如何回覆此查詢？

Question

$result = mysql_query("SELECT 

    car.id  car_id, 


FROM 
    car 

WHERE car.id= $id "); 

我如何迴應上面的查詢？

Answer 1

您直接將查詢傳遞給函數。它存儲在一個變量，如果你想呼應它：

$query = "SELECT 

    car.id  car_id, 


FROM 
    car 

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

Answer 2

以上回答將回聲查詢的結果，如果你只是想呼應的查詢字符串本身，它首先存儲爲一個字符串，通過串入的mysql_query功能，呼應字符串：

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

Answer 3

$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"; 
}