NetEase PHP Most Frequently Asked Latest Interview Questions Answers
How To Merge Values Of Two Arrays Into A Single Array?
You can use the array_merge() function to merge two arrays into a single array.
array_merge() appends all pairs of keys and values of the second array to the end of the first array. Here is a PHP script on how to use array_merge():
<?php
$lang = array("Perl", "PHP", "Java",);
$os = array("i"=>"Windows", "ii"=>"Unix", "iii"=>"Mac");
$mixed = array_merge($lang, $os);
print("Merged:\n");
print_r($mixed);
?>
This script will print:
Merged:
Array
(
[0] => Perl
[1] => PHP
[2] => Java
[i] => Windows
[ii] => Unix
[iii] => Mac
)
How To Randomly Retrieve A Value From An Array?
If you have a list of favorite greeting messages, and want to randomly select one of them to be used in an email, you can use the array_rand() function. Here is a PHP example script:
<?php
$array = array("Hello!", "Hi!", "Allo!", "Hallo!", "Coucou!");
$key = array_rand($array);
print("Random greeting: ".$array[$key]."\n");
?>
This script will print:
Random greeting: Coucou!
Can You Define An Argument As A Reference Type?
You can define an argument as a reference type in the function definition. This will automatically convert the calling arguments into references. Here is a PHP script on how to define an argument as a reference type:
<?php
function ref_swap(&$a, &$b) {
$t = $a;
$a = $b;
$b = $t;
}
$x = "PHP";
$y = "JSP";
print("Before swapping: $x, $y\n");
ref_swap($x, $y);
print("After swapping: $x, $y\n");
?>
This script will print:
Before swapping: PHP, JSP
After swapping: JSP, PHP
Can You Pass An Array Into A Function?
You can pass an array into a function in the same as a normal variable. No special syntax needed. Here is a PHP script on how to pass an array to a function:
<?php
function average($array) {
$sum = array_sum($array);
$count = count($array);
return $sum/$count;
}
$numbers = array(5, 7, 6, 2, 1, 3, 4, 2);
print("Average: ".average($numbers)."\n");
?>
This script will print:
Average: 3.75
How Arrays Are Passed Through Arguments?
Like a normal variable, an array is passed through an argument by value, not by reference. That means when an array is passed as an argument, a copy of the array will be passed into the function. Modipickzyng that copy inside the function will not impact the original copy. Here is a PHP script on passing arrays by values:
<?php
function shrink($array) {
array_splice($array,1);
}
$numbers = array(5, 7, 6, 2, 1, 3, 4, 2);
print("Before shrinking: ".join(",",$numbers)."\n");
shrink($numbers);
print("After shrinking: ".join(",",$numbers)."\n");
?>
This script will print:
Before shrinking: 5,7,6,2,1,3,4,2
After shrinking: 5,7,6,2,1,3,4,2
As you can see, original variables were not affected.
Can You Define An Array Argument As A Reference Type?
You can define an array argument as a reference type in the function definition. This will automatically convert the calling arguments into references. Here is a PHP script on how to define an array argument as a reference type:
<?php
function ref_shrink(&$array) {
array_splice($array,1);
}
$numbers = array(5, 7, 6, 2, 1, 3, 4, 2);
print("Before shrinking: ".join(",",$numbers)."\n");
ref_shrink($numbers);
print("After shrinking: ".join(",",$numbers)."\n");
?>
This script will print:
BBefore shrinking: 5,7,6,2,1,3,4,2
After shrinking: 5
What Is The Scope Of A Variable Defined In A Function?
The scope of a local variable defined in a function is limited with that function. Once the function is ended, its local variables are also removed. So you can not access any local variable outside its defining function. Here is a PHP script on the scope of local variables in a function:
<?php
?>
function myPassword() {
$password = "U8FIE8W0";
print("Defined inside the function? ". isset($password)."\n");
}
myPassword();
print("Defined outside the function? ". isset($password)."\n");
?>
This script will print:
Defined inside the function? 1
Defined outside the function?
How To Pad An Array With The Same Value Multiple Times?
If you want to add the same value multiple times to the end or beginning of an array, you can use the array_pad($array, $new_size, $value) function. If the second argument, $new_size, is positive, it will pad to the end of the array. If negative, it will pad to the beginning of the array. If the absolute value of $new_size if not greater than the current size of the array, no padding takes place. Here is a PHP script on how to use array_pad():
<?php
$array = array("Zero"=>"PHP", "One"=>"Perl", "Two"=>"Java");
$array = array_pad($array, 6, ">>");
$array = array_pad($array, -8, "---");
print("Padded:\n");
print(join(",", array_values($array)));
print("\n");
?>
This script will print:
Padded:
---,---,PHP,Perl,Java,>>,>>,>>
How To Join Multiple Strings Stored In An Array Into A Single String?
If you multiple strings stored in an array, you can join them together into a single string with a given delimiter by using the implode() function. Here is a PHP script on how to use implode():
<?php
$date = array('01', '01', '2006');
$keys = array('php', 'string', 'function');
print("A formated date: ".implode("/",$date)."\n");
print("A keyword list: ".implode(", ",$keys)."\n");
?>
This script will print:
A formated date: 01/01/2006
A keyword list: php, string, function
How To Define A User Function?
You can define a user function anywhere in a PHP script using the function statement like this: "function name() {...}". Here is a PHP script example on how to define a user function:
<?php
function msg() {
print("Hello world!\n");
}
msg();
?>
This script will print:
Hello world!
How To Invoke A User Function?
You can invoke a function by entering the function name followed by a pair of parentheses. If needed, function arguments can be specified as a list of expressions enclosed in parentheses. Here is a PHP script example on how to invoke a user function:
<?php
function hello($f) {
print("Hello $f!\n");
}
hello("Bob");
?>
This script will print:
Hello Bob!
How To Return A Value Back To The Function Caller?
You can return a value to the function caller by using the "return $value" statement. Execution control will be transferred to the caller immediately after the return statement. If there are other statements in the function after the return statement, they will not be executed. Here is a PHP script example on how to return values:
<?php
function getYear() {
$year = date("Y");
return $year;
}
print("This year is: ".getYear()."\n");
?>
This script will print:
This year is: 2006
How To Pass An Argument To A Function?
To pass an argument to a function, you need to:
•Add an argument definition in the function definition.
•Add a value as an argument when invoking the function.
Here is a PHP script on how to use arguments in a function():
<?php
function f2c($f) {
return ($f - 32.0)/1.8;
}
print("Celsius: ".f2c(100.0)."\n");
print("Celsius: ".f2c(-40.0)."\n");
?>
This script will print:
Celsius: 37.777777777778
Celsius: -40
How Variables Are Passed Through Arguments?
Like more of other programming languages, variables are passed through arguments by values, not by references. That means when a variable is passed as an argument, a copy of the value will be passed into the function. Modipickzyng that copy inside the function will not impact the original copy. Here is a PHP script on passing variables by values:
<?php
function swap($a, $b) {
$t = $a;
$a = $b;
$b = $t;
}
$x = "PHP";
$y = "JSP";
print("Before swapping: $x, $y\n");
swap($x, $y);
print("After swapping: $x, $y\n");
?>
This script will print:
Before swapping: PHP, JSP
After swapping: PHP, JSP
As you can see, original variables were not affected.
How To Pass Variables By References?
You can pass a variable by reference to a function by taking the reference of the original variable, and passing that reference as the calling argument. Here is a PHP script on how to use pass variables by references:
<?php
function swap($a, $b) {
$t = $a;
$a = $b;
$b = $t;
}
$x = "PHP";
$y = "JSP";
print("Before swapping: $x, $y\n");
swap(&$x, &$y);
print("After swapping: $x, $y\n");
?>
This script will print:
Before swapping: PHP, JSP
After swapping: JSP, PHP
As you can see, the function modified the original variable.
Note that call-time pass-by-reference has been deprecated. You need to define arguments as references.
How To Merge Values Of Two Arrays Into A Single Array?
You can use the array_merge() function to merge two arrays into a single array.
array_merge() appends all pairs of keys and values of the second array to the end of the first array. Here is a PHP script on how to use array_merge():
<?php
$lang = array("Perl", "PHP", "Java",);
$os = array("i"=>"Windows", "ii"=>"Unix", "iii"=>"Mac");
$mixed = array_merge($lang, $os);
print("Merged:\n");
print_r($mixed);
?>
This script will print:
Merged:
Array
(
[0] => Perl
[1] => PHP
[2] => Java
[i] => Windows
[ii] => Unix
[iii] => Mac
)
 |
| NetEase PHP Most Frequently Asked Latest Interview Questions Answers |
How To Randomly Retrieve A Value From An Array?
If you have a list of favorite greeting messages, and want to randomly select one of them to be used in an email, you can use the array_rand() function. Here is a PHP example script:
<?php
$array = array("Hello!", "Hi!", "Allo!", "Hallo!", "Coucou!");
$key = array_rand($array);
print("Random greeting: ".$array[$key]."\n");
?>
This script will print:
Random greeting: Coucou!
Can You Define An Argument As A Reference Type?
You can define an argument as a reference type in the function definition. This will automatically convert the calling arguments into references. Here is a PHP script on how to define an argument as a reference type:
<?php
function ref_swap(&$a, &$b) {
$t = $a;
$a = $b;
$b = $t;
}
$x = "PHP";
$y = "JSP";
print("Before swapping: $x, $y\n");
ref_swap($x, $y);
print("After swapping: $x, $y\n");
?>
This script will print:
Before swapping: PHP, JSP
After swapping: JSP, PHP
Can You Pass An Array Into A Function?
You can pass an array into a function in the same as a normal variable. No special syntax needed. Here is a PHP script on how to pass an array to a function:
<?php
function average($array) {
$sum = array_sum($array);
$count = count($array);
return $sum/$count;
}
$numbers = array(5, 7, 6, 2, 1, 3, 4, 2);
print("Average: ".average($numbers)."\n");
?>
This script will print:
Average: 3.75
How Arrays Are Passed Through Arguments?
Like a normal variable, an array is passed through an argument by value, not by reference. That means when an array is passed as an argument, a copy of the array will be passed into the function. Modipickzyng that copy inside the function will not impact the original copy. Here is a PHP script on passing arrays by values:
<?php
function shrink($array) {
array_splice($array,1);
}
$numbers = array(5, 7, 6, 2, 1, 3, 4, 2);
print("Before shrinking: ".join(",",$numbers)."\n");
shrink($numbers);
print("After shrinking: ".join(",",$numbers)."\n");
?>
This script will print:
Before shrinking: 5,7,6,2,1,3,4,2
After shrinking: 5,7,6,2,1,3,4,2
As you can see, original variables were not affected.
Can You Define An Array Argument As A Reference Type?
You can define an array argument as a reference type in the function definition. This will automatically convert the calling arguments into references. Here is a PHP script on how to define an array argument as a reference type:
<?php
function ref_shrink(&$array) {
array_splice($array,1);
}
$numbers = array(5, 7, 6, 2, 1, 3, 4, 2);
print("Before shrinking: ".join(",",$numbers)."\n");
ref_shrink($numbers);
print("After shrinking: ".join(",",$numbers)."\n");
?>
This script will print:
BBefore shrinking: 5,7,6,2,1,3,4,2
After shrinking: 5
What Is The Scope Of A Variable Defined In A Function?
The scope of a local variable defined in a function is limited with that function. Once the function is ended, its local variables are also removed. So you can not access any local variable outside its defining function. Here is a PHP script on the scope of local variables in a function:
<?php
?>
function myPassword() {
$password = "U8FIE8W0";
print("Defined inside the function? ". isset($password)."\n");
}
myPassword();
print("Defined outside the function? ". isset($password)."\n");
?>
This script will print:
Defined inside the function? 1
Defined outside the function?
How To Pad An Array With The Same Value Multiple Times?
If you want to add the same value multiple times to the end or beginning of an array, you can use the array_pad($array, $new_size, $value) function. If the second argument, $new_size, is positive, it will pad to the end of the array. If negative, it will pad to the beginning of the array. If the absolute value of $new_size if not greater than the current size of the array, no padding takes place. Here is a PHP script on how to use array_pad():
<?php
$array = array("Zero"=>"PHP", "One"=>"Perl", "Two"=>"Java");
$array = array_pad($array, 6, ">>");
$array = array_pad($array, -8, "---");
print("Padded:\n");
print(join(",", array_values($array)));
print("\n");
?>
This script will print:
Padded:
---,---,PHP,Perl,Java,>>,>>,>>
How To Join Multiple Strings Stored In An Array Into A Single String?
If you multiple strings stored in an array, you can join them together into a single string with a given delimiter by using the implode() function. Here is a PHP script on how to use implode():
<?php
$date = array('01', '01', '2006');
$keys = array('php', 'string', 'function');
print("A formated date: ".implode("/",$date)."\n");
print("A keyword list: ".implode(", ",$keys)."\n");
?>
This script will print:
A formated date: 01/01/2006
A keyword list: php, string, function
How To Define A User Function?
You can define a user function anywhere in a PHP script using the function statement like this: "function name() {...}". Here is a PHP script example on how to define a user function:
<?php
function msg() {
print("Hello world!\n");
}
msg();
?>
This script will print:
Hello world!
How To Invoke A User Function?
You can invoke a function by entering the function name followed by a pair of parentheses. If needed, function arguments can be specified as a list of expressions enclosed in parentheses. Here is a PHP script example on how to invoke a user function:
<?php
function hello($f) {
print("Hello $f!\n");
}
hello("Bob");
?>
This script will print:
Hello Bob!
How To Return A Value Back To The Function Caller?
You can return a value to the function caller by using the "return $value" statement. Execution control will be transferred to the caller immediately after the return statement. If there are other statements in the function after the return statement, they will not be executed. Here is a PHP script example on how to return values:
<?php
function getYear() {
$year = date("Y");
return $year;
}
print("This year is: ".getYear()."\n");
?>
This script will print:
This year is: 2006
How To Pass An Argument To A Function?
To pass an argument to a function, you need to:
•Add an argument definition in the function definition.
•Add a value as an argument when invoking the function.
Here is a PHP script on how to use arguments in a function():
<?php
function f2c($f) {
return ($f - 32.0)/1.8;
}
print("Celsius: ".f2c(100.0)."\n");
print("Celsius: ".f2c(-40.0)."\n");
?>
This script will print:
Celsius: 37.777777777778
Celsius: -40
How Variables Are Passed Through Arguments?
Like more of other programming languages, variables are passed through arguments by values, not by references. That means when a variable is passed as an argument, a copy of the value will be passed into the function. Modipickzyng that copy inside the function will not impact the original copy. Here is a PHP script on passing variables by values:
<?php
function swap($a, $b) {
$t = $a;
$a = $b;
$b = $t;
}
$x = "PHP";
$y = "JSP";
print("Before swapping: $x, $y\n");
swap($x, $y);
print("After swapping: $x, $y\n");
?>
This script will print:
Before swapping: PHP, JSP
After swapping: PHP, JSP
As you can see, original variables were not affected.
How To Pass Variables By References?
You can pass a variable by reference to a function by taking the reference of the original variable, and passing that reference as the calling argument. Here is a PHP script on how to use pass variables by references:
<?php
function swap($a, $b) {
$t = $a;
$a = $b;
$b = $t;
}
$x = "PHP";
$y = "JSP";
print("Before swapping: $x, $y\n");
swap(&$x, &$y);
print("After swapping: $x, $y\n");
?>
This script will print:
Before swapping: PHP, JSP
After swapping: JSP, PHP
As you can see, the function modified the original variable.
Note that call-time pass-by-reference has been deprecated. You need to define arguments as references.
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