[英]PHP Mysqli code not displaying query results
我想從查詢結果中輸出數據。 該查詢在另一個PHP頁面中使用了print_r(json_encode($regions))
,但未輸出任何內容。 我在php中沒有錯誤,我在mysqli代碼中做錯了什么嗎?
//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
}
?>
在您的SQL中,where子句引用region_id,在這種情況下,它是在兩個表(sales和region)中定義的,如果您需要這兩個表,則需要限定要使用哪個表中的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));
但是由於您不使用結果中來自區域的任何列,因此可以刪除連接...
$sql=mysqli_query($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));
同樣如Barmar所說,請刪除查詢的第二次執行,否則可能會失敗並停止腳本。
同樣在檢查$_GET['region_id']
,這應該是更多的情況,如果未設置則什么也不做。 僅將其設置為false會引起更多問題。
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