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Closed 11 years ago.
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What is the n+1 selects problem?
what is N+1 type queries ?
i don't know what is this?
there is nothing like N+1 type query
areyou talking about this -->
What is SELECT N+1?
see here:
What is SELECT N+1?
:)
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What is the meaning of <> in mysql query?
(7 answers)
Closed 3 years ago.
I know this is a really beginner question, but I don't know what the meaning of this. I'm reading the existing system code. I don't know what does this symbol means <> in Query Builder.
Here is the sample code:
$builder = DB::table('product');
if (isset($product->type)) {
$builder->where('product.type', '<>', $product->type);
}
Thanks!
In MySQL syntax, this is the same as not equal or !=.
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T-SQL dynamic pivot
(5 answers)
Why is processing a sorted array faster than processing an unsorted array?
(26 answers)
Closed 4 years ago.
I have spent an hour already on this problem.
I want to dynamically generate columns based on the values from the column AttendanceDate.
I have found some similar questions, but unfortunately the examples were too complicated for me to comprehend.
Data:
Expected output:
This can be done with the stuff method as mention in comments or with a while exists implementation:
http://rextester.com/FPU47008
Closed. This question needs debugging details. It is not currently accepting answers.
Edit the question to include desired behavior, a specific problem or error, and the shortest code necessary to reproduce the problem. This will help others answer the question.
Closed 4 years ago.
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I Couldn't find the value that has 'r' in the third position using LIKE operation in SQL.
could anyone help me, please?
I think it would be better to write something like this.
SELECT <column>, FROM <table> WHERE <condition> LIKE "__r%";
EDIT:
This is tested and functional for SQL Server and Oracle DBMSs.
SELECT * FROM <table> WHERE substring(<column>,3,1) = 'r'
Try this. This is using SQL Server which will yield the result you are looking for.
You want to get the r from 'scr' with position 3
Try this
SELECT * FROM Customers
WHERE CustomerName LIKE '__r%';
position is count of _.
This question already has an answer here:
Relational Algebra check for error
(1 answer)
Closed 7 years ago.
Is there a way to display a sql LIKE-operator in relational algebra?
right now I have done it like this:
Ďfirstname LIKE ('B%')person
But I'm not sure if this is the right way to go.
Thanks in advance for the help.
Here is a similar question covered earlier on this site regarding how to use LIKE in relational algebra. In that case, the professor had indicated that it was ok to use LIKE just as in SQL.
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How to search for a comma separated value
(3 answers)
Closed 8 years ago.
I have data in table in below format
id brand_ids
--------------
2 77,2
3 77
6 3,77,5
8 2,45,77
--------------
(Note the brand ids will be stored like comma separated values, this is common for values in this field)
Now i am trying to get a query which is capable of querying out only rows which have '77'
in it..
I know i can use LIKE command in three formats LIKE '77,%' OR LIKE '%,77,%' OR LIKE '%,77' with or condition to achieve it. But i hope this will increase the load time of the sql.
is there any straight forward method to achieve this? if so please suggest.
Thanks,
Balan
A strict answer to your question would be: no. Your suggestion of using LIKE is your best option with this data model. However, as mentioned, it is highly suggested that you use a more normalized model.