在PhP中顯示多行

Question

我能夠從數據庫中檢索並顯示一行數據，但是我想要檢索多行數據。我所擁有的代碼已經搞砸了，但我認爲我只是在正確的領域進行這項工作。

我的代碼：

$query = "SELECT * FROM hotel"; 
$results = $conn->query($query); 
    while($hotel = $results->fetch()) { echo "<br> id: ". $hotel['hotel_id']. " - Name: ". $hotel['name']. " - Address: " . $hotel['address'] . " - Postcode: " . 
$hotel['postcode']. " - Town: " . $hotel['town']. " - Description: ". $hotel['description']. " - rating: ". $hotel['rating']. " - image: " . <img src='". $hotel['image']. "'>""; 
    } 
$conn=NULL; 
?> 

Answer 1

我看到你正在使用PDO。你可以簡單地遍歷這樣的結果。

while($results_disp = $results->fetch(PDO::FETCH_OBJ)) 
{ 
//access rows here in this manner 
echo "Hotel Id :- " . $results_disp->id; 
} 

Answer 2

while($hotel = $results->fetch()) { 
    echo "<br> id: ". $hotel['id']. " - Name: ". $hotel['name']. " - Address: " . $hotel['address'] . " - Postcode: " . $hotel['postcode']. 
    " - Town: " . $hotel['town']. " - Description: ". $hotel['description']. " - rating: ". $hotel['rating']. " - image: " . $hotel['image']; 
} 

這種替換while循環。

Answer 3

有幾件事你的代碼錯誤：

$query = "SELECT * FROM hotel"; 
$results = $conn->query($query); 
$hotel = $results->fetch(); 

在這一點上，加載了第一條記錄爲$hotel。如果你想遍歷記錄集，你不需要這樣做。

$hotel[$id] = $hotel['hotel_id']; 
$hotel[$name] = $hotel['name']; 
$hotel[$address] = $hotel['address']; 
$hotel[$postcode] = $hotel['postcode']; 
$hotel[$town] = $hotel['town']; 
$hotel[$description] = $hotel['description']; 
$hotel[$rating] = $hotel['rating']; 
$hotel[$image] = $hotel['image']; 

這些行有些多餘。您已經獲得了$hotel中該行的詳細信息，並且您正在使用尚未定義的數組鍵添加額外的條目。

/* 
$id = $hotel['hotel_id']; 
$name = $hotel['name']; 
$address = $hotel['address']; 
$postcode = $hotel['postcode']; 
$town = $hotel['town']; 
$description = $hotel['description']; 
$rating = $hotel['rating']; 
$image = $hotel['image']; 
*/ 
while($hotel = $results->fetch()) { 
    echo "<br> id: ". $hotel[$id] = $hotel[id]. " - Name: ". $hotel[$name]. " - Address: "  . $hotel[$address] . " - Postcode: " . $hotel[$postcode]. 
    " - Town: " . $hotel[$town]. " - Description: ". $hotel[$description]. " - rating: ". $hotel[$rating]. " - image: " . $hotel[$image]; 
} 

在你的循環中，你試圖錯誤地訪問數組鍵。 $hotel[$id]正在使用$id作爲關鍵字，並且該變量不存在。 $hotel[id]正在使用不帶引號的密鑰，PHP會猜測它是$hotel['id']，但會抱怨。所以你可以直接引用它來避免警告。

所以，你可以簡化你的代碼到：

$query = "SELECT * FROM hotel"; 
$results = $conn->query($query); 

while($hotel = $results->fetch()) { 
    echo "<br> id: ". $hotel['id'] . " - Image: <img src='". $hotel['image']. "'>" .... 
}