attribute`, `DateChanged`, `HRUser`, `OldValue`, `NewValue`) VALUES ('Candidates', '".$_GET["cid"]."', '".$row["Field"]."', '".date("d/m/Y h:i:sa")."', '$login_session', '$row4[$tempname]', '$_POST[$tempname]')";
                    if (mysqli_query($con,$sql4) === TRUE) {
                    } else {
                        echo "Error: " . $sql4 . "<br>" . mysqli_error($con);
                    }
                }
            }

Answer 1

使用mysqli_affected_rows()。

mysqli_affected_rows返回先前MySQL操作中受影响的行数

因此，您可以判断是否已根据以前的数据进行了更改。

mysqli_affected_rows php.net reference

将结果导入文本字段

1 个答案: