javascript - node-sql-parser ， parser.astify(sqlquery) 不返回准确的输出

Question

我正在尝试使用node-sql-parser在nodejs中解析sql查询。

const {Parser} = require('node-sql-parser');
const parser = new Parser();
const ast = parser.astify(sqlQuery)

使用的sqlery如下

SELECT * FROM SUBSCRIPTION WHERE COLUMN_1 = 'NORM' AND COLUMN_2 IN ('11','12','13','14','15','16','17','18','42','43','44','45','101','102','103','104','128','129','130','131','159') AND (COLUMN_3 IS NOT NULL OR COLUMN_4 = 1) AND COLUMN_5 <> 1 AND (COLUMN_6 <> '8088' AND COLUMN_7 <>  '9088') 

在解析时得到输出

{ with: null,
  type: 'select',
  options: null,
  distinct: null,
  columns: '*',
  from: [ { db: null, table: 'SUBSCRIPTION', as: null } ],
  where:
   { type: 'binary_expr',
     operator: 'AND',
     left:
      { type: 'binary_expr',
        operator: 'AND',
        left: [Object],
        right: [Object] },
     right:
      { type: 'binary_expr',
        operator: 'AND',
        left: [Object],
        right: [Object],
        parentheses: true } },
  groupby: null,
  having: null,
  orderby: null,
  limit: null }

我无法迭代left:[Object]。 有什么方法可以解析这个并只获取运算符列表吗？

Answer 1

您将其记录为对象，这就是您在其中看到对象的原因。您需要在控制台之前对对象进行 JSON.stringify(ast) 才能查看所有内容。

关于查找所有运算符，您可以使用此代码

const sqlQuery = "SELECT * FROM SUBSCRIPTION WHERE COLUMN_1 = 'NORM' AND COLUMN_2 IN ('11','12','13','14','15','16','17','18','42','43','44','45','101','102','103','104','128','129','130','131','159') AND (COLUMN_3 IS NOT NULL OR COLUMN_4 = 1) AND COLUMN_5 <> 1 AND (COLUMN_6 <> '8088' AND COLUMN_7 <>  '9088')";
const {Parser} = require('node-sql-parser');
const parser = new Parser();
const ast = parser.astify(sqlQuery)

function getOperator(obj, result = []) {
  let value;
  Object.keys(obj).forEach(key => {
    value = obj[key];
    if (value instanceof Object) {
      getOperator(value, result);
    } else if (key === "operator") {
      result.push(value);
    }
  });
  return result;
}

console.log(getOperator(ast));

// output will be -  ["AND", "AND","AND","AND","=","IN","OR","IS NOT","=","<>","AND","<>","<>"]