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Babys Day Out Babys Day Out 1994 DVDRip DiVX111

 

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Babys Day Out Babys Day Out 1994 DVDRip DiVX111
Oct 17, 2018
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Babys Day Out Babys Day Out 1994 DVDRip DiVX111
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References

Category:1972 births
Category:Living people
Category:Romanian film directors
Category:Romanian screenwriters
Category:Male screenwriters
Category:Romanian male writers
Category:People from Bucharest
Category:Cinema of Romania14)/(2 – 0). Solve -4*n – 2 = h for n.
-1
Suppose -4*h = -4*d + 4, -4*h – 2*d + 8 = -0*d. Solve h*i = 2*i – 4 for i.
2
Let a(p) = -p**3 + 6*p**2 + 6*p + 7. Let t be a(7). Suppose -w + 3*w = t. Let c = 3 – 2. Solve -3*h + 4 + c = w for h.
1
Suppose -5*o + 60 = 20. Suppose 9*m – o*m = 30. Suppose -m = -3*z – 2*z. Solve -j = z*j for j.
0
Let v be 6/(-4)*(-16)/(-6). Let m = v + 5. Let y be -4*(m/(-2))/3. Solve 0 = -2*n + y*n – 4 for n.
-4
Suppose 9 = -2*s – 15. Let x = s + 19. Suppose -3*f – 2*f = 0. Solve f = -7*b

 

mQ:

How to extract last day’s weekday from a string using sed?

I am trying to write a script to extract last day’s weekday name from a string using sed. For example, “Wednesday”, “Tuesday”, “Saturday”, “Thursday”
I need to extract the last one. I have tried the following:
$ echo “The last weekday of this month is Thursday” | sed -e’s/\w\w$//’
Thursday

$ echo “The last weekday of this month is Wednesday” | sed -e’s/\w\w$//’
Wednesday

But it seems it doesn’t work for other examples like “The week is Thursday”.
Could anyone help? Thanks.

A:

Use
sed -n’s/.*\([^ ]*\)\(.*\).*/\1\2/p’

See the online demo
See the output:
The week is Thursday
Friday
Friday
Thursday
Wednesday
Wednesday
Tuesday
Saturday
Thursday

A:

You can use awk for this task:
$ echo “The last weekday of this month is Thursday” | awk ‘{print $NF}’
Thursday

Here’s the explanation of the above code:

NF – Number of fields in current record
$NF – Last field in current record (see answer above)

UPDATE:
Explanation of the below script in detail (comments are inline):

NF – Number of fields in current record
$ – end of input
\(…\) – capturing group which will be used later in a regex
[^ ]* – any 0 or more chars other than whitespaces
\(…\) – capturing group which will be used later in a regex
.* – any 0 or more chars
\) – end of capturing group
\1 – first capturing group’s value
.* – any 0 or more chars
\2 – second capturing group’s value
p – print the result

The output:
$ echo “The last weekday of this month is Thursday” | awk ‘{print $NF}’
Thursday

A:

$¬†grep -Po ‘.*?([^ ]*)\s.*?(\s\1)?’
54b84cb42d

 

 

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