Question

这是我的代码：

$gid = (int) stripslashes($_POST['id']);
echo $gid;

$db = dbConnect();
$test = $db->query('update games set played = played + 1 where id = "$gid"');
echo $db->error;
echo $db->errno;
die();
}

它从终端工作正常，它正确打印出$ gid，并且不返回任何错误。我错过了一些非常明显的东西吗？

Answer 1

您将查询括在单引号中。在单引号中，变量插值（也称为替换）不会发生。

简单示例：

$who = 'harry potter';
echo 'hi "$who"'; // prints hi "$who"
echo "hi '$who'"; // prints hi 'harry potter'

将您的代码更改为：

$test = $db->query("update games set played = played + 1 where id = '$gid'");

同样来自行$gid = (int) stripslashes($_POST['id']);，明确$gid是一个整数，并且不需要在查询中将其括在引号中。所以我们有：

$test = $db->query("update games set played = played + 1 where id = $gid");

Answer 2

$test = $db->query("UPDATE `games` SET `played` = `played` + 1 WHERE `id` = '" . $gid . "'");

Answer 3

单引号中的任何内容都不会被解析。做id="' . $gid . '"

或者将整个事物用双引号括起来并在$ gid周围加上单引号。

Answer 4

尝试

$test = $db->query
("
    UPDATE `games` 
    SET
       `played` = (`played` + 1)
    WHERE `id` ='" . $gid . "'
");

PHP MySQL查询不起作用，但可以在终端上运行

4 个答案: