I have a PHP code that looks like this: class A { public function __construct() { $this->b = new B(function($x) { return $x + 1; }); } }; class B { public function __construct($dataProcessingFunction) { $this->dataProcessingFunction = $dataProcessingFunction; } public function processData($data) { $f = $this->dataProcessingFunction;
我有一个如下所示的PHP代码： class A { public function __construct() { $this->b = new B(function($x) { return $x + 1; }); } }; class B { public function __construct($dataProcessingFunction) { $this->dataProcessingFunction = $dataProcessingFunction; } public function processData($data) { $f = $this->dataProcessingFunction; return $f($d
I'd like to be able to include a tag when making a spot request via PHP. When creating on-demand instances, you can create the instance, then use it's instance to issue the following: $ec2->create_tags($instance_id, array( array('Key' => 'Name', 'Value' => 'MyTestMachine'), )); However, when issuing a spot bid, the instance isn't started right away, so you'd
我希望能够通过PHP进行现场请求时包含一个标签。 在创建按需实例时，您可以创建该实例，然后使用它的实例来执行以下操作： $ec2->create_tags($instance_id, array( array('Key' => 'Name', 'Value' => 'MyTestMachine'), )); 但是，在发布竞价时，实例不会立即启动，因此您必须创建一个监视器标记来处理此问题......除非您可以在请求阶段添加标记。 我还没有找到任何文件来显示它会如何或看起来像它是否
I'm trying to store a one-to-many relationship in my mySQL database and I want to avoid duplicate entries if at all possible. I read on the internet and saw that 'ON DUPLICATE KEY UPDATE' is an option. Would this work for my situation? CODE: function insert_options($uid, $array) { if(!is_array($array)) { return false; } $db = db_connect(); foreach($array
我试图在mySQL数据库中存储一对多关系，并且我想尽可能避免重复条目。 我在网上阅读，看到'重复密钥更新'是一个选项。 这是否适合我的情况？ 码： function insert_options($uid, $array) { if(!is_array($array)) { return false; } $db = db_connect(); foreach($array as $a) { $sql = 'INSERT INTO newsletter_coupon_codes_options (uid, option_name, value) VALUES (?, ?,
Is there a good object-relational-mapping library for PHP? I know of PDO/ADO, but they seem to only provide abstraction of differences between database vendors not an actual mapping between the domain model and the relational model. I'm looking for a PHP library that functions similarly to the way Hibernate does for Java and NHibernate does for .NET. Look into Doctrine. Doctrine 1.2 im
有没有一个好的PHP对象关系映射库？ 我知道PDO / ADO，但他们似乎只提供了数据库供应商之间差异的抽象，而不是域模型和关系模型之间的实际映射。 我正在寻找一个PHP库，其功能与Hibernate为Java和NHibernate为.NET所做的功能类似。 研究教义。 Doctrine 1.2实现了Active Record。 Doctrine 2+是一个DataMapper ORM。 另外，请查看Xyster。 它基于Data Mapper模式。 另外，请看DataMapper vs. Active Record。 尝
I read this question (Should I use bindValue() or execute(array()) to avoid SQL injection?) and found that both execute( array() ) and bindParam() both prevent SQL injections when working with PDO and prepared statements. However I also recall reading somewhere that execute() automatically treats every variable as a string, regardless if it's an integer. My question is surely this is a sec
我读过这个问题（我应该使用bindValue（）还是execute（array（））来避免SQL注入？），并发现execute（array（））和bindParam（）在处理PDO和预处理语句时都会阻止SQL注入。 不过，我还记得在某处execute（）会自动将每个变量视为一个字符串，无论它是一个整数。 我的问题肯定是除了字符串变量以外，这是一个安全问题？ 例如，如果我通过准备好的查询来获取用户的id，那么通过execute执行的参数将被视为字符串而不是整数
I have the following Prepared Statement: SELECT `e1`.`field_value` AS `county`, `e2`.`field_value` AS `venue_type`, `l`.`ID` AS `listing_id`, `l`.`ID` AS `id`, `l`.`user_ID` AS `user_id`, IF(`l`.`user_ID` > 1, 'b', 'a') AS `flag` , COUNT(`img`.`ID`) AS `img_num` FROM ( `listingsDBElements` `e1`, `listingsDBElements` `e2`, `listingsDB` `l
我有以下准备声明： SELECT `e1`.`field_value` AS `county`, `e2`.`field_value` AS `venue_type`, `l`.`ID` AS `listing_id`, `l`.`ID` AS `id`, `l`.`user_ID` AS `user_id`, IF(`l`.`user_ID` > 1, 'b', 'a') AS `flag` , COUNT(`img`.`ID`) AS `img_num` FROM ( `listingsDBElements` `e1`, `listingsDBElements` `e2`, `listingsDB` `l` ) LEFT JOIN `listing
I'm reading about PDO and prepared statement, focused in Postgres. As I understood how ps is implemented in PHP, I can register a named query, then define variables for it and run it multiple times. But what happens when the use of where clauses depend on parameters. For exemple, if !empty($param3) then I add a where clause, but if it is empty that clause won't be used in the query.
