如何在 php 中顯示（回顯）選定的值？

Question

如何顯示來自 mysql_query 的值或將值保存在變量中？ 目前我得到資源 ID #4。

$currentpointsquery = "SELECT user_points FROM points WHERE user_id = '$user_id';";
$currentpointsquery2 = mysql_query($currentpointsquery);
echo(currentpointsquery2);

Answer 1

嘗試一下：

$currentpointsquery = "SELECT user_points FROM points WHERE user_id = '$user_id'";
$currentpointsquery2 = mysql_query($currentpointsquery);
$currentpoints = mysql_fetch_array($currentpointsquery2);
echo $currentpoints['user_points'];

Answer 2

echo($currentpointsquery2);

在這里，您嘗試打印一個 mysql資源變量，而不是它引用的資源中包含的值。 mysql_fetch_array()函數從記錄集中返回一行作為關聯數組和/或數字數組。 所以嘗試

echo $currentpoints[0];

或者

 echo $currentpoints['user_points'];