我在运行此功能时遇到问题。当它运行时,它完全符合我的要求,除了在我的 like_requests 表中,request_id 不是链接到变量的 mysql 查询结果,$select
而是资源 ID #22。我认为当您尝试回显结果时会出现资源 ID,但我没有使用回显。代码有什么问题?
function update_likes($band_requested, $new_likes, $session_user_id) {
$select = mysql_query("SELECT `primary_id` FROM `requests` WHERE
`user_requester_id` = '$session_user_id' AND `person_requested` =
'$band_requested'");
$sql_2 = "INSERT INTO `like_requests` (user_id, request_id) VALUES
('$session_user_id', '$select')";
mysql_query($sql_2);
}
$band_requested = 'rally done';
$new_likes = 239;
$the_session_user_id = 3;
update_likes($band_requested, $new_likes, $the_session_user_id);
使用更正答案进行更新 这是在 David 的帮助下更正的代码。
function update_likes($band_requested, $new_likes, $session_user_id)
{
$select = mysql_query("SELECT `primary_id` FROM `requests` WHERE `user_requester_id` =
'$session_user_id' AND `person_requested` = '$band_requested'");
$row = mysql_fetch_row($select);
$request_id = $row[0];
$sql_2 = "INSERT INTO `like_requests` (user_id, request_id) VALUES ('$session_user_id',
'$request_id')";
mysql_query($sql_2);
}