传递给 PDOStatement::bindParam() 的参数名称的前导冒号是可选的吗?
将 :name 形式的命名参数传递给 PDOStatement::bindParam()，不管是否使用前导冒号似乎都有效.
When passing named parameters of the form :name to PDOStatement::bindParam(), it seems to work whether or not the leading colon is used.
即要么这样:
$statement->bindParam(':name', $var);
或者这个:
$statement->bindParam('name', $var);
似乎有效.
这里是 PDOStatement::bindParam()
参数
参数标识符.对于使用命名的准备好的语句占位符，这将是 :name. 形式的参数名称.使用问号占位符准备好的语句，这将是1-参数的索引位置.
Parameter identifier. For a prepared statement using named placeholders, this will be a parameter name of the form :name. For a prepared statement using question mark placeholders, this will be the 1-indexed position of the parameter.
这是否意味着可以省略冒号?
Does this mean the colon can be left off?
推荐答案
不，因为文档没有提到这一点，我认为可以安全地假设这种行为不受官方支持，不应依赖.
No, since the documentation doesn't mention this I think it's safe to assume that this behaviour isn't officially supported and shouldn't be relied upon.
但是，它确实可以正常工作(至少在 PHP 5.3.24 中) - 如果缺少冒号，内部会在参数中添加一个冒号(参见 PHP 5.3.24 中的ext/pdo/pdo_stmt.c:363源代码).
However, it does actually happen to work (in PHP 5.3.24 at least) - internally a colon will be added to the parameter if it's missing (see ext/pdo/pdo_stmt.c:363 in the PHP 5.3.24 source code).
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