如何回应此查询？

Question

$result = mysql_query("SELECT

  car.id     car_id,


FROM
    car

WHERE car.id= $id ");

我如何回应上面的查询？

Answer 1

您将查询直接传递给该函数。如果要回显它，请将其存储在变量中：

$query = "SELECT

  car.id     car_id,


FROM
    car

WHERE car.id= $id ";
echo $query;
$result = mysql_query($query);

Answer 2

$query = "SELECT car.id,car_id FROM car WHERE car.id= $id ";
$result = mysql_query($query)
while ($car_details = mysql_fetch_array($result)){
  echo "$car_details[id], $car_details[car_id]\n";
}

Answer 3

如果您只想回显查询字符串本身，首先将其存储为字符串，将字符串传递给mysql_query函数，并回显字符串，上面的答案将回显查询结果：

$sql = "SELECT car.id car_id, FROM car WHERE car.id= $id";
echo $sql;    
$result = mysql_query($sql);