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Creating Unique Names


Creating Unique Names

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Creating Unique Names
Tag : python , By : Matt
Date : November 24 2020, 01:01 AM

like below fixes the issue One thing you could do is to compute a unique hash of the files, e.g. with MD5 or SHA1 (or any other), cf. this article. For a large number of files this can become quite slow, though.
But you don't really see to touch the files in this piece of code. For generating some unique id, you could use uuid and put this somewhere in the name.

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PHP: Creating unique directory names


Tag : php , By : Bobblegate
Date : March 29 2020, 07:55 AM
this one helps. Fixed it - i was confusing the vars within the loop.
Posting for future reference:
if(!empty($_POST['newdir'])) {
    $dir_base= preg_replace("/[^a-zA-ZÀ-ÿ0-9\s]/",'',$_POST['newdir']);
    $dir_path=$pattern.$dir_base;
    $dir_path_lwr = strtolower($dir_path);

    $old_dirs = glob($pattern.'*', GLOB_ONLYDIR);
    $old_dirs_lwr = array_map('strtolower', $old_dirs);

    $i = 1;
    $cond = true;
    while($cond) {

        if(in_array($dir_path_lwr, $old_dirs_lwr)) {
            $i++;
            $new_base=$dir_base.' '.$i.'';
            $dir_path= $pattern.$new_base;
            $dir_path_lwr = strtolower($dir_path);
        }
        elseif(!in_array($dir_path_lwr, $old_dirs_lwr)) {
            mkdir($dir_path, 0755);
            $cond = false;
        }
    }
}

Creating unique variable names in PHP


Tag : php , By : Grant
Date : March 29 2020, 07:55 AM
it helps some times I have a script which can be used multiple times per script, each instance of the script requires some php stuff to be done, I need to create a unique name for each instance variable e.g. , You really should use arrays, as MightyPork suggested...:
$typedbefore = array();
$typedbefore[$text] = "value";

Creating objects from a class with unique names in Python3


Tag : python , By : user181706
Date : March 29 2020, 07:55 AM
this will help You can use globals() function to get the dictionary of global namepsace, and then create your variables using string.
Example -
In [19]: class Enemy:
   ....:        pass
   ....:

In [20]: gbl = globals()

In [21]: for i in range(1,6):
   ....:     gbl['enemy' + str(i)] = Enemy()
   ....:

In [22]: enemy1
Out[22]: <__main__.Enemy at 0x4903a58>

In [23]: enemy2
Out[23]: <__main__.Enemy at 0x4903a90>

In [24]: enemy3
Out[24]: <__main__.Enemy at 0x4903b00>

In [25]: enemy4
Out[25]: <__main__.Enemy at 0x4903b70>

In [26]: enemy5
Out[26]: <__main__.Enemy at 0x4903be0>

Creating two dictionary variables within an iteration with unique names


Tag : pandas , By : Verbal
Date : March 29 2020, 07:55 AM
it should still fix some issue The problem is that return gives you just one value -- the last dictionary that's created. You can use yield to generate a succession of dictionaries like this:
def slicing(stock, sliced_data, num_of_days):  
    time_delta = dt.timedelta(days =num_of_days)
    for i in stock:   # stock name
        vars()['mini_dfs' + i] = {}
        for date in sliced_data[i].index: 
            start_date = date
            end_date = date + time_delta
            vars()['mini_dfs' + i][date] = stocks[i].loc[start_date:end_date]
        yield vars()['mini_dfs' + i]  
my_list = [i for i in slicing(['AAPL','OPK'], High_volume, 1)]

Creating view error column names must be unique


Tag : sql , By : yatil
Date : March 29 2020, 07:55 AM
I wish this helpful for you Hi I am trying to resolve an error with creating view. My view is as follows: , Don't use *. List out all the columns:
CREATE VIEW [dbo].[vw_Employee_Data] AS
    SELECT e.*, m.EmpID AS MGREmpID, m.BaseLID AS MGRBaseLID
    FROM [dbo].Employee e INNER JOIN
         [dbo].Manager m
         ON e.NewEmp = m.EmpID;
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