cant use like clause in android app

Question

I am working on a database with sqllite in an android app I want to retrieve sm data using a like clause ex:

Cursor c = myDB.query(MY_DATABASE_TABLE, null,          


        

Answer 1

a simple way is to use the concatenate operator ||

"SongName LIKE '%'||?||'%'"

so there is no need to edit the match_str.

Cursor c = myDB.query(MY_DATABASE_TABLE, null, " SongName LIKE '%'||?||'%'" , 
       new String[]{match_str}, null, null,"SongHit  DESC");

from https://stackoverflow.com/a/46420813/908821

Answer 2

If you wish to use query() instead of rawQuery(), you can do it like so:

// searchString is the string to search for
final String whereClause = "MyColumnName" + " like ?";
String[] whereArgs = new String[]{"%" + searchString + "%"};

Cursor cursor = sqLiteDatabase.query("MyTableName",
        null, // columns
        whereClause, // selection
        whereArgs, // selectionArgs
        null, // groupBy
        null, // having
        null, // orderBy
        null); // limit

Answer 3

hi Are you getting any exception while executing the above query. Try by removing the "=" symbol infront of the question mark in the like condition. Else try using

db.rawQuery(sql, selectionArgs)

Answer 4

This works perfectly for me...

"Songname LIKE ? "...

cursor = database.rawQuery(query, new String[]{"%" + searchTerm + "%"});

Answer 5

Try this:

String[] args = new String[1];
args[0] = "%"+song+"%";
Cursor friendLike = db.rawQuery("SELECT * FROM songs WHERE SongName like ?", args);

Answer 6

This:

" SongName LIKE '%"+"=?"+"%'"

Will end up looking like this when the SQL interpreter sees it:

" SongName LIKE '%=?%'"

And that will match any SongName that contains a literal "=?" and I don't think that's anything like what you want.

A % matches any sequence of characters in an SQL LIKE, it is essentially the same as .* in a regular expression; so, if you want to match at the beginning then you don't want a leading %. Also, your ? placeholder needs to be outside the single quotes or it will be interpreted as a literal question mark rather than a placeholder.

You want something more like this:

String[] a = new String[1];
a[0]       = match_str + '%';
Cursor   c = myDB.rawQuery("SELECT * FROM songs WHERE SongName LIKE ?", a);

If you wanted to be really strict you'd also have to escape % and _ symbols inside match_str as well but that's left as an exercise for the reader.