如何在php中显示（回显）所选值？

Question

如何显示mysql_query中的值或将值保存在变量中？目前我获得资源ID＃4。

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

Answer 1

试一试：

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

Answer 2

echo($currentpointsquery2);

在这里，您尝试打印mysql resource variable，而不是它引用的资源中包含的值。 mysql_fetch_array()函数从记录集返回一行作为关联数组和/或数字数组。所以试试

echo $currentpoints[0];

或

 echo $currentpoints['user_points'];