如何过滤json编码数组计数？

Question

在我的示例中，我有一个来自AJAX的PHP脚本，可对名为“ Richard”的每个人进行汇总。 我想通过计算与某些输入值的年龄相匹配的名叫Richard的人来进一步修改它，如$theirage所示。

具体来说，我想将这两个年龄相加，仅在我的数组中显示等于20的年龄。 $row[age]对应于我数据库中的值age ，而$theirage是用户输入。

最后，在这种情况下，我只想统计等于20的总年龄。

    $theirage = $_GET['theirage'];
    $query = mysqli_query($con, "SELECT count(*) as cnt FROM members WHERE
    name = 'Richard'
    ");
    $row = mysqli_fetch_assoc($query);
    $bow = $row['age'] + $theirage;
    $if ($bow = 20){
    $person = $row['cnt'];} 

echo json_encode( array(
    'person' => $person
) );    

我该怎么办？

Answer 1

我建议您查询members表以计算符合条件的所有人：特定名称'Richard'和特定年龄$_GET['theirage'] ：

<?php
$theirAge = $_GET['theirage'];

// This is important: since we're using $theirAge in query string, we need to escape it
$theirAge = mysqli_escape_string($theirAge);

// Let's query all members whose name is Richard and age is equal to $_GET['theirage']
$query = mysqli_query($con, "SELECT count(*) FROM members WHERE name = 'Richard' AND age + $theirAge = 20");

$row = $query->fetch_row();

$totalRichardsOfGivenAge = $row[0]; // here is your result.

// And now let's return JSON
echo json_encode($totalRichardsOfGivenAge)