Question

当我尝试运行查询时，出现500内部错误。我想将query2结果作为数据插入query。

查询：

exports.create = async (req, res) => {
  try {
  const connection = await mysql.createConnection(config.mysql.credentials);
  var query2='select MAX(orders) from pdb_product';
  const query = `insert into pdb_product (product_code, description, active, third_party, 
  orders) values ("${req.body.product_code}", "${req.body.description}", ${(req.body.active == 'on' ? 1 : 0)}, ${(req.body.third_party == 'on' ? 1 : 0)}, ${query2});`;
  await connection.query(query);
  res.redirect('/products');
  } catch (e) {
    utils.error500(req, res, e.message);
  }
  };

我得到的错误：

您的SQL语法有误；请查看与您的MySQL服务器版本相对应的手册以获取正确的语法，以在第1行“从pdb_product选择MAX（订单）”附近使用

我该如何解决？

Answer 1

在query2参数之前，您缺少'（'括号：

const query = `insert into pdb_product (product_code, description, active, third_party, 
  orders) 
values ("${req.body.product_code}", 
    "${req.body.description}", 
     ${(req.body.active == 'on' ? 1 : 0)}, 
     ${(req.body.third_party == 'on' ? 1 : 0)}, 
     (${query2}));`;

如何将一个查询结果插入到另一查询作为输入

1 个答案: