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2 years ago

What is a commonly publicized password sql injection string?

1 answer:

7 0

The strings "or 1=1" and ""or ''=''" can be commonly used to trick an SQL WHERE clause into becoming true.

So if you specify ' or ''=' as a password, you can log in if the query string would be:

select username,pass from users where username='you' and password='' or ''='' limit 0,1;

You might be interested in

Arrays are described as immutable because they are two dimensional. are arranged sequentially. can be reordered. cannot be chang

Vlada [557]

Answer:

Arrays are described as immutable because they cannot be changed once they are defined. (D on Edge)

Explanation:

It's in the notes and I just took the test (2020)

6 0

2 years ago

8.Change the following IP addresses from binary notation to dotted-decimal notation: a.01111111 11110000 01100111 01111101 b.101

Andrews [41]

Answer:

a. 01111111 11110000 01100111 01111101 dotted decimal notation:

(127.240.103.125)

b. 10101111 11000000 11111000 00011101 dotted decimal notation: (175.192.248.29)

c. 11011111 10110000 00011111 01011101 dotted decimal notation:

(223.176.31.93)

d. 11101111 11110111 11000111 00011101 dotted decimal notation:

(239.247.199.29)

a. 208.34.54.12 class is C

b. 238.34.2.1 class is D

c. 242.34.2.8 class is E

d. 129.14.6.8 class is B

a.11110111 11110011 10000111 11011101 class is E

b.10101111 11000000 11110000 00011101 class is B

c.11011111 10110000 00011111 01011101 class is C

d.11101111 11110111 11000111 00011101 class is D

Explanation:

8 a. 01111111 11110000 01100111 01111101

we have to convert this binary notation to dotted decimal notation.

01111111 = 0*2^7 + 1*2^6 + 1*2^5 + 1*2^4 + 1*2^3 + 1*2^2 + 1*2^1 + 1*2^0

= 0 + 1*64 + 1*32 + 1*16 + 1*8 + 1*4 + 1*2 + 1

= 64 + 32 + 16 + 8 + 4 + 2 + 1

= 127

11110000 = 1*2^7 + 1*2^6 + 1*2^5 + 1*2^4 + 0*2^3 + 0*2^2 + 0*2^1 + 0*2^0

= 1*128 + 1*64 + 1*32 + 16 + 0 + 0 + 0 + 0

= 128 + 64 + 32 + 16

= 240

01100111 = 0*2^7 + 1*2^6 + 1*2^5 + 0*2^4 +0*2^3 + 1*2^2 + 1*2^1 + 1*2^0

= 0 + 1*64 + 1*32 + 0 + 0 + 4 + 2 + 1

= 64 + 32 + 4 + 2 + 1

= 103

01111101 = 0*2^7 + 1*2^6 + 1*2^5+ 1*2^4 +1*2^3 +1*2^2 + 0*2^1 + 1*2^0

= 0 + 1*64 + 1*32 + 1*16 + 1*8 + 1* 4 + 0 + 1

= 64 + 32 + 16 + 8 + 4 + 1

= 125

So the IP address from binary notation 01111111 11110000 01100111 01111101 to dotted decimal notation is : 127.240.103.125

b) 10101111 11000000 11111000 00011101

10101111 = 1*2^7 + 0*2^6 + 1*2^5 + 0*2^4 + 1*2^3 + 1*2^2 + 1*2^1 + 1*2^0

= 175

11000000 = 1*2^7 + 1*2^6 + 0*2^5 + 0*2^4 + 0*2^3 + 0*2^2 + 0*2^1 + 0*2^0

= 192

11111000 = 1*2^7 + 1*2^6 + 1*2^5 + 1*2^4 +1*2^3 + 0*2^2 + 0*2^1 + 0*2^0

= 248

00011101 = 0*2^7 + 0*2^6 + 0*2^5 + 1*2^4 +1*2^3 + 1*2^2 + 0*2^1 + 1*2^0

= 29

So the IP address from binary notation 10101111 11000000 11111000 00011101 to dotted decimal notation is : 175.192.248.29

c) 11011111 10110000 00011111 01011101

11011111 = 1*2^7 + 1*2^6 + 0*2^5 + 1*2^4 +1*2^3 + 1*2^2 + 1*2^1 + 1*2^0

= 223

10110000 = 1*2^7 + 0*2^6 + 1*2^5 + 1*2^4 +0*2^3 + 0*2^2 + 0*2^1 + 0*2^0

= 176

00011111 = 0*2^7 + 0*2^6 + 0*2^5 + 1*2^4 +1*2^3 + 1*2^2 + 1*2^1 + 1*2^0

= 31

01011101 = 0*2^7 + 1*2^6 + 0*2^5 + 1*2^4 +1*2^3 + 1*2^2 + 0*2^1 + 1*2^0

= 93

So the IP address from binary notation 11011111 10110000 00011111 01011101 to dotted decimal notation is :223.176.31.93

d) 11101111 11110111 11000111 00011101

11101111 = 1*2^7 + 1*2^6 + 1*2^5 + 0*2^4 +1*2^3 + 1*2^2 + 1*2^1 + 1*2^0

= 239

11110111 = 1*2^7 + 1*2^6 + 1*2^5 + 1*2^4 +0*2^3 + 1*2^2 + 1*2^1 + 1*2^0

= 247

11000111 = 1*2^7 + 1*2^6 + 0*2^5 + 0*2^4 +0*2^3 + 1*2^2 + 1*2^1 + 1*2^0

= 199

00011101 = 0*2^7 + 0*2^6 + 0*2^5 + 1*2^4 +1*2^3 + 1*2^2 + 0*2^1 + 1*2^0

= 29

So the IP address from binary notation 11101111 11110111 11000111 00011101 to dotted decimal notation is : 239.247.199.29

9. In order to the find the class check the first byte of the IP address which is first 8 bits and check the corresponding class as follows:

Class A is from 0 to 127

Class B is from 128 to 191

Class C is from 192 to 223

Class D is from 224 to 239

Class E is from 240 to 255

a. 208.34.54.12

If we see the first byte of the IP address which is 208, it belongs to class C as class C ranges from 192 to 223.

b. 238.34.2.1

If we see the first byte of the IP address which is 238, it belongs to class D as Class D ranges from 224 to 239.

c. 242.34.2.8

If we see the first byte of the IP address which is 242, it belongs to class E as Class E ranges from 240 to 255.

d. 129.14.6.8

If we see the first byte of the IP address which is 129, it belongs to class B as Class B ranges from 128 to 191.

10. In order to find the class of the IP addresses in easy way, start checking bit my bit from the left of the IP address and follow this pattern:

0 = Class A

1 - 0 = Class B

1 - 1 - 0 = Class C

1 - 1 - 1 - 0 = Class D

1 - 1 - 1 - 1 = Class E

a. 11110111 11110011 10000111 11011101

If we see the first four bits of the IP address they are 1111 which matches the pattern of class E given above. So this IP address belongs to class E.

b. 10101111 11000000 11110000 00011101

If we see the first bit is 1, the second bit is 0 which shows that this is class B address as 1 0 = Class B given above.

c. 11011111 10110000 00011111 01011101

The first bit is 1, second bit is 1 and third bit is 0 which shows this address belongs to class C as 110 = Class C given above.

d. 11101111 11110111 11000111 00011101

The first bit is 1, the second bit is also 1 and third bit is also 1 which shows that this address belongs to class D.

3 0

2 years ago

Jefferson is interested in starting his own business. He plans to borrow money from the local bank in order to finance the busin

sweet [91]

A business plan and a financial plan to show the both his plan to make money, and how much money he will make, spend, use.

8 0

2 years ago

Read 2 more answers

Write a loop that displays all possible combinations of two letters where the letters are 'a', or 'b', or 'c', or 'd', or 'e'. T

Dafna11 [192]

Answer:

for (char outerChar='a'; outerChar<='e'; outerChar++){

for (char innerChar='a'; innerChar<='e'; innerChar++){

cout << outerChar << innerChar << "\n";

}

3 0

1 year ago

Write a code segment to change the name of the Thing object something such that the new name consists of the old name with one c

ololo11 [35]

Answer:

import random

def name_change( name: list ) -> list :

choice = random.choice( name )

choice_index = name.index( choice )

name.pop( choice_index )

return name

Explanation:

The python source code above makes use of the random package to select an item from a list in the function "name_change" and the index of that item is gotten fro the list and popped from the list "name", then the name is returned.

To call the function, assign it to a variable and add a list ( must be a list ) as its argument then print the result.

6 0

1 year ago