Question

我正在使用PHP版本7.0.32和MySQL 5.6

我创建了一个简单的查询生成器，该生成器确定如何根据给定的参数构建查询。我正在使用PDO连接到我的数据库并执行查询。查询本身很好，我将在下面提供示例。当我直接在数据库中运行查询时，会得到正确的返回结果。但是，每当我运行代码并通过PHP查询数据库时，似乎所有没有排序规则集的列值都将变为NULL。

令人困惑的部分是，这仅在一个特定查询中发生。我所有其他查询都在列中返回了正确的值，其他查询中的一个包含了工作不正常的查询的结果集，但在这种情况下却给出了正确的值。

工作查询：

SELECT * FROM fm_submission AS sub WHERE sub.archived = :archived ORDER BY sub.dateSubmitted LIMIT :start, :limit;

SELECT * FROM fm_submission AS sub LEFT JOIN fm_submission_saved AS saved ON sub.submissionID = saved.submissionID WHERE sub.archived = :archived AND saved.submissionID IS NOT NULL ORDER BY sub.dateSubmitted LIMIT :start, :limit;

SELECT * FROM fm_submission AS sub WHERE sub.archived = :archived ORDER BY sub.dateSubmitted;

不起作用的查询：

SELECT * FROM fm_submission AS sub LEFT JOIN fm_submission_saved AS saved ON sub.submissionID = saved.submissionID WHERE sub.archived = :archived AND saved.submissionID IS NULL ORDER BY sub.dateSubmitted LIMIT :start, :limit;

正确的结果集将返回接近如下的值：

[“ formID”] =>字符串（1）“ 1” [“ personnelID”] => int（0）[“ storeName”] =>字符串（6）“ Albany”

要使一个查询不正确的结果集不起作用，如下：

[“ formID”] =>字符串（1）“ 1” [“ personnelID”] => NULL [“ storeName”] =>字符串（11）“ River Ridge”

如果研究此问题，我发现当Collation和php charset不能一起工作时，可能会发生这种情况。我已经确保所有设置都相同，但是对于INT，FLOATS，DATE等类型，我无法做到这一点。

对于该主题的任何帮助以及任何解决方案，我将不胜感激。询问我是否还有其他信息可以帮助您。

Answer 1

我找到了问题，原来是一个小问题，没有报告错误。

SELECT * FROM fm_submission AS sub LEFT JOIN fm_submission_saved AS saved ON sub.submissionID = saved.submissionID WHERE sub.archived = :archived AND saved.submissionID IS NULL ORDER BY sub.dateSubmitted LIMIT :start, :limit;

上述查询的问题在于SELECT中*的含糊性。

SELECT sub.* FROM fm_submission AS sub LEFT JOIN fm_submission_saved AS saved ON sub.submissionID = saved.submissionID WHERE sub.archived = :archived AND saved.submissionID IS NULL ORDER BY sub.dateSubmitted LIMIT :start, :limit;

将sub引用添加到*允许数据从正确的位置提取并返回该数据。我不确定为什么没有错误，并且返回的某些数据没有问题。我想了解为什么它部分起作用，而不是失败或失败。

PHP MYSQL查询在带有值的列中返回null

1 个答案: