我浏览了所有其他StackOverflow(和谷歌)的帖子,都有同样的问题,但似乎没有一篇能解决我的问题.
我正在使用PDO和PHP.
我的代码:
$vals = array(
':from' => $email,
':to' => $recipient,
':name' => $name,
':subject' => $subject,
':message' = >$message
);
print_r($vals);
try {
$pdo = new PDOConfig();
$pdo->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$sql = "SELECT * FROM messages WHERE `message` LIKE :message";
$q = $pdo->prepare($sql);
$q->execute(array(':message' => $vals[':message']));
$resp = $q->fetchAll();
foreach ($resp as $row) {
throw new Exception('Please do not post the same message twice!');
}
$sql = "INSERT INTO messages (from, to, name, subject, message) VALUES (:from, :to, :name, :subject, :message)";
$q = $pdo->prepare($sql);
$q->execute($vals);
}
catch(PDOException $e) {
echo $e->getMessage();
}
第一张照片
Array ( [:from] => abc@gmail.com
[:to] => lala@me.com
[:name] => abc
[:subject] => abc
[:message] => abc )
预期为(无为空)
但它会输出错误
SQLSTATE[42000]:语法错误或访问冲突:1064您的SQL语法有错误;查看与您的MySQL服务器版本对应的手册,了解使用"from、to、name、subject、message)值('abc@gmail.com', 'lala@me.com"在1号线
不知道怎么解决这个问题.有什么 idea 吗?