PHP中JSON响应混入DB类echo消息如何屏蔽？仅返回指定JSON

我有如下PHP代码：

<?php
header("Access-Control-Allow-Origin: *");

include_once('../includes/mysql_connect.php');
include_once('../db/tables/search.php');

// SAVE TO DB
$search = new search($mysql);
//$search->drop();
//$search->create();
$search->save(array("shop" => "test store", "searchterm" => $_POST['search']));


// LOG TO FILE
$file = fopen('search-data.txt', 'a+');
$result = fwrite($file, "
search query data: " . print_r($_POST['search'], true) );
    
// SEND RESPONSE
$response = array('response' => 'received', 'data' => $_POST['search'] );
echo json_encode($response);

当我查看浏览器控制台时，看到如下内容：

<br>INSERT QUERY SUCCESSFUL: insert_id=69{"response":"received","data":"product"}

其中“INSERT QUERY SUCCESSFUL”来自执行INSERT操作的DB类，该类会echo各类消息。我希望保留DB类中的这些消息，但浏览器控制台仅显示最后两行代码生成的JSON响应，该怎么做？

解决方法 #

方法1：用输出缓冲捕获并丢弃DB类的输出 #

在调用DB类方法前开启输出缓冲，执行完后清空缓冲，再输出JSON：

<?php
header("Access-Control-Allow-Origin: *");

include_once('../includes/mysql_connect.php');
include_once('../db/tables/search.php');

// 开启输出缓冲，捕获后续所有echo内容
ob_start();

// SAVE TO DB
$search = new search($mysql);
//$search->drop();
//$search->create();
$search->save(array("shop" => "test store", "searchterm" => $_POST['search']));

// 清空缓冲（不输出捕获的内容）
ob_end_clean();

// LOG TO FILE
$file = fopen('search-data.txt', 'a+');
$result = fwrite($file, "
search query data: " . print_r($_POST['search'], true) );
    
// SEND RESPONSE
$response = array('response' => 'received', 'data' => $_POST['search'] );
echo json_encode($response);

方法2：微调DB类的输出方式 #

如果可以对DB类做小改动，把echo调试消息改成写入日志文件，既保留信息又不干扰响应：
将DB类中类似这样的代码：

echo "<br>INSERT QUERY SUCCESSFUL: insert_id=" . $insert_id;

替换为：

file_put_contents('db-debug.log', "INSERT QUERY SUCCESSFUL: insert_id=" . $insert_id . PHP_EOL, FILE_APPEND);

方法3：重定向输出到日志文件 #

在调用DB类方法前，把PHP标准输出重定向到日志文件，执行完成后恢复输出：

<?php
header("Access-Control-Allow-Origin: *");

include_once('../includes/mysql_connect.php');
include_once('../db/tables/search.php');

// 保存当前输出句柄
$original_stdout = fopen('php://stdout', 'w');
// 重定向输出到日志文件
$log_file = fopen('db-debug.log', 'a');
fclose(STDOUT);
STDOUT = $log_file;

// SAVE TO DB
$search = new search($mysql);
//$search->drop();
//$search->create();
$search->save(array("shop" => "test store", "searchterm" => $_POST['search']));

// 恢复原输出句柄
fclose(STDOUT);
STDOUT = $original_stdout;

// LOG TO FILE
$file = fopen('search-data.txt', 'a+');
$result = fwrite($file, "
search query data: " . print_r($_POST['search'], true) );
    
// SEND RESPONSE
$response = array('response' => 'received', 'data' => $_POST['search'] );
echo json_encode($response);

内容的提问来源于stack exchange，提问作者user1532669