将更新的数据插入另一个表

时间:2019-01-11 09:52:11

标签: php mysqli

我的更新脚本运行良好。但是我想做一个日志文件。我想要的是,当我更新数据时,它也会与 Table1 $ orderId 一起插入另一个

我的代码在这里运行良好。请帮助我如何从同一表格将此数据插入另一个表。期待专家的帮助。我是编码的初学者。

这是表单代码HTML

<div class="form-group">
    <label for="due" class="col-sm-3 control- 
    label">Due Amount</label>
    <div class="col-sm-9">
        <input type="text" class="form-control" id="due" name="due" disabled="true" />
    </div>
</div>
<!--/form-group-->
<div class="form-group">
    <label for="payAmount" class="col-sm-3 
    control-label">Pay Amount</label>
    <div class="col-sm-9">
        <input type="text" class="form-control" id="payAmount" name="payAmount" />
    </div>
</div>
<!--/form-group-->
<div class="form-group">
    <label for="clientContact" class="col-sm-3 
    control-label">Payment Type</label>
    <div class="col-sm-9">
        <select class="form-control" name="paymentType" id="paymentType">
            <option value="">~~SELECT~~</option>
            <option value="1">Cheque</option>
            <option value="2">Cash</option>
            <option value="3">Credit Card</option>
        </select>
    </div>
</div>
<!--/form-group-->
<div class="form-group">
    <label for="clientContact" class="col-sm-3 
    control-label">Payment Status</label>
    <div class="col-sm-9">
        <select class="form-control" name="paymentStatus" id="paymentStatus">
            <option value="">~~SELECT~~</option>
            <option value="1">Full Payment</option>
            <option value="2">Advance Payment</option>
            <option value="3">Credit Payment</option>
        </select>
    </div>
</div>
<!--/form-group-->
</div>



 <?php
 require_once 'core.php';
 $valid['success'] = array('success' => 
 false, 'messages' => array());
 if($_POST) {	
 $orderId 					= 
 $_POST['orderId'];
 $payAmount 				= 
 $_POST['payAmount']; 
 $paymentType 			= 
 $_POST['paymentType'];
 $paymentStatus 		= 
 $_POST['paymentStatus'];  
 $paidAmount        = $_POST['paidAmount'];
 $grandTotal        = $_POST['grandTotal'];
 $updatePaidAmount = $payAmount + 
 $paidAmount;
 $updateDue = $grandTotal - 
 $updatePaidAmount;
 $sql = "UPDATE orders SET paid =
'$updatePaidAmount', due = '$updateDue', 
 payment_type = '$paymentType', 
 payment_status = '$paymentStatus' WHERE 
 order_id = {$orderId}";
 if($connect->query($sql) === TRUE) {
 $valid['success'] = true;
 $valid['messages'] = "Successfully 
 Update";	
 } else {
 $valid['success'] = false;
 $valid['messages'] = "Error while 
 updating product info";
 }
 $connect->close();
 echo json_encode($valid);
 } // /if $_POST

0 个答案:

没有答案