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zalisa [80]
4 years ago
10

The mutating-table error is raised when a trigger attempts to execute a. an INSERT, UPDATE, or DELETE while another user is upda

ting the same table b. a SELECT statement on the same table that the trigger is based on c. a SELECT statement while another user is updating the same table d. an INSERT, UPDATE, or DELETE on the same table that the trigger is based on
Computers and Technology
1 answer:
maw [93]4 years ago
8 0

Answer:

d. an INSERT, UPDATE, or DELETE on the same table that the trigger is based on

Explanation:

In Computer programming, the mutating-table error is raised when a trigger attempts to execute an INSERT, UPDATE, or DELETE on the same table that the trigger is based on.

The mutating-table error occurs when a row-level trigger attempts to read or write on the same table that the trigger is based on.

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Define an iterative function named append_ordered; it is passed two arguments: a linked list (ll) whose values are ordered from
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Answer:

A python programming language (code) was used for this exercise.

The code is shown below in the explanation section

Explanation:

Solution

Python code:

class LN:

def __init__(self,value,next=None):

self.value = value

self.next = next

  def list_to_ll(self,l):

if l == []:

return None

front = rear = LN(l[0])

for v in l[1:]:

rear.next = LN(v)

rear = rear.next

return front

  def str_ll(self,ll):

answer = ''

while ll != None:

answer += str(ll.value)+'->'

ll = ll.next

return answer + 'None'

def append_ordered(self,ll,v):

ln1=LN(v); #create a node with the new value by creating an instance of the LN class

if ll is None: #if the linked list is empty

ll=ln1 #return the new created node

elif(ll.value >= ln1.value): #if the value to be add is smallest, append the linked list to it

ln1.next=ll

ll=ln1

else:

current=ll; #crate a temporary pointer to iterate over linked list

#iterate till the next node value is smaller than the new value and list is not over

while (current.next is not None and current.next.value < ln1.value):

current=current.next

ln1.next=current.next #save the current node in the next of the new node

current.next=ln1 #save new node in the next of current node

return ll

ln=LN(0); #create an instance of class LN

ll1=ln.list_to_ll([1,2,4,5]); #create a linked list of the list

answer=ln.str_ll(ll1);    

print(answer); #print the linked list

ll2=ln.append_ordered(ll1,3); #append the value 3 in the linked list

answer=ln.str_ll(ll2)

print(answer); #print the new linked list

Now,

  • If Linked list is empty then make the node as head and return it.
  • If value of the node to be inserted is smaller than value of head node, then insert the node at start and make it head.
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  • Insert the node (10) after the appropriate node (8) found in step 3.

Note: Kindly find the output python code attached to the question below.

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Answer:

Explanation:

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