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Boolean Field in Oracle
...r using Y and NULL as the values. This makes for a very small (read fast) index that takes very little space.
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How to get the first five character of a String
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You can use Substring(int startIndex, int length)
string result = str.Substring(0,5);
The substring starts at a specified character position and has a
specified length. This method does not modify the value of the current
instance. Instead, it re...
How to overlay images
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<div class="container" style="position: relative">
<img style="z-index: 32; left: 8px; position: relative;" alt="bottom image" src="images/bottom-image.jpg">
<div style="z-index: 100; left: 72px; position: absolute; top: 39px">
<img alt="top image" src="images/top-image.jpg">...
Using LIMIT within GROUP BY to get N results per group?
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For me something like
SUBSTRING_INDEX(group_concat(col_name order by desired_col_order_name), ',', N)
works perfectly. No complicated query.
for example: get top 1 for each group
SELECT
*
FROM
yourtable
WHERE
id IN (SELECT
S...
SQL Server 2005 How Create a Unique Constraint?
...t to do it from a Database Diagram:
right-click on the table and select 'Indexes/Keys'
click the Add button to add a new index
enter the necessary info in the Properties on the right hand side:
the columns you want (click the ellipsis button to select)
set Is Unique to Yes
give it an appropriate...
C# Iterate through Class properties
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// the index of each item in fieldNames must correspond to
// the correct index in resultItems
var fieldnames = new []{"itemtype", "etc etc "};
for (int e = 0; e < fieldNames.Length - 1; e++)
{
newRecord
.GetType()
...
Facebook database design?
...s to make a composite primary key. A unique key, absolutely. The clustered index on that unique key, definitely. But I'd also put some sort of non-composite identity as the PK with a nonclustered index. That would allow other tables that need a "friend relationship ID" FK to easily tie to this table...
Shuffling a list of objects
...erm = list(range(len(list_one)))
random.shuffle(perm)
list_one = [list_one[index] for index in perm]
list_two = [list_two[index] for index in perm]
Numpy / Scipy
If your lists are numpy arrays, it is simpler:
import numpy as np
perm = np.random.permutation(len(list_one))
list_one = list_one[per...
TypeError: ObjectId('') is not JSON serializable
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from bson import json_util
import json
@app.route('/')
def index():
for _ in "collection_name".find():
return json.dumps(i, indent=4, default=json_util.default)
This is the sample example for converting BSON into JSON object. You can try this.
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Get top 1 row of each group
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@domanokz: no, it's not a subquery. If you have correct indexes then millions shouldn't be a problem. There are only 2 set based ways anyway: this and the aggregate (Ariel's solution). So try them both...
– gbn
Jul 27 '11 at 9:30
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