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从Rest Get调用解析where子句查询

[英]Parse where clause query from Rest Get call

I am receiving a where clause from Rest API Get and I should convert properties, So I need to convert this string to java object with logical and .... for example : 我从Rest API Get接收到where子句,并且应该转换属性,因此我需要将此字符串转换为具有逻辑和....的java对象,例如:

String where = "prop = 1 and prop2 = 'ssdf' or date > 20121204";

Is there any java library that converts where clause to java object with separate condition and operator ? 是否有任何Java库可以将where子句转换为具有单独条件和运算符的java对象?

Finally I found this library : https://github.com/JSQLParser/JSqlParser 终于我找到了这个库: https : //github.com/JSQLParser/JSqlParser

Seems like very good for query parsing 看起来非常适合查询解析

Using JSqlParser one could use an already implemented utility method to parse conditions like in your where clause: 使用JSqlParser可以使用一种已经实现的实用程序方法来解析条件,例如在where子句中:

String where = "prop = 1 and prop2 = 'ssdf' or date > 20121204";
Expression expr = CCJSqlParserUtil.parseCondExpression(where);
System.out.println(expr);

This works with JSqlParser V0.9.x ( https://github.com/JSQLParser/JSqlParser ) 这适用于JSqlParser V0.9.xhttps://github.com/JSQLParser/JSqlParser

Then you have an object tree with separated operators, ands, ors, values. 然后,您将获得一个对象树,其中包含分开的运算符ands,ors值。 This one you could traverse using the ExpressionVisitorAdapter, eg 您可以使用ExpressionVisitorAdapter遍历这一行,例如

ExpressionVisitorAdapter visitor = new ExpressionVisitorAdapter() {
    @Override
    public void visit(Column column) {
        System.out.println(column);
    }
};

expr.accept(visitor);

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