我正在尝试检查现有字段是否已更改并识别它,以便稍后将其添加到更改表中。有关如何这样做的任何想法?
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>';
}
答案 0 :(得分:1)
首先,我认为你错过了".$var."
这一行:
$sql2 = "UPDATE Employees SET ".$row['Field']."= '$_POST[$tempname]' WHERE AFNumber='".$_GET["af"]."'";
它应该是这样的:
$sql2 = "UPDATE Employees SET ".$row['Field']."= '".$_POST[$tempname]."' WHERE AFNumber='".$_GET["af"]."'";
您可以先执行选择查询,以根据要更新的数据进行区分
// get the rows that will be changed
$sqlOldData = "SELECT * FROM Employees WHERE AFNumber='".$_GET["af"]."' AND (".$row['Field']." NOT LIKE '".$_POST[$tempname]."')";
然后更新表格。
问:但有一个问题,至于将它集成到代码中,请给我任何帮助,我只是从这个领域开始:
$sql3 = "INSERT INTO Changes (Table, AFNumber, Attribute,DateChanged,HRUser,OldValue,NewValue) VALUES ('Employees', '".$_GET["af"]."', '".$row["Field"]."', '".date('dd/m/Y HH:mm:ss')."', '$login_session', '', '$_POST[$tempname]')";
注意:首先,您再次错过了一些字符串突破:
'$login_session'
- &gt; '".$login_session."'
'$_POST[$tempname]'
- &gt; '".$_POST[$tempname]."'
所以你得到:
$sql3 = "INSERT INTO Changes (Table, AFNumber, Attribute,DateChanged,HRUser,OldValue,NewValue) VALUES ('Employees', '".$_GET["af"]."', '".$row["Field"]."', '".date('dd/m/Y HH:mm:ss')."', '".$login_session."', '', '".$_POST[$tempname]."')";
A:采用$resultOldData
是$sqlOldData
这应该有效:
while($rowOldData = mysqli_fetch_array($result))
{
$sql3 = "INSERT INTO Changes (Table, AFNumber, Attribute,DateChanged,HRUser,OldValue,NewValue) VALUES ('Employees', '".$_GET["af"]."', '".$row["Field"]."', '".date('dd/m/Y HH:mm:ss')."', '".$login_session."', '".$rowOldData[$row['Field']]."', '".$_POST[$tempname]."')";
mysqli_query($con,$sql3);
}