Question

我创建了一个带有用户输入的html表单，并且应该返回相关的发票。我没有收到任何错误，我已经检查了我的HTML代码，以确保isset的条件是正确的。我还在PHPAdmin中测试了查询（它返回了我要求的信息）。但是，发票不会打印到屏幕上，也不会出现错误或PHP代码。我在屏幕上唯一得到的是标题（html未显示）和超链接（html未显示）。下面是相关的PHP代码部分。提前谢谢。

if( isset($_POST["submit"] )) {
if (!($stmt =$mysqli->prepare("SELECT DISTINCT INVOICE.date, CUSTOMER.Name, CUSTOMER.Lname, CUSTOMER.Phone, CUSTOMER.Address, CUSTOMER.email, INVOICE.Invoice_num, INVOICE.Itm_num, INVOICE.Itm_Price, INVOICE.Discount, INVOICE.Total 
FROM CUSTOMER, INVOICE
      WHERE CUSTOMER.Phone = INVOICE.Cust_phone AND CUSTOMER.Lname = INVOICE.Cust_lname AND
(CUSTOMER.Name = ? OR  CUSTOMER.Lname = ? OR INVOICE.date = ? OR INVOICE.Invoice_num = ? OR INVOICE.Itm_num = ?)
GROUP BY date;"))) {                                
    print "Prepare failed: (" . $mysqli->errno . ") " . $mysqli->error;
          }

if (!$stmt->bind_param("sssis",$_POST['fname'], $_POST['lname'], $_POST['date'], $_POST['invoice_num'], $_POST['item_num'])) {
    print "Binding parameters failed: (" . $stmt->errno . ") " . $stmt->error;
}

if (!$stmt->execute()) {
    print "Execute failed: (" . $stmt->errno . ") " . $stmt->error;
    }

$stmt->store_result();

if (!$stmt->bind_result($date, $Fname, $Lname, $Phone, $Address, $Email, $Invoice_num, $Item_num, $Itm_Price, $Discount,$Total)) {      
    print "Binding output parameters failed: (" . $stmt->errno . ") " . $stmt->error;
}

if ($stmt->num_rows == 0){  
    print "No results were found for the following search <p>".$_POST['fname'].$_POST['lname'].$_POST['date'].$_POST['invoice_num'].$_POST['item_num']."</p>";
}

else {
    print "<table border=2 cellpadding=4>

        <tr bgcolor=white>
        <th>Date</th>
        <th>First Name</th>         
        <th>Last Name</th>
        <th>Phone</th>          
        <th>Address</th>
        <th>Email</th>          
        <th>Invoice #</th>
        <th>Item #</th>
        <th>Price</th>
        <th>Discount (if applicable)</th>
        <th>Total</th>          
        </tr>";

    while ($stmt->fetch()) {                    
                    print "<tr><td>".$date."</td>
                    <td>".$Fname."</td>
                    <td>".$Lname."</td>
                    <td>".$Phone."</td> 
                    <td>".$Address."</td>
                    <td>".$Email."</td>
        <td>".$Invoice_num."</td>
                    <td>".$Itm_num."</td>
                    <td>".$Itm_Price."</td>
                    <td>".$Discount."</td>
                    <td>".$Total."</td></tr>";
    }

    print "</table>";
}
$stmt->free_result();    
}

任何建议都将不胜感激。

Answer 1

如果没有从您的代码打印，我猜想，根据isset，$_POST["submit"]未设置或者为空。

要验证这一点，请在if (isset(...))之后立即发出回音。如果您没有看到回声，请检查您的$_POST变量。它可能拼写错误，甚至作为GET请求而不是POST发送。

没有错误，但也没有打印

1 个答案: