表doens不显示更新的列(table doens't show the updated column)

在我的代码中，一旦更新按钮被点击，表格不会刷新并显示更新的列。

这是我的代码。

<?php if(isset($_POST['update'])){ $Project = $_POST['Project']; $No = $_POST['No']; $SubID = $_POST['SubID']; $RequestAmount = $_POST['RequestAmount']; $PaidAmount = $_POST['PaidAmount']; $AmountToPay = $_POST['AmountToPay']; $State = $_POST['State']; //UPDATE Query of SQL $sql = "UPDATE memo SET Project='$Project',No='$No',SubID='$SubID',RequestAmount='$RequestAmount',PaidAmount='$PaidAmount',AmountToPay='$AmountToPay',State='$State' WHERE No='$No' AND SubID='$SubID'" or die("Failed to query database" .mysqli_error()); $result = $link->query($sql); } ?>

in my code, the table doesn't refresh and show the updated column once the update button is clicked.

here is my code.

<?php if(isset($_POST['update'])){ $Project = $_POST['Project']; $No = $_POST['No']; $SubID = $_POST['SubID']; $RequestAmount = $_POST['RequestAmount']; $PaidAmount = $_POST['PaidAmount']; $AmountToPay = $_POST['AmountToPay']; $State = $_POST['State']; //UPDATE Query of SQL $sql = "UPDATE memo SET Project='$Project',No='$No',SubID='$SubID',RequestAmount='$RequestAmount',PaidAmount='$PaidAmount',AmountToPay='$AmountToPay',State='$State' WHERE No='$No' AND SubID='$SubID'" or die("Failed to query database" .mysqli_error()); $result = $link->query($sql); } ?>

最满意答案

这是因为您在加载页面后更新数据库。

你可以做这样的事情：

<?php if(isset($_POST['update'])) { $Project = $_POST['Project']; $No = $_POST['No']; $SubID = $_POST['SubID']; $RequestAmount = $_POST['RequestAmount']; $PaidAmount = $_POST['PaidAmount']; $AmountToPay = $_POST['AmountToPay']; $State = $_POST['State']; //UPDATE Query of SQL $sql = "UPDATE memo SET Project='$Project',No='$No',SubID='$SubID',RequestAmount='$RequestAmount',PaidAmount='$PaidAmount',AmountToPay='$AmountToPay',State='$State' WHERE No='$No' AND SubID='$SubID'" or die("Failed to query database" .mysqli_error()); $result = $link->query($sql); header("Refresh: 0;"); die(); } ?>

我添加了行

header("Refresh: 0;"); die();

它只是在更新后刷新页面。希望这有助于，如果没有，那么抱歉。

It is because you update database after loading the page.

You can do something like that:

<?php if(isset($_POST['update'])) { $Project = $_POST['Project']; $No = $_POST['No']; $SubID = $_POST['SubID']; $RequestAmount = $_POST['RequestAmount']; $PaidAmount = $_POST['PaidAmount']; $AmountToPay = $_POST['AmountToPay']; $State = $_POST['State']; //UPDATE Query of SQL $sql = "UPDATE memo SET Project='$Project',No='$No',SubID='$SubID',RequestAmount='$RequestAmount',PaidAmount='$PaidAmount',AmountToPay='$AmountToPay',State='$State' WHERE No='$No' AND SubID='$SubID'" or die("Failed to query database" .mysqli_error()); $result = $link->query($sql); header("Refresh: 0;"); die(); } ?>

I added lines

header("Refresh: 0;"); die();

It simply refreshes the page after update. Hope this helps, if not, then sorry.

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