Question

我正在创建一些对象的数组，例如下面的代码：

$eventos = DB::select('SELECT *
    FROM eventos
    ORDER BY timestamp DESC'); 
    foreach ($eventos as $evento){
        $produtos = DB::select("SELECT * 
        FROM produtos
        WHERE transaction_id = '$evento->transaction_id'"); 
        $eventos = array(
            'timestamp' => $evento->timestamp,
            'revenue' => $evento->revenue,
            'transaction_id' => $evento->transaction_id,
            'store_name' => $evento->store_name,
            'products' => array()
            ); 
            foreach ($produtos as $produto){ 
                $products[] = array(
                    'name' => $produto->nome,
                    'price' => $produto->preco,
                    );
            }
        $eventos['products'] = $products;
        echo json_encode($eventos);
    }
}

但是我的JSON无效：根据验证器，出现错误“多个JSON根元素”。应该是

]

}，

{

不是

]

} {

我该如何实现？

 {

   "timestamp":"2016 1",
   "revenue":120,
   "transaction_id":334,
   "store_name":"Uai",
   "products":[

      {

         "name":"Black Shoes",
         "price":99
      }
   ]
}{  //ERROR here
   "timestamp":"2016 2",
   "revenue":250,
   "transaction_id":6665,
   "store_name":"M Mall",
   "products":[

      {

         "name":"Blue Shirt",
         "price":100
      },
      {

         "name":"Pink Pants",
         "price":150
      }
   ]
}

Answer 1

您不会回传一系列JSON文档，那是无效的。

相反，您需要组装一个最终单数数组，并将其编码为JSON：

$final = [ ];
$eventos = DB::select('SELECT *
    FROM eventos
    ORDER BY timestamp DESC'); 
    foreach ($eventos as $evento){
        $produtos = DB::select("SELECT * 
        FROM produtos
        WHERE transaction_id = '$evento->transaction_id'"); 
        $data = array(
            'timestamp' => $evento->timestamp,
            'revenue' => $evento->revenue,
            'transaction_id' => $evento->transaction_id,
            'store_name' => $evento->store_name,
            'products' => array()
            ); 
            foreach ($produtos as $produto){ 
                $products[] = array(
                    'name' => $produto->nome,
                    'price' => $produto->preco,
                    );
            }
        $data['products'] = $products;
        $final[] = $data;
    }
}

echo json_encode($final);

我不知道DB::Select是什么，但是如果它只是mysqli或PDO之类的脆弱包装，那么如果它不支持占位符值，就会出现问题。请勿直接在查询中插入数据。而是使用?之类的占位符，并分别绑定值。

您的数据库驱动程序（无论它是什么）可能已经支持此功能，因此请立即进行检查以避免将来出现潜在的问题。

多个JSON根元素-PHP

1 个答案: