PHP ZF2使用SQL函数选择where子句

问题描述

我需要比较DATETIME中的DATE格式.

I need to compare DATE format from a DATETIME.

在ZF 2.3.5之前的版本中，以下代码可以正常工作:

Prior ZF 2.3.5, the following code was working fine:

    $select->where('DATE(arrival_date) <= DATE(NOW())');
    $select->where('DATE(departure_date) >= DATE(NOW())');
    $select->where('enable = true');

在ZF 2.4.2+中，它不再起作用，并产生以下错误:

With ZF 2.4.2+ it does not work anymore and produce the following error:

无法从Zend \ Db \ Sql \ Predicate \ PredicateInterface接口继承或继承常量TYPE_IDENTIFIER

我尝试了以下操作(无错误).问题是"arrival_date"是DATETIME格式，我只需要与DATE进行比较.

I have tried the following (without error). The problem is that "arrival_date" is a DATETIME format, and I need to compare with a DATE only.

    $where = new Zend\Db\Sql\Where();
    $where->lessThanOrEqualTo('arrival_date', date('Y-m-d'));
    $where->greaterThanOrEqualTo('arrival_date', date('Y-m-d'));
    $where->equalTo('enable', true);
    $select->where($where);

理想情况下，此代码应该可以工作，但不能:

Ideally, this code should work, but it does not:

        $select->where(new Zend\Db\Sql\Predicate\Expression('DATE(arrival_time) <= ?', 'DATE(NOW())'));

我迷路了，知道吗?

推荐答案

只需使用像这样的表达式:

Just use Expression like this:

    $select->where(
        array(
            new \Zend\Db\Sql\Predicate\Expression("DATE(arrival_date) <= DATE(NOW())")
        )
    );

不需要?就像你尝试过的一样.和 ?您的参数将被转义，并且不能与表达式一起使用.仅适用于参数，例如PHP datetime或类似的东西.

No need of ? like you tried. With ? you params will be escaped and it won't work with an expression. Only with params, like a PHP datetime or something like this.

经过2.4.2测试

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