从数据库提取数据到文本区域

4

我正在将当前数据与更新后的数据进行比较,以检查信息是否发生更改,并将更改添加到一个名为"changes"的新表中:

if (isset($_POST['submit'])) 
{
    $sql = "SHOW COLUMNS FROM Employees";
    $result = mysqli_query($con,$sql);
     while($row = mysqli_fetch_array($result)){
                $tempname = $row['Field'];
                $sql2 = "UPDATE Employees SET ".$row['Field']."= '$_POST[$tempname]' WHERE AFNumber='".$_GET["af"]."'";
                $result2 = mysqli_query($con,$sql2);
                if ($con->query($sql2) === TRUE) {
                } else {
                    echo "Error: " . $sql2 . "<br>" . $con->error;
                    echo '<script>swal("Error", "Something went wrong '.$con->error.'", "error");</script>';
                }

    $sqlOldData = "SELECT * FROM Employees WHERE AFNumber='".$_GET["af"]."' AND (".$row['Field']." NOT LIKE '".$_POST[$tempname]."')";
    $result3 = $con->query($sqlOldData);
    if ($result3->num_rows > 0) {

    while($row3 = $result3->fetch_assoc()) {
        $sql3 = "INSERT INTO Changes (Table, AFNumber, Attribute,DateChanged,HRUser,OldValue,NewValue)
VALUES ('Employees', '".$_GET["af"]."', '".$row["Field"]."', '".date('dd/m/Y HH:mm:ss')."', '$login_session', '.$row3[0]', '$_POST[$tempname]')";
        if ($con->query($sql3) === TRUE) {
        echo "New record created successfully";
        } else {
            echo "Error: " . $sql3 . "<br>" . $con->error;
        }
    }
    } else {
        echo "0 results";
    }

}

现在我想提取有关更改的信息,例如日期、用户...并以以下形式放入textarea标签中:
<textarea name="changes" rows="50" cols="59" disabled>
  12/07/2015 - User:"Mike"  Changed:"Actual Location" From: "blabla" to "bla"
</textarea>

但我不确定如何做到这一点,有什么帮助吗...


@gabe3886 说实话,我刚开始学习 PHP-SQL 连接,我相信我应该使用选择数据库代码,但我不太确定如何在文本区域中使用它。$sql = "SELECT id, firstname, lastname FROM MyGuests"; $result = $conn->query($sql);if ($result->num_rows > 0) { // 输出每一行的数据 while($row = $result->fetch_assoc()) { echo "id: " . $row["id"]. " - Name: " . $row["firstname"]. " " . $row["lastname"]. "<br>"; } } else { echo "0 results"; } $conn->close(); ?> - mikeb
2个回答

3

如果您不了解数据(例如AFNumber是什么),我建议简单地从Changes表中查询所有内容,并以所需形式显示:

$changes = $con->query("SELECT * FROM Changes WHERE Table = 'Employees'");
if ($changes->num_rows > 0) {
    echo '<textarea name="changes" rows="50" cols="59" disabled>' . "\n";
    while ($row = $changes->fetch_assoc()) {
        echo sprintf('%s - User:%s  Chnaged:"%s" From: "%s" to "%s"',
            $row['DateChanged'], $row['HRUser'], $row['Attribute'],
            $row['OldValue'], $row['NewValue']) . "\n";
    }
    echo "</textarea>";
}

为了显示数据,在这里我简单地使用了echo,但是使用某些模板系统对解决方案的核心概念并没有太大影响。

亲爱的,我遇到了一个问题:在C:\xampp\htdocs\p\login\data\employee.php文件的第127行,出现了“Parse error: syntax error, unexpected 'employees' (T_STRING), expecting ',' or ';'” 的错误提示。 - mikeb
$changes = $con->query("SELECT * FROM Changes WHERE Table = 'employees'");$changes = $con->query("SELECT * FROM Changes WHERE Table = 'employees'"); - mikeb
这行代码在语法上看起来没有问题,你还是收到错误信息吗? - aorcsik
你在第127行之前有没有漏打一个单引号?你用的是什么代码编辑器?它支持语法高亮吗?高亮显示看起来正常吗? - aorcsik
让我们在聊天中继续这个讨论。点击此处进入聊天室 - mikeb

0

从您的评论中,您有以下代码:

$sql = "SELECT id, firstname, lastname FROM MyGuests"; 
$result = $conn->query($sql); 
if ($result->num_rows > 0) 
{ 
// output data of each row 
    while($row = $result->fetch_assoc()) 
    { 
        echo "id: " . $row["id"]. " - Name: " . $row["firstname"]. " " . $row["lastname"]. "<br>"; 
    } 
} 
else 
{ 
    echo "0 results"; 
} 
$conn->close();

如果您想以您目前的方式将信息放入文本区域,请使用以下代码:

echo '<textarea name="changes" rows="50" cols="59" disabled>id:' . $row['id' . ' - Name: ' . $row['firstname'] . ' ' .  $row['lastname'] . '</textarea>';

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