[英]Using a SELECT query to get data from two different tables in database
我真的很難理解我的查詢出了什么問題。 當我使用var_dump($ sqli); 它只是回顯整個查詢,其中沒有任何數據。
我想創建一個查詢,以便當用戶搜索城市時,返回的搜索結果將是與該城市關聯的景點。 我意識到我的代碼存在一些錯誤,但是,如果我完成了查詢,那么我將在那里調試其余的代碼。 如果有人對我的查詢為什么不起作用有任何想法,我將不勝感激。
require_once('config1.php');
error_reporting(E_ALL);
$output = '';
if(isset($_POST['search'])){
$searchq = $_POST['search'];
$sqli = 'SELECT attraction_name, lat, long, cost FROM zz_attractions WHERE city_id IN SELECT city_id FROM zz_city WHERE city_name LIKE %searchq%' or die("could not search");
var_dump($sqli);
$result = mysqli_query($conn, $sqli);
$count = mysqli_num_rows($result);
if ($count == 0) {
$output = 'there was no search results';
} else {
while ($row = mysql_fetch_array($sqli)) {
$attraction_name = $row['attractionname'];
$lat = $row['latitude'];
$long = $row['longitude'];
$cost = $row['cost'];
$output .= '<div>'.$attraction_name.' '.$lat.' '.$long.' '.$cost.'</div>';
}
}
}
您當前的查詢失敗,因為您需要此處顯示的括號:
SELECT attraction_name, lat, long, cost
FROM zz_attractions
WHERE city_id IN (SELECT city_id
FROM zz_city
WHERE city_name
LIKE '%searchq%')
但是更好的方法是內部聯接:
SELECT a.attraction_name, a.lat, a.long, a.cost
FROM zz_attractions a
INNER JOIN zz_city c ON a.city_id = c.city_id
WHERE c.city_name LIKE '%searchq%'
只是觀察,僅此而已...
從來沒有完全理解為什么PHP編碼人員傾向於將每個SQL查詢都作為一行
$query = '
SELECT a.attraction_name, a.lat, a.long, a.cost
FROM zz_attractions a
INNER JOIN zz_city c ON a.city_id = c.city_id
WHERE c.city_name LIKE %searchq%
';
echo $query;
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