VB experts come in to guide!!!!

And this paragraph.

with combo1

do while not

i = i + 1

additem trim(mador!user_id) 'The username in the database is added to the combo box.

The pointer moves down. loopend with

There's no need to use with, it's just a property.

Changed to: do while not

i = i + 1

trim(mador!user_id) 'The username in the database is added to the combo box.

The pointer moves down. loop doesn't need to use with at all

dim mador as 'Declare recordset variables.

Amend the sentence to: dim mador as new'Declare recordset variables.

The problem may be with the executesql function.

You didn't open the database.

The form load module is changed as follows:

private sub form_load()dim i as integer

i = 0txtsql = "select * from user_form" 'Set up query commands.

set mador = executesql(txtsql, msgtext) 'Query the user.

do while not

i = i + 1

trim(mador!user_id) 'The username in the database is added to the combo box.

The pointer moves down. okf = false 'Set the marker.

countn = 0

end sub

Give it a try!

Single selection 2 a, 0 3 b、good 4 .a、pi*2*r6. c、p and q mod 2=0 and y mod 2=0 and z mod 2=0

8. b、word 9.No program 10 、s=1!+2!+3! +10!Multiple choices. c

ba，b，c

ca, b, d judgment.

(10*x+sqr(3*y)/(x*y)

b+sqr(b^2+4*a*c)/(4*a*c)1/(1/r1+1/r2+1/r3)

sin(45* (

Logarithmic arithmetic: With the log function, log(n) returns the natural logarithmic value of n (double). To calculate the logarithmic value of base x with n as the base, write log(x) log(n).

At first arr(1) = 0

In this case, arr(1) = 1 (change).

When i=1 arr(1)=arr(1)+1=1 , arr(2)=arr(1)+1=2 , arr(3)=arr(1)+1=2

At this point arr(2)=3 (change).

When i=2 arr(1)=arr(2)+1=3 , arr(2)=arr(2)+1=3 , arr(3)=arr(2)+1=4

At this point, arr(3)=5 (change).

When i=3 arr(1)=arr(3)+1=5 , arr(2)=arr(3)+1=5 , arr(3)=arr(3)+1=5

The result is: 5

i=0 cycle, a(0)=1, t=1, a(1)=a(0)=1 a(4)=1, output a(0)=1;

i=1 cycle, a(1)=2, t=2, a(2)=a(1)=2 a(4)=1, output a(1)=2;

i=2 cycle, a(2)=3, t=3, output a(2)=3; and make a(2)=a(0)=1; a(4)=1, output a(2)=1;

i=3 cycle, a(3)=4, t=4, a(4)=a(3)=4, a(4)=a(0)=1, output a(3)=4;

i=4 cycle, a(4)=4, t=4, a(3)=a(2)=1, a(4)=1 (direct assignment), output a(4)=1;

Therefore, it is 123141Communicate with each other, and ask you and other masters to criticize and correct inappropriately.

1. The output is: 3 5 7

2. The output is: 12

3. The value of x is: 25