如何根據包含相同值的另一行來計算SUM值？

Question

我試圖讓根據date 用戶number of gusts可能能夠通過輸入date

我的表格來搜索number of guests的日期是：

<form action="index.php" method="post" class="booking_reference"> 
    <input type="text" placeholder="date" required name="search_text"> 
    <span class="arrow"> 
     <input type="submit" value="CHECK" name="search" class="search_button"> 
    </span> 
</form> 

我的動作部分是：

if (isset($_POST['search'])){ 
$serch_text = $_POST["search_text"]; 
$con=mysqli_connect("localhost","root","","guest"); 
    // Check connection 
    if (mysqli_connect_errno()) { 
     die ("Failed to connect to MySQL: " . mysqli_connect_error()); 
    }  
    $searching_date = $serch_text; 
    $searchroute = "select * from guest_tbl where date = '$searching_date'"; 
    $result = $con->query($searchroute); 
    $row = $result->fetch_assoc(); 
    if ($row == 0) { echo '<div style="color: red;">No one found for the date: <b>' . $serch_text . '</b>. Please refine your date</div>' ; 
    } else {   
     echo '<h3>Total guest for the day : '. $serch_text .'</h3>';   
     $total_guests = mysqli_query($con, 'SELECT date SUM(nmbr_of_guests) FROM guest WHERE date = "'.$serch_text.'"'); // selecting and summing the value of all the 'nmbr_of_guests' rows in same date - $serch_text  
     echo $total_guests; 
     mysqli_close($con); 
    } 
} 

我在我的數據庫中有以下幾行：

期待的結果是：

1. 如果我在搜索字段中$echo $total_guests 輸入2015年6月30日應回聲 - （因爲2 + 3）
2. 如果我在搜索字段輸入2015-06-25$echo $total_guests 應該回顯 - （因爲nmbr_of_guests列剛6 日期）
3. ，如果我進入任何其他日期在搜索領域，它應該呼應沒有 一個發現的日期：任何其他日期。請優化 你的日期。

第三是精細我得到的紅線文本時未找到該日期。

但是，當我進入2015年6月30日或2015年6月25日我得到只是空白。 這意味着$total_guests;什麼都不返回/空白。

怎樣才能讓$total_guests;顯示日期（本例中爲5或6）的nmbr_of_guests行的值的總和？

---

注意：這是一箇舊網站。我將完成此操作後，將所有MySQLi更換爲PDO。這是一個很大的文件，需要時間來替換找到MySQLi的所有行。

Answer 1

逗號缺少日期和總和...查詢應該是─

$total_guests = mysqli_query($con, 'SELECT date, SUM(nmbr_of_guests) FROM guest WHERE date = "'.$serch_text.'"'); // selecting and summing the value of all the 'nmbr_of_guests' rows in same date - $serch_text 

更新

你的代碼 -

echo '<h3>Total guest for the day : '. $serch_text .'</h3>';   
     $total_guests = mysqli_query($con, 'SELECT date SUM(nmbr_of_guests) FROM guest WHERE date = "'.$serch_text.'"'); // selecting and summing the value of all the 'nmbr_of_guests' rows in same date - $serch_text  
     echo $total_guests; 
     mysqli_close($con); 

嘗試查看按如下─

$total_guests = mysqli_query($con, 'SELECT SUM(nmbr_of_guests) FROM guest WHERE date = "'.$serch_text.'"'); // selecting and summing the value of all the 'nmbr_of_guests' rows in same date - $serch_text  

echo '<h3>Total guest for the date of '. $serch_text .' are '.$total_guests.'</h3>';   
    mysqli_close($con); 

Answer 2

能不能請你這樣

$total_guests = mysqli_query($con, 'SELECT date,nmbr_of_guests FROM guest_tbl WHERE date = "'.$serch_text.'"'); 

分配一個可變的 「0」 和計數的值

$total = 0; 
foreach ($total_guests as $guests){ 
$total = $total + $guests->nmbr_of_guests; 
} 
echo $total;