I want to learn PHP well for the fun of making the yellow FILM.

Last Update:2016-08-05 Source: Internet

Author: User

Tags learn php

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The new project was laravel, so I finally started to happily learn the best language in the WORLD.Learning makes me happy, I love learning:) (i Use this sentence to express, touch my attitude to it for two days, ps: I always believe that Javascript will eventually rule the world! ）This article mainly explains two questions:1, PHP Reference Problem (&);2, php Object Inheritance Problem.first, PHP references:Add & symbols to variables or functions, objects, and so on, with different names accessing the same variable Content.<pre>1<?2 $a= "ABC";3 $b=&$a;4 Echo $a;//Output here: ABC5 Echo $b;//Output here: ABC6 $b= "EFG";7 Echo $a;//here a $ A value becomes EFG so the output EFG8 Echo $b;//output EFG here9?></pre>Line 3rd $b quoted the $a, which means that $b and $a point to the same variable (two variables pointing to the same content), $b have the value of $a, and can change the $b by changing the $a value, which is why 7, 8 rows output EFG.Look again at an example, well, see examples also make me happy:)<pre>1<?PHP2 function&Test ()3 {4 Static $b= 0;//declare a static variable5 $b=$b+1;6 Echo $b;7 return $b;8 }9 Ten $a=test ();//This statement will output a value of 1 for the $b one $a=5; a $a=test ();//This statement will output a value of 2 for the $b - - $a=&test ();//This statement will output a value of 3 for the $b the $a=5; - $a=test ();//This statement will output a value of 6 for the $b -?></pre>Line 10th, $a call the test method, because there is no add &, that is, is not a reference, so just return the function value, assigned to the $a, output 1.Line 12th, Although $a in line 11th, is assigned a value of 5, but because it is still not a reference, so there is no ability to change the value of the $b test method, the call can only return the original function value, because in line 10th when the $b is equal to 1, so the call output value is 2.Lines 14th and 15 call the function by $a = &test (), which is to point the memory address of the $b variable in the return $b to the same place as the memory address of the $a variable. This time changes the value of the $a and also changes the value of the $B. The 16th row then outputs 6.The second problem is the problem with the PHP object inheritance:<pre>1 /**2 * Define MyClass3 */4 classMyClass5 {6 //declaring a public constructor7 public function__construct () {}8 9 //to declare a public methodTen public functionmypublic () {} one a //declaring a protected method - protected functionