duanmie9741 2016-12-08 09:48
浏览 108
已采纳

如何从mysql显示平均值

How to show the average of a column in mysql?

Below is my code which i have tried so far :

<?php
  if (isset($_GET["age"]));
  $age = ($_GET["age"]);
  include($_SERVER["DOCUMENT_ROOT"] . "/includes/config.php");

  // Input
  $sql = "SELECT AVG(column_name) FROM table_name";

  // Check age
  if ($age > 99 or $age < 5) {
    echo ("We only store data of people between the age of 5 and 99.");
    if (!mysqli_query($conn, $sql)) {
      die('Error: ' . ((is_object($GLOBALS["___mysqli_ston"])) ? mysqli_error($GLOBALS["___mysqli_ston"]) : (($___mysqli_res = mysqli_connect_error()) ? $___mysqli_res : false)));
    }
  } 
  else {
    echo ("We got it!");
  }

  // Close connection
  ((is_null($___mysqli_res = mysqli_close($conn))) ? false : $___mysqli_res);

  die();
?>

But how to exactly define a variable to the result of the AVG with a maximum of 2 decimals?

I want to used and show it into another file (so I will include this one).

What I have right now

<?php
if (isset($_GET["age"]));
$age = ($_GET["age"]);
include($_SERVER["DOCUMENT_ROOT"] . "/3/includes/config.php");
include($_SERVER["DOCUMENT_ROOT"] . "/3/includes/opendb.php");

// My own created code
$sql = $conn->query("SELECT ROUND(AVG(price AS FLOAT), 2) FROM data WHERE age= '$age'");
$data = $sql->mysqli_fetch_assoc();
$avg_data = $data['price'];
echo $avg_data;

// This below is from an other post but don't know how it works and if it is good.
$ratings = $conn->query("SELECT AVG(price) avg_rating FROM data form_id = '" . $age . "'");
$data2 = $ratings->mysqli_fetch_assoc();
$avg_rating = $data2['avg_rating'];
echo $avg_rating;

die();

?>
  • 写回答

3条回答 默认 最新

  • dousao9569 2016-12-08 13:59
    关注

    How I fixed it:

    <?php
    
    if (isset($_GET["age"])) {
    $age = ($_GET["age"]);
    
    include($_SERVER["DOCUMENT_ROOT"] . "/3/includes/config.php");
    
    $con=mysqli_connect($dbhost,$dbuser,$dbpass,$dbname);
    
    if (mysqli_connect_errno($con)) {
        echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }
    
    $result = mysqli_query($con,"SELECT AVG(price) FROM data WHERE age= '$age'") or die("Error: " . mysqli_error($con));
    
    while($row = mysqli_fetch_array($result)) {
       echo $row['AVG(price)'];
       echo number_format($row['AVG(price)'], 2);
    }
    
    die();
    }
    else {
    echo 'Something went wrong, try again.';
    }
    
    ?>
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论
查看更多回答(2条)

报告相同问题?

悬赏问题

  • ¥15 求差集那个函数有问题,有无佬可以解决
  • ¥15 【提问】基于Invest的水源涵养
  • ¥20 微信网友居然可以通过vx号找到我绑的手机号
  • ¥15 寻一个支付宝扫码远程授权登录的软件助手app
  • ¥15 解riccati方程组
  • ¥15 display:none;样式在嵌套结构中的已设置了display样式的元素上不起作用?
  • ¥15 使用rabbitMQ 消息队列作为url源进行多线程爬取时,总有几个url没有处理的问题。
  • ¥15 Ubuntu在安装序列比对软件STAR时出现报错如何解决
  • ¥50 树莓派安卓APK系统签名
  • ¥65 汇编语言除法溢出问题