Question

我正在尝试从MySQL获取数据并以JSON格式显示它这是部分PHP代码

$sql = "SELECT item, cost, veg, spicy_level FROM food1";
$result = $conn->query($sql);

    while($row = $result->fetch_assoc()) {

    echo  json_encode($row),"<br/>";}

＆GT？; 我正在输出

{"item":"dosa","cost":"20","veg":"0","spicy_level":"1"}
{"item":"idli","cost":"20","veg":"0","spicy_level":"2"}

但我需要它

food1:[
{"item":"dosa","cost":"20","veg":"0","spicy_level":"1"},
{"item":"idli","cost":"20","veg":"0","spicy_level":"2"}
]

任何人都可以指导我吗？我认为我得到的是对象格式，我需要以数组格式输出，即[＆amp; ]。这个json和php非常新。

Answer 1

您可以将查询结果封装在数组中并在打印后填充;

$sql = "SELECT item, cost, veg,     spicy_level FROM food1";
$result = $conn->query($sql);
$a = array();
while($row = $result->fetch_assoc()) {
  if($a['food1'] ==null)
     $a['food1'] = array():
  array_push($a['food1'],$row);}


  echo json_encode($a);
?></i>

Answer 2

您的代码应为：

$sql = "SELECT item, cost, veg, spicy_level FROM food1";
$result = $conn->query($sql);

$food['food1'] = array();

while($row = $result->fetch_assoc()) {
    $food['food1'][] = $row;    
}

echo  json_encode($food);

Answer 3

每次循环都不要调用json_encode。将所有行放入一个数组中，然后对其进行编码。

$food = array();
while ($row = $result->fetch_assoc()) {
    $food[] = $row;
}
echo json_encode(array('food1' => $food));

MySQL - ＆gt; PHP数组 - ＆gt; Json需要以数组加对象格式输出

3 个答案: