超薄JSON输出 [英] Slim JSON Outputs

问题描述

我正在将Slim框架与PHP结合使用，以为我的应用程序创建RESTful API.但是，我认为该框架将具有某种方式来创建更简单的JSON输出，而不仅仅是exit($jsonEncodedVariable);.

I am using the Slim framework with PHP to create a RESTful API for my app. However, I assumed that the framework would have some way of creating easier JSON outputs rather than just exit($jsonEncodedVariable);.

我是否缺少框架中的内容，还是我需要为每个方法使用json_encode() ... exit($json) ...?

Am I missing something in the framework, or do I need to use json_encode()... exit($json)... for every method?

所有数据都从我的MySQL数据库中取出，然后将根据调用的REST请求放入JSON数组中.

All of the data is taken out of the my MySQL database and would then be put into a JSON array depending on what REST request was called.

例如，如果请求/api/posts/all，我将exit()所有帖子的JSON数组，每个帖子都为其自己的键"value" : key赋值.

For example, if /api/posts/all was requested, I would exit() a JSON array of all the posts which each value for its own key, "value" : key.

我的问题是，有没有一种简单的方法，可以使用苗条的框架来实现exit()的JSON代码而不是将其作为纯文本退出?

My question is, is there an easy way, using the slim framework, for exit()'ing JSON code instead of exiting it as plain text?

推荐答案

header("Content-Type: application/json");
echo json_encode($result);
exit;

提示:使用Slim用于开发REST API的PHP框架

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