Question

我使用PHP从json中的mysql表中检索数据，即

$table_first = 'recipe';
$query = "SELECT * FROM $table_first";
$resouter = mysql_query($query, $conn);


$set=array();
while ($link = mysql_fetch_array($resouter, MYSQL_ASSOC)){
foreach ($link as $fieldname => $fieldvalue){
    $set[]= $fieldvalue;}
 $query2="SELECT ingredients.ingredient_id,ingredients.ingredient_name,ingredients.ammount FROM ingredients where rec_id = ".$link['rec_id'];
$result2 = mysql_query($query2, $conn);

 while ($rs = mysql_fetch_array($result2, MYSQL_ASSOC)){
     foreach($rs as $fieldname =>$fieldvalue){
         $set[]=$fieldvalue;
     }

 }

}
echo json_encode($set);

代码的结果是

["14","Spaghetti with Crab and Arugula","http:\/\/www","","2010-11-11 14:35:11","localhost\/pics\/SpaghettiWithCrabAndArugula.jpg",
"7","13 ounces spaghetti","10 kg",
"8","1 pound crabmeat","10"]

注意：成分id在图像标记之后开始。 7是成分id后跟两个字段“成分txt和数量”，然后8是与食谱id相关的另一成分id。就像我的结果中没有（{）open或（}）关闭括号。

我想要做的是以正确的json格式输出它。即

[
 {
  "rec_id": "14",
  "name":"Spaghetti with Crab and Arugula",
  "overview":"http:\/\/www",
  "category":"category",
                "time":"2010-11-11 14:35:11",
                "image":"localhost\/pics\/SpaghettiWithCrabAndArugula.jpg"
  "ingredients":
   {
    "ingredient":
     [                       {"ingredient_id":"7","ingredient_name":"13ounces spaghetti","amount":"10kg" },
{ "ingredient_id": "8", "ingredient_name": "1 pound crabmeat","amount":"10kg" },

     ]
   }]

和食谱id 15相同.......

所以怎么能得到这个.... !!任何建议

Answer 1

你输出的是完全有效的json。

那么看看你如何构建$set ......看看问题了吗？

你只是将标量值推送到一个数组上，所以当你对它进行json编码时，你会得到一长串的标量，没有结构就不足为奇了。您的代码正在积极破坏您想要的结构！

我可以为你修复你的代码，但我不会这样做。你需要查看你的查询和循环，并弄清楚发生了什么。

你基本上想做这样的事情：

$result = array();
$recipes = mysql_query('....');
while($recipe = mysql_fetch_assoc($recipes)){
    $result[$recipe['id']] = $recipe;

    $ingredients = mysql_query('...');
    while($ingredient = mysql_fetch_assoc($ingredients)){
        $result[$recipe['id']] = $ingredient;
    }
}

var_dump($result); //or echo json_encode($result);

看到差异？

如何提取数据json格式

1 个答案: