Problem using Sql LIKE in Angularjs

Question

I have been a few days having trouble searching my app using AngularJS and today I was able to discover the reason: During the consultation SQL with LIKE, it is simply not receiving the parameter as it should.

My job is like this:

self.searchAll = function(text) {
  var parameters = [text];

  return DBA.query("SELECT id, place_name FROM tblPlaces WHERE place_name LIKE '%(?)%'", parameters)
    .then(function(result) {
      return DBA.getAll(result);
  });
}

I think because of (?) be glued to the %%, it does not recognize that there should fit the text parameter. I decided to do some tests and put some direct text in the query, without coming by parameter and it worked.

I also tried to concatenate the text, to see if this way would work, because if it worked I could try to pass the parameter this way. But also didn’t work, it returns me an error of Sqlite.

Then I had two questions: Is there the possibility of doing this search using LIKE and receiving this text by parameter the way it is there in the function? Or it would be better if I made a general SELECT, turned it into a JSON object and made use of the Angular filter?

Obs: This search I’m trying to do is on a screen that lists several categories and the user will be able to search for products in general.

Answer 1

I just tested something that hadn’t even crossed my mind: %% with the parameter already in the variable parameters. And it worked!

self.searchAll = function(nameSearch) {
  var parameters = ["%"+nameSearch+"%"];

  return DBA.query('SELECT id, place_name FROM tblPlaces WHERE place_name LIKE ?', parameters)
    .then(function(result) {
      return DBA.getAll(result);
});
}