JPA Like Integer运算符 [英] JPA Like Operator with Integer

问题描述

有人可以告诉我，为什么这不起作用：

Can someone tell me, why this is not working:

criteria.add(cb.like((myentity.<Integer>get("integerid")).as(String.class), "2%"))

我收到以下错误：

The object [2%], of class [class java.lang.String], from mapping [org.eclipse.persistence.mappings.DirectToFieldMapping[integerid-->MYENTITY.INTEGERID]] with descriptor [RelationalDescriptor(org.example.model.MyEntity --> [DatabaseTable(MYENTITY)])], could not be converted to [class java.lang.Integer]

是唯一可以改变<$的解决方案c $ c> myinteger 属性为模型中的字符串，以便我可以使用之类的运算符？

Is the only solution to change the myinteger property to a string in the model, in order that I can use the like operator?

BR

推荐答案

JPA不支持带有整数的like（），只支持字符串。有些数据库确实支持整数，有些数据库不支持。

JPA does not support like() with Integers, only Strings. Some databases do support like with integer and others do not.

EclipseLink应该允许在Integer中使用like（）（假设你的数据库支持它）。你用的是什么版本？可能需要使用> = 2.1。如果它在最新版本上失败，那么请记录一个错误。

EclipseLink should allow usage of like() with Integer (provided your database supports it). What version are you using? May need to use >= 2.1. If it fails on the latest version, then please log a bug.

您还可以使用CHAR，TO_CHAR或CONVERT将整数转换为字符串函数取决于您的数据库。条件API支持函数（）API来调用本机函数。

You can also convert the integer to a string using a "CHAR, "TO_CHAR" or "CONVERT" function depending on your database. The criteria API supports a function() API to call a native function.

请注意，as（）API不适用于从Integer转换to String，用于转换为子类实体。

Note that the as() API is not intended for converting from Integer to String, it is for casting to a subclass entity.

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