rawQuery, getting cursor to point to row with STRING in column

Question

I've been learning Java, using Eclipse for Android, for 2 months and have just started learning about Databases. I know I am doing something wrong here and the answer is simple but I just can't get it. And I know this type of question has been asked 1M times before. Sorry to ask it again.

I have a table called TABLE_NAME:

|-----|-------|-------|
|._id.|.ITEM..|.LAST..|
|..1..|...A...|.......|
|..2..|...B...|...L...|<--I want the cursor to point to this row which has "L" in column LAST 
|..3..|...C...|.......|
|-----|-------|-------|

I want the cursor to point to the row with the "L" in the "LAST" column. I am using the following code:

cursor = db.rawQuery("SELECT * FROM " +TABLE_NAME +" WHERE "+LAST+" LIKE '%"L"%'", null);

but I get a red line under the last bit of my code. Can anyone help?

Just figured out how to answer my own question:

Thanks everyone for the help. After changing '%"L"%'" to '%L%'" the new code is:

cursor = db.rawQuery("SELECT * FROM " +TABLE_NAME +" WHERE "+LAST+" LIKE '%L%'", null);

The I wrote: cursor.moveToFirst();<--- telling the cursor to go to the row the query found and then:

THESTRING =cursor.getString(cursor.getColumnIndex(LAST));<--- getting the info in the row

Thanks again everyone.

what red line? You have missed some of these -> } ??

mihail
– mihail

2012-06-28 09:32:02 +00:00
Commented Jun 28, 2012 at 9:32 — mihail
– mihail, Commented Jun 28, 2012 at 9:32

barrel · Accepted Answer · 2012-06-28 09:31:44Z

1

LIKE '%"L"%'"

Should be

LIKE '%L%'"

answered Jun 28, 2012 at 9:31

barrel

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5 Comments

Peter James Over a year ago

Thanks Barrel that solved the red line problem. I knew it had to be simple. It now runs but the pointer points to an index that is out of bounds, but I'll try and figure out why. thanks again

Peter James Over a year ago

Thanks everyone for the help. The code looks like this and works perfectly: cursor = db.rawQuery("SELECT * FROM " +TABLE_NAME +" WHERE "+LAST+" LIKE '%L%'", null); cursor.moveToFirst(); THESTRING =cursor.getString(cursor.getColumnIndex(LAST)); Thanks again everyone.

barrel Over a year ago

Glad I could help.It would be great if you can mark this post as answered as well :-)

Peter James Over a year ago

How do I make it as answered?

barrel Over a year ago

Just below the arrow flashes that allow you to give points, you can also accept an answer

Nermeen · Accepted Answer · 2012-06-28 09:31:35Z

0

It should be:

cursor = db.rawQuery("SELECT * FROM " +TABLE_NAME +" WHERE "+LAST+" LIKE '%"+L+"%'", null);

with the + around L variable..

answered Jun 28, 2012 at 9:31

Nermeen

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Comments

idiottiger · Accepted Answer · 2012-06-28 09:32:44Z

0

try this: you forget the +

 db.rawQuery("SELECT * FROM " + TABLE_NAME + " WHERE " + LAST + " LIKE '%" + L + "%'", null);

answered Jun 28, 2012 at 9:32

idiottiger

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Comments

Uku Loskit · Accepted Answer · 2012-06-28 09:37:11Z

0

There's syntactically this error:

cursor = db.rawQuery("SELECT * FROM " +TABLE_NAME +" WHERE "+LAST+" LIKE '%"+L+"%'", null);

You are missing two + for concatenation. Or like this if L is an actual value not a variable

cursor = db.rawQuery("SELECT * FROM " +TABLE_NAME +" WHERE "+LAST+" LIKE '%L%'", null);

edited Jun 28, 2012 at 9:37

answered Jun 28, 2012 at 9:30

Uku Loskit

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2 Comments

Peter James Over a year ago

Thanks Uku, I knew it had to be simple. I ran it and got an out of bounds error, when I tried to read what the cursor points, but I'm happy with getting % % problem solved. Thanks again.

Peter James Over a year ago

This is the best answer because it gives both cases:1: when L is a variable and 2: when L is an actual value.

Dheeresh Singh · Accepted Answer · 2012-06-28 09:46:59Z

0

|..2..|...B...|...L...|<--I want the cursor to point to this row which has "L" i

it should be

cursor = db.rawQuery("SELECT * FROM " +TABLE_NAME +" WHERE "+LAST+" LIKE '%L%'", null);

answered Jun 28, 2012 at 9:46

Dheeresh Singh

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