Question

123

Using dynamic search parameters with Sequelize.js

rated 0 times [ 127] [ 4] / answers: 1 / hits: 24870 / 13 Years ago, fri, february 17, 2012, 12:00:00

I'm trying to follow the Sequelize tutorial on their website.

I have reached the following line of code.

Project.findAll({where: [id > ?, 25]}).success(function(projects) {

  // projects will be an array of Projects having a greater id than 25

})

If I tweak it slightly as follows

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);

    }

});

everything works fine. However when I try to make the search parameter dynamic as follows

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);

    }

});

It no longer returns any results. How can I fix this?

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tyreem

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3 Years ago

answered 13 Years ago katharinek · Accepted Answer

I think you would do that like this:

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

So if you were doing this with an actual variable you'd use:

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);

    }

});