PHP Mysqli代码未显示查询结果

Question

我想从查询结果中输出数据。 该查询在另一个PHP页面中使用了print_r(json_encode($regions)) ，但未输出任何内容。 我在php中没有错误，我在mysqli代码中做错了什么吗？

//connecting to database
    <?php
    require_once('DbConnection.php');   
    //querying the database 
    $region_id = isset( $_GET['region_id'] )? $_GET['region_id']: false;
    $sql=mysqli_query($connection,"SELECT sales.region_id, sales.image_name,                 sales.price, sales.location, sales.Terms, sales.Contacts
    FROM sales INNER JOIN region ON sales.region_id=region.region_id  where region_id = $region_id") or die(mysqli_error($connection));
    $result = mysqli_query($connection,"SELECT sales.region_id, sales.image_name, sales.price, sales.location, sales.Terms, sales.Contacts FROM sales INNER JOIN region ON sales.region_id=region.region_id  where region_id = $region_id");             
    while ($row = mysql_fetch_assoc($sql)) {
             ?>
    <div class="col-md-4">
    <div class="thumbnail">
    <a href="<?php echo "http://" . $_SERVER['SERVER_NAME'] ?>/photo/imageuploads/<?php echo $row["image_name"]; ?>">
    <img src="<?php echo "http://" . $_SERVER['SERVER_NAME'] ?>/photo/imageuploads/<?php echo $row["image_name"]; ?>" alt="Lights" style="width:100%">
    <div class="caption">
    Image Name:<?php echo $row["image_name"]; ?>
    Price:<?php echo $row["price"]; ?>
    Location`enter code here`:<?php echo $row["location"]; ?>
    Terms:<?php echo $row["Terms"]; ?>
    Contacts:<?php echo $row["Contacts"]; ?>
    </div>
     </a>
    </div>
    </div>
    <?php
    }
    ?>

Answer 1

在您的SQL中，where子句引用region_id，在这种情况下，它是在两个表（sales和region）中定义的，如果您需要这两个表，则需要限定要使用哪个表中的region_id

$sql=mysqli_query($connection,"SELECT sales.region_id, sales.image_name, 
                                sales.price, sales.location, sales.Terms, sales.Contacts
                   FROM sales     
                   INNER JOIN region ON sales.region_id=region.region_id  
                   where region.region_id = $region_id") or die(mysqli_error($connection));

但是由于您不使用结果中来自区域的任何列，因此可以删除连接...

$sql=mysqli_query($connection,"SELECT sales.region_id, sales.image_name, 
                                sales.price, sales.location, sales.Terms, sales.Contacts
                   FROM sales   
                   where region_id = $region_id") or die(mysqli_error($connection));

同样如Barmar所说，请删除查询的第二次执行，否则可能会失败并停止脚本。

同样在检查$_GET['region_id'] ，这应该是更多的情况，如果未设置则什么也不做。 仅将其设置为false会引起更多问题。