Question

我正在尝试从头开始创建一个公共REST API，而不使用任何框架，我创建了当用户在地址末尾输入时将其作为JSON响应检索所有数据的地方“...？ list = all“但是，我想通过ID检索行或检索最受欢迎的项目。我知道我可能不得不在查询字符串中更改它，但是如何创建检索它的地址的结尾？喜欢？ID = 05（从数据库中检索行ID 5）或TOP = 15（检索最受欢迎的15）。如果有更好或更有效的方法来重写我的代码，请随时告诉我。

<?php
    // Check Connection
if(!($db = mysqli_connect($server, $user, $password, $database))) {
    die('SQL ERROR: Connection failed: '.mysqli_error($db));
    exit;
    }
return $db;
}

    //call the passed in function
if(function_exists($_GET['list'])) {
    $_GET['list']();
} 

    //methods
function all(){
    $db = getMysqlDB();
    $query = "SELECT * FROM books;";
    $books_sql = mysqli_query($db, $query);
    $books = array();
    while($book = mysqli_fetch_array($books_sql)) {
        $books[] = $book;
    }
    $books = json_encode($books);
    echo $_GET['jsoncallback'].$books;
 }
?>

Answer 1

基于PHP的REST API的一个简单的开始是：

<?php

// get the HTTP method, path and body of the request
$method = $_SERVER['REQUEST_METHOD'];
$request = explode('/', trim($_SERVER['PATH_INFO'],'/'));
$input = json_decode(file_get_contents('php://input'),true);

// connect to the mysql database
$link = mysqli_connect('localhost', 'user', 'pass', 'dbname');
mysqli_set_charset($link,'utf8');

// retrieve the table and key from the path
$table = preg_replace('/[^a-z0-9_]+/i','',array_shift($request));
$key = array_shift($request)+0;

// escape the columns and values from the input object
$columns = preg_replace('/[^a-z0-9_]+/i','',array_keys($input));
$values = array_map(function ($value) use ($link) {
  if ($value===null) return null;
  return mysqli_real_escape_string($link,(string)$value);
},array_values($input));

// build the SET part of the SQL command
$set = '';
for ($i=0;$i<count($columns);$i++) {
  $set.=($i>0?',':'').'`'.$columns[$i].'`=';
  $set.=($values[$i]===null?'NULL':'"'.$values[$i].'"');
}

// create SQL based on HTTP method
switch ($method) {
  case 'GET':
    $sql = "select * from `$table`".($key?" WHERE id=$key":''); break;
  case 'PUT':
    $sql = "update `$table` set $set where id=$key"; break;
  case 'POST':
    $sql = "insert into `$table` set $set"; break;
  case 'DELETE':
    $sql = "delete `$table` where id=$key"; break;
}

// excecute SQL statement
$result = mysqli_query($link,$sql);

// die if SQL statement failed
if (!$result) {
  http_response_code(404);
  die(mysqli_error());
}

// print results, insert id or affected row count
if ($method == 'GET') {
  if (!$key) echo '[';
  for ($i=0;$i<mysqli_num_rows($result);$i++) {
    echo ($i>0?',':'').json_encode(mysqli_fetch_object($result));
  }
  if (!$key) echo ']';
} elseif ($method == 'POST') {
  echo mysqli_insert_id($link);
} else {
  echo mysqli_affected_rows($link);
}

// close mysql connection
mysqli_close($link);

source

有关更完整的实施，请查看：

https://github.com/mevdschee/php-crud-api

披露：我是作者。

使用MySQL和PHP格式化创建REST API作为JSON响应

1 个答案: