为什么这段代码不能像我期望的那样获取查询？

Question

如果我在sql

之后运行此操作

SELECT date, SUM(nmbr_of_guests) FROM guest_tbl WHERE date = "2015-06-30"

我得到了这个：

这就是我想要获取特定日期行nmbr_of_guests的总值的NUMBER（SUM）。 我现在需要做的是，当通过表格动态输入日期时，我想获得相同的结果（在屏幕截图中显示）。

创建表单后，我将其设置为action

形式

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;">We could not find the Reference ID: <b>' . $serch_text . '</b>. Please refine your reference ID</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_tbl WHERE date = "'.$serch_text.'"');               
        echo $total_guests;
        mysqli_close($con);
    }
}

现在，如果我选择通过表单搜索2015-06-30，我将获得Catchable fatal error: Object of class mysqli_result could not be converted to string in C:\xampp\htdocs\guest\index.php on line 26

第26行是echo $total_guests;

如果我更改echo $total_guests;

要

echo $total_guests->fetch_assoc();

我正在Notice: Array to string conversion in C:\xampp\htdocs\guest\index.php on line 25 Array

现在第25行是echo $total_guests->fetch_assoc();

如何获取特定日期的总值（nmbr_of_guests之和）？

Answer 1

正如错误所说，fetch_assoc返回一个数组，所以print_r($total_quests->getch_assoc())并查看数组的结构。你可能必须拿起第一个结果，但我建议你使用foreach循环来做。