我正在阅读有关PDO并准备好声明，重点放在Postgres上。 正如我理解ps是如何在PHP中实现的，我可以注册一个命名查询，然后为它定义变量并多次运行它。 但是当where子句的使用取决于参数时会发生什么。 例如，如果!empty($param3)那么我添加一个where子句，但如果它是空的，那么该子句将不会在查询中使用。 对于这种情况，最好的情况是缩小记录列表的筛选表单：如果未使用筛选器，则查询甚至不会使用WHERE，而可能会使用所
Should PDO prepared statements be freed up after use? And if so, how? Specifically I'm asking about MySQL - how can you, and should you, call DEALLOCATE PREPARE though PDO. (Edit: To clarify, this question is not referring to emulated prepares, but real prepares. ) Also - will this free the results set (when large)? Explanation: I have seen code along the lines of $stmnt = $db->p
PDO准备好的陈述应该在使用后释放出来吗？ 如果是这样，怎么样？ 具体来说，我正在询问关于MySQL的内容 - 你应该怎样，并且应该通过PDO调用DEALLOCATE PREPARE 。 （编辑：为了澄清，这个问题不是指模拟的准备，而是真正的准备。） 另外 - 这将释放结果集（当大）？ 说明： 我已经看到了代码 $stmnt = $db->prepare($sql); $stmnt->execute($aParams); $stmnt = null; 这导致我想知道这是什么，何时以及如果uns
I experience an issue when working with PDO prepared statements and bit data types When retrieving the result from the DB with PDO prepared statement an empty result set is returnd, but the correct results are returned if the query is done by a normal sql query If the field is_demo in the where clause is omitted it works as it should!? DB table structure CREATE TABLE IF NOT EXISTS `user` (
使用PDO预处理语句和bit数据类型时遇到问题 当使用PDO准备语句从数据库检索结果时，会返回一个空结果集，但如果查询是通过正常的SQL查询完成的，则会返回正确的结果 如果where子句中的is_demo字段被忽略，它的工作原理应该如此！ 数据库表结构 CREATE TABLE IF NOT EXISTS `user` ( `id` smallint(5) unsigned NOT NULL AUTO_INCREMENT, `client_id` smallint(5) unsigned NOT NULL, `is_admin` bit(1) NOT NULL,
Using PHP v. 5.2.14 and PDO-MySQL extension. I am new to prepared statements. Need to create a search form (so user input) and a working query of the "Select all X where X like…" variety. Code and Results: $sql = 'SELECT COUNT(*) as num_books from t_books where title LIKE :search_term'; // Later access as num_books $prep = $dbh->prepare($sql); $num = $prep->execute(array(':s
使用PHP v。5.2.14和PDO-MySQL扩展。 我是新准备的陈述。 需要创建一个搜索表单（以便用户输入）以及“选择所有X，其中X像......”类型的工作查询。 代码和结果： $sql = 'SELECT COUNT(*) as num_books from t_books where title LIKE :search_term'; // Later access as num_books $prep = $dbh->prepare($sql); $num = $prep->execute(array(':search_term' => '%'.$search_term. '%')); $total=$num->fetchCo