Question

我有一个简单的mysql表，其中包含＆＃39; ratee＆＃39;作为用户和评级是给予该用户的评级。我想显示该用户被评级的次数以及这些评级的平均值。前者如下面的代码所示，但后者没有。我哪里出错了？我正在使用PHP。

    //working
$sql = "SELECT * FROM ratings WHERE ratee='" . $user1 . "'";
$result = mysqli_query($conn, $sql);
$ratingsqty = mysqli_num_rows($result);
echo $ratingsqty;

//not working 1
$sql = "SELECT * FROM ratings WHERE ratee='" . $user1 . "'";
$result = mysqli_query($conn, $sql);
$rating = mysqli_avg($result);
echo $rating;


    //not working 2
$sql = "SELECT avg(rating) FROM ratings WHERE ratee='" . $user1 . "'";
$rating = mysqli_query($conn, $sql);
echo $rating;

Answer 1

mysqli_query()返回一个Mysql结果对象。因此，你不能只是回应它。您仍需要mysqli_fetch_row()或类似内容。

Answer 2

你的问题是你甚至懒得去完成一个基本的mysqli教程或阅读mysqli文档。

不工作1：没有mysqli_avg()这样的功能。周期。
不工作2：mysqli_query()返回mysqli_result object，而不是平均值。您可以使用mysqli_result object的fetch _ *（）方法之一来获取MySQL计算的实际平均值：

请参阅以下代码：

$sql = "SELECT avg(rating) FROM ratings WHERE ratee='" . $user1 . "'"; //in real life code pls take some measures to prevent sql injection attacks
$result = mysqli_query($conn, $sql); //in real life code pls check for execution errors
$rating = $result->fetch_array(MYSQLI_NUM)
echo $rating[0];

Answer 3

试试这个

 $sql = "SELECT * FROM ratings WHERE ratee='" . $user1 . "'";
$result = mysqli_query($conn, $sql);
$rating = mysqli_fetch_array($result,MYSQLI_ASSOC);

现在$ rating是一个数组，你可以打印你的表格数据使用：echo $rating['ratee'];或echo $rating['rating'];

平均试试这个：

 $sql = "SELECT avg(rating) as av FROM ratings WHERE ratee='" . $user1 . "'";
    $result = mysqli_query($conn, $sql);
    $rating = mysqli_fetch_array($result,MYSQLI_ASSOC);
    echo $rating['av'];

Answer 4

对你来说也许为时已晚，但也许其他人会受益：

从MySQL获得一行结果：

$val = fetch_one_record("SELECT some_field FROM some_table LIMIT 1);

function fetch_one_record($sql) {
    $conn = db_connect();
    $result = array_shift($conn->query($sql)->fetch_assoc());
    $conn->close();
    return $result;
}

从PHP中的MYSQL平均查询返回一个结果

4 个答案: