函数返回和

Question

我有一个SQL表，在这个表中，我存储了用户每天的取水量。我要为当前日期选择用户的摄入量。我想我做得对，只是不知道怎么回号码：

function intake_so_far(){
    $connection=get_db();
    $email=$_SESSION['email'];
    $datetoday=date("Y-m-d");
    $sql="SELECT SUM(intake_amount) FROM intake WHERE intake_email='$email' AND date(intake_date)='$datetoday'";
    $q=mysqli_query($connection,$sql);
}

如何将SUM of (intake_amount)返回为$intake_so_far

希望你得到我想要的，非常感谢提前！

Answer 1

您可以在(intake_amount)中得到$intake_so_fa之和如下：

function intake_so_far()
{
    $connection = get_db();
    $email=$_SESSION['email'];
    $datetoday=date("Y-m-d");
    $sql = "SELECT SUM(intake_amount) AS intake_so_far FROM intake WHERE intake_email = '$email' 
    AND DATE(intake_date) ='$datetoday'";
    $res = mysqli_query($connection,$sql);
    $row = mysqli_fetch_array($res);
    $intake_so_far = $row['intake_so_far'];
    return $intake_so_far; // return sum of this value
}

您可以使用mysqli_fetch_array和mysqli_fetch_assoc函数来获取值。

在本例中，我使用了mysql_fetch_array，因为mysqli_fetch_array()函数将结果行作为关联数组、数字数组或两者都获取。(从W3School获取行)

mysqli_fetch_assoc将只返回结果中的关联数组。

Answer 2

function intake_so_far(){
    $connection=get_db();
    $email=$_SESSION['email'];
    $datetoday=date("Y-m-d");
    $sql="SELECT SUM(intake_amount) AS intake_so_far FROM intake WHERE intake_email='$email' AND date(intake_date)='$datetoday'";
    $res=mysqli_query($connection,$sql);
    $row = mysqli_fetch_assoc($res);
    return $row['intake_so_far'];
}

Answer 3

function intake_so_far(){
$connection=get_db();
$email=$_SESSION['email'];
$datetoday=date("Y-m-d");
$sql="SELECT SUM(intake_amount) FROM intake WHERE intake_email='$email' AND date(intake_date)='$datetoday'";
$q=mysqli_query($connection,$sql);
$getSum = mysqli_fetch_array($q, MYSQLI_ASSOC);
return $getSum[0];
}

http://php.net/manual/en/mysqli-result.fetch-array.php查阅了有关mysqli_fetch_array的教程。