我是一个PHP新手,我一整天都在努力解决这个问题,但无济于事,我提供的代码可以获得一个' id'从一个页面打开一个新页面,我使用if(isset)_GET方法来调用新页面中的id,同时想要连接两个表来从中选择数据并在同一页面上显示。我写了下面的代码
如果有人告诉我错误的地方以及如何做到这一点,我将不胜感激
<?php
session_start();
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN"
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html lang="en">
<body class="home">
<!-- Intro -->
<div class="container text-center">
</div>
<!-- /Intro-->
<?php
require_once 'includes/connect.php';
if(isset($_GET['id'])) {
$full_name = $_GET['id'];
$select_query = "SELECT quoteform.quote_id, quoteform.gender, quoteform.feedback_bad, quoteform.feedback_ok, quoteform.size, quoteform.accessory,quoteform.quantity, quoteform.full_name, quoteform.phone_number, quoteform.email, quoteform.location, quoteform.quote_date,quoteform.order_number, price_products.price_withfabric FROM quoteform, price_products WHERE qouteform.feedback_bad = price_products.product_name OR quoteform.feedback_ok = price_products.product_name AND full_name='$full_name' LIMIT 1";
$run_query = mysqli_query($con, $select_query);
?>
<!-- Highlights - jumbotron -order -->
<div class="jumbotron-order">
<div class="container">
<?php
if(mysqli_num_rows($run_query) > 0)
{
while($row = mysqli_fetch_array($run_query))
{
?>
<h3 class="text-center thin"> - Order Form - </h3></br>
<h4 class=" thin"> Hello <?php echo $row["full_name"];?></h4></br>
<h4 class=" thin"> Order No. <?php echo $row["order_name"];?> </h4>
<div class="row">
<div class="col-md-8 col-sm-8">
<div class=" panel">
<table class="table">
<tr>
<th scope="row"> Outfit</th>
<th><?php echo $row["feedback_bad"];?> </th>
</tr>
<tr>
<th scope="row"> Size</th>
<th><?php echo $row["size"];?></th>
</tr>
<tr>
<th scope="row"> Price </th>
<th> <?php echo $quantity; ?> X <?php echo $price_withfabric; ?> </th>
<th> = Amount</th>
</tr>
<tr>
<th scope="row"> Accessory <<- Add More Button ->> </th>
</tr>
<tr>
<td scope="row"> <?php echo $accessory; ?> </td>
<th> Qty X price = </th>
<th> = Amount</th>
</tr>
<tr>
<th scope="row"> Click to select Service Charge </th>
<th> Display Service Name </th>
<th> = Amount</th>
</tr>
<tr>
<th scope="row"> Click to select Delivery </th>
<th> Display Delivery </th>
<th> = Amount</th>
</tr>
<tr>
<th scope="row"> </th>
<th scope="row"> Total </th>
<th> Display Amount Here!!!</th>
</tr>
<tr>
<th scope="row"> Do you have a Coupon Code? </th>
<th scope="row"> coupon input space </th>
<th> The amount to be deducted from Total</th>
</tr>
<tr>
<th scope="row"> </th>
<th scope="row"> Amount to be Charged </th>
<th> Display Amount Here!!!</th>
</tr>
</table>
<div class="text-center"><button class="btn btn-success btn-lg">Preview and Pay</button> <button class="btn btn-danger btn-lg">Cancel Order</button></div>
</div>
</div>
</div>
<!-- /row -->
<?php } } ?>
</div>
</div>
<?php } ?>
答案 0 :(得分:-1)
我们可以使用一些代码,但看起来您正在执行查询,将该查询的结果存储在变量中,然后将该变量传递给mysqli_num_rows
。查询可能失败了。当mysqli_query()
失败时,它返回布尔值FALSE,而不是mysqli_result
的实例。输出查询,因为它发送到MySQL&amp;我打赌看起来有些不对劲。
答案 1 :(得分:-1)
更改此行:
if(mysqli_num_rows($run_query) > 0)
对此:
if(!empty($run_query) && mysqli_num_rows($run_query) > 0)
这样,如果$run_query
的失败查询值为false
,则不会打印数据。
现在你可能想要做一些事情 else 如果那个查询失败了,那么就像这样进行另一次检查:
if(empty($run_query)) {
// do something here if the query failed
}