Question

我在项目中使用angular2-query-builder。此外，后端项目使用Java，Spring boot和Rulebook作为规则引擎，以触发由angular2-query-builder生成的规则。因此，我需要从angular2-query-builder的输出中做出一个逻辑表达式。

这是由查询生成器生成的类型脚本对象。

var members = [
  { name: "Sue", hobby: ["Swimming", "Running", "Dancing"] },
  { name: "Sam", hobby: ["Cars", "Travelling"] },
  { name: "John", hobby: ["Reading", "Swimming"] },
  { name: "Rob", hobby: ["Running", "Coding"] },      
];
var swimmers = members.filter(({hobby}) => hobby.includes('Swimming')).map(({name}) => name);
console.log(swimmers)

我在后端创建的映射模型类：

  query = {
    condition: 'and',
    rules: [
      {field: 'age', operator: '<=', value: '16'},
      {field: 'gender', operator: '=', value: 'm'}
    ]
  };

我需要从对象中获取RuleSet { String field; String value; String operator; String entity; String condition; RuleSet[] rules; //...getters and setters}。

Answer 1

从ramin，typeScript实现中引用相同的内容

首先将模型定义为：

export class RuleSet {
    field?  : string;
    value? : string | string[] | number;
    operator? : string;
    entity ?: string;
    condition ?: string;
    rules ?: RuleSet[];
}

然后定义解码功能：

  //params : [{field: 'age', operator: '<=', value: '16'},{field: 'gender', operator: 
  //'=', value: 'm'}],'and'
  decodeQuery(rules : RuleSet[], condition : string) {
    for (var i = 0; i < rules.length; i++) {
        var item :RuleSet = rules[i];
        if (item.rules == null) {
            this.expression = this.expression + this.mergeString("(", item.field, item.operator, item.value, ")");
            if (i != rules.length - 1) {
                this.expression = this.expression + " " + condition + " ";
            }
        } else {
            condition = item.condition;
            this.expression = this.expression + "(";
            this.decodeQuery(item.rules, condition);
            this.expression = this.expression + ")";
        }
    }
  }

  mergeString(bracket1,field,operator,value,bracket2) {
    var mergedString = bracket1 + " " + field + " " + operator + " " + value + " " + bracket2 ;
    return mergedString;
  }

称呼它为：

expression : string = "";
this.decodeQuery(query.rules,query.condition);

结果输出如下：

(age <= 16) && (gender = 'm')

Answer 2

制定一种递归方法以遍历规则

private void decode(RuleSet[] rules, List<String> expressions, String condition) {
        for (int i = 0; i < rules.length; i++) {
            RuleSet item = rules[i];
            if (item.rules == null) {
                expressions.add(merge("(", item.field, item.operator, item.value, ")"));
                if (i != rules.length - 1) {
                    expressions.add(condition);
                }
            } else {
                condition = item.getCondition();
                expressions.add("(");
                decode(item.rules, expressions, condition);
                expressions.add(")");
            }
        }
    }
public String merge(String... strings) {
    StringBuilder result = new StringBuilder();
    for (String item : strings) {
        if (item != null) {
            result.append(item);
        }
    }
    return result.toString();
}

然后这样称呼它：

List<String> expressions = new ArrayList<>();
decode(querySchema.ruleSet.rules, expressions, querySchema.ruleSet.condition);

通过将表达式的所有项添加到字符串对象中。将会生成逻辑表达式。

将angular2-query-builder查询转换为逻辑表达式

2 个答案: