Q & A programming language.-php Tutorial

Answers to white programming. 1. declare a number of five members (each member contains the member id, name, age, gender, and registration time information), and then present the data in a table, if the age is less than 18 ~ 30 young people, 30 ~ 60 shows middle-aged, 60 or more shows elderly, and count the number of people of different age groups

2. use the for loop to create a 9-9 Multiplication table

3. use the while loop ~ Locate the prime number between 500 --> array

4. $ year indicates that the leap year outputs 366; otherwise, the output is 365.

Reply to discussion (solution)

Job?
1.

 1, 'name' => 'A', 'age' => 17, 'addtime' => date ('Y-m-D ')), array ('id' => 2, 'name' => 'B', 'age' => 19, 'addtime' => date ('Y-m-D'), array ('id' => 3, 'name' => 'C ', 'age' => 31, 'addtime' => date ('Y-m-D'), array ('id' => 4, 'name' => 'D', 'age' => 62, 'addtime' => date ('Y-m-D ')), array ('id' => 5, 'name' => 'e', 'age' => 18, 'addtime' => date ('Y-m-D'),); echo'
 '; Echo'
 
 
  
  
'; Echo'
  
  '; $ Agegroup = array (); foreach ($ arr as $ v) {$ agetype = checkage ($ v ['age']); if (isset ($ agegroup [$ agetype]) {$ agegroup [$ agetype] ++;} else {$ agegroup [$ agetype] = 1;} echo' 
   
     '; Echo' 
    '; Echo' 
    '; Echo' 
    '; Echo' 
    '; Echo' 
    '; Echo' 
   ';} Echo' 
  
 
   
 
    
Id
 
    
Name
 
    
Age
 
    
Addtime
 
    
Age Group
 
   
'. $ V ['id'].'
'. $ V ['name'].'
'. $ V ['age'].'
'. $ V ['addtime'].'
'. $ Agetype .'
 
 
'; Echo' statistical result
'; Foreach ($ agegroup as $ k => $ v) {echo $ k.': '. $ v .'
';} Function checkage ($ age) {if ($ age <18) {return 'minor';} elseif ($ age >=18 & $ age <30) {return 'youth ';} elseif ($ age >=30 & $ age <60) {return 'middle-aged';} else {return 'elder ';}}?>

2.

echo '
 
 
  
  ';for($i=1; $i<=9; $i++){    echo '
  
  
     '; for($j=1; $j<=9; $j++){ echo ' 
    '; } echo ' 
   ';}echo ' 
  
 
   
'.$i.'*'.$j.'='.($i*$j).'
 
 
';

3.

4.

echo checkyear('2016');function checkyear($year){    if($year%4==0 && $year%100!=0 || $year%400==0){        return 366;    }else{        return 365;    }}