为什么mysql数据库在这种情况下没有更新？

Question

我有这个 PHP 脚本：

require_once('global.php'); //connects to db and various functions used
//select the user's information
$statement = $db->prepare("SELECT * FROM user WHERE id=1");
$statement->execute();
$result = $statement->fetchObject();
//get the chat
$string = $result->chats;

$from = $_REQUEST['from'];
$msg = $_REQUEST['msg'];
$sent = $_REQUEST['sent'];

//see what we should do with the recieved data
if($string == "") {
    //there isnt any chats right now, we must add the first ever contact
    $string = "<ResultSet><chats><chat><messages><sms><from>{$from}</from><msg>{$msg}</msg><sent>{$sent}</sent></sms></messages><contact>{$from}</contact></chat></chats></ResultSet>";
    //send the data back to the user's row in the database
    $statement = $db->prepare("UPDATE user SET chats='{$string}' WHERE id=1");
    $statement->execute();
} else if($from == $result->name) {
    //the user is sending a message to a contact. we now need to get the "to" value.
    $to = trim(str_replace("_", " ", $_REQUEST['to']));
    //add the sms to the contact's chat
    $string = str_replace("</sms></messages><contact>{$to}</contact>", "</sms><sms><from>{$from}</from><msg>{$msg}</msg><sent>{$sent}</sent></sms></messages><contact>{$to}</contact>", $string);
    //send the data back to the user's row in the database
    $statement = $db->prepare("UPDATE user SET chats='{$string}' WHERE id=1");
    $statement->execute();
} else if(strstr($string, "<contact>".$from."</contact>")) {
    //The contact that sent the message already exists in this user's row, add the message to the contact's chat
    $string = str_replace("</sms></messages><contact>{$from}</contact>", "</sms><sms><from>{$from}</from><msg>{$msg}</msg><sent>{$sent}</sent></sms></messages><contact>{$from}</contact>", $string);
    //send the data back to the user's row in the database
    $statement = $db->prepare("UPDATE user SET chats='{$string}' WHERE id=1");
    $statement->execute();
} else {
    //Person who sent the message doesnt exist in the chats, add him.
    $string = str_replace("</chats>", "<chat><messages><sms><from>{$from}</from><msg>{$msg}</msg><sent>{$sent}</sent></sms></messages><contact>{$from}</contact></chat></chats>", $string);
    //send the data back to the user's row in the database
    $statement = $db->prepare("UPDATE user SET chats='{$string}' WHERE id=1");
    $statement->execute();
}

问题出在这个 else if 代码中：

else if($from == $result->name) {
//the user is sending a message to a contact. we now need to get the "to" value.
$to = trim(str_replace("_", " ", $_REQUEST['to']));
//add the sms to the contact's chat
$string = str_replace("</sms></messages><contact>{$to}</contact>", "</sms><sms><from>{$from}</from><msg>{$msg}</msg><sent>{$sent}</sent></sms></messages><contact>{$to}</contact>", $string);
//send the data back to the user's row in the database
$statement = $db->prepare("UPDATE user SET chats='{$string}' WHERE id=1");
$statement->execute();
}

我确信代码正在通过这个 else 运行，我已经回显并确认了。当我使用 $string = str_replace() 时，我打印了 $string 并且它确实被替换了。但是，当我将数据提交到数据库中的行时，刷新数据库时什么也没有发生。为什么它对 else 不起作用，但对 if 语句的其余部分（以及它之前的 if）起作用？

这对我来说没有意义。我尽力对代码进行适当的注释，如果您需要解释，请询问。

Answer 1

如果 $db 不是 MySQLi 或 PDO 的实例，而是数据库包装器，请查看实例化为 $db 的类并确保它不会启动事务。如果启动了事务，您将需要提交事务，以便应用/保存对数据库的更改。

Answer 2

我不知道为什么，但我必须更改我的代码，以便我在执行函数而不是查询中发送变量。在查询中我会输入“？”我想要变量的地方，在execute()函数中我会放置一个像这样的数组，

$statement->execute(array($string, 1));

这似乎已经解决了我的问题。