运行此代码时出现此错误。
Parse error: syntax error, unexpected 'users_kudos_to_posts' (T_STRING)
代码。
if ($number > 0) {
$arr=explode(",",`$upvoted);
//If I comment out everything below I still get the error. But if I comment out
//the above code, then the error goes away.
if (in_array($primary_id, $arr)) {
array_push($arr, $primary_id);
$new_string = implode(",", $arr);
//the line below is where the parse error is pointing too.
if ($stmt2 = $mysqli->prepare("UPDATE `users_kudos_to_posts` SET `upvoted` = ?
WHERE `user_id` = ?")) {
$stmt2->bind_param('si', $new_string, $session_user_id);
$stmt2->execute();
$stmt2->close();
}
}
}
疯狂的是,除了列的更改之外,我在同一页代码中有其他多个准备好的语句,它们使用相同的sql语句,但它们仍然有效。
此外,我在phpmyAdmin中运行了sql语句,它很有效。
问题是这一行:
$arr=explode(",",`$upvoted);
你有一个多余的背景,不属于那里。应该是:
$arr=explode(",",$upvoted);
StackOverflow的语法高亮显示(呵呵),在`$upvoted
:前面的explode
语句中有一个额外的backtick
$arr=explode(",",`$upvoted);