在Sequelize.js中使用动态搜索参数

Question

我正试图在他们的网站上关注Sequelize教程.

我已达到以下代码行.

Project.findAll({where: ["id > ?", 25]}).success(function(projects) {
  // projects will be an array of Projects having a greater id than 25
})

如果我稍微调整它如下

Project.findAll({where: ["title like '%awe%'"]}).success(function(projects) {
    for (var i=0; i<projects.length; i++) {
        console.log(projects[i].title + " " + projects[i].description);
    }
});

一切正常.但是,当我尝试使搜索参数动态如下

Project.findAll({where: ["title like '%?%'", 'awe']}).success(function(projects) {
    for (var i=0; i<projects.length; i++) {
        console.log(projects[i].title + " " + projects[i].description);
    }
});

它不再返回任何结果.我怎样才能解决这个问题？

Answer 1

现在在Sequelize你可以尝试这个

{ where: { columnName: { $like: '%awe%' } } }

有关更新的语法,请参见http://docs.sequelizejs.com/en/latest/docs/querying/#operators

Answer 2

我想你会这样做:

where: ["title like ?", '%' + 'awe' + '%']

因此,如果您使用实际变量进行此操作,则使用:

Project.findAll({where: ["title like ?", '%' + x + '%']}).success(function(projects) {
    for (var i=0; i<projects.length; i++) {
        console.log(projects[i].title + " " + projects[i].description);
    }
});

Answer 3

请尝试此代码

const Sequelize = require('sequelize');
const Op = Sequelize.Op;
{ where: { columnName: { [Op.like]: '%awe%' } } }