Question

在我的示例中，我有一个PHP脚本，来自AJAX，记录了每个名为'Richard'的人。我想通过计算名为Richard的人来进一步修改这个，这些人匹配某个输入值的年龄，如$theirage所示。

具体来说，我想总结两个时代，并且只显示我的数组中等于20的那些年龄。 $row[age]对应于我的数据库中的值age，而$theirage是用户输入。

最后，我想仅计算等于20的总和年龄，在这种情况下。

    $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
) );

我怎么能这样做？

Answer 1

我建议您查询members表，以计算符合条件的所有人：具体名称'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)

如何过滤json编码数组？

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