PHP 脚本 (mysqli) 不回显什么? [英] PHP script (mysqli) doesn't echo's something?

问题描述

为什么这段代码什么都不做?他没有回应结果.此脚本的目标是回显列的平均值(小数点后 2 位).

Why is this code doing nothing? He isn't echoing the result out.. The goal of this script is to echo the average of a column (on 2 decimals).

<?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['price'];

die();

?>

推荐答案

您已经有效地将 echo 语句放在循环之后:

You have effectively put your echo statement after the loop:

while($row = mysqli_fetch_array($result));
                                         ^ this is the problem
echo $row['price'];

... 等同于:

while($row = mysqli_fetch_array($result)) {}
echo $row['price'];

在循环之后 $row 将是 false 所以什么都不会被回显出来.

And after the loop $row will be false so nothing will be echoed out.

你想要:

while($row = mysqli_fetch_array($result))
    echo $row['price'];

或者(更好，因为它会避免这些类型的错误......):

or (better as it will avoid these kinds of mistakes...):

while($row = mysqli_fetch_array($result)) {
    echo $row['price'];
}

并且您应该使用准备好的语句来避免您现在遇到的 sql 注入问题.

And you should be using a prepared statement to avoid the sql injection problem you have now.

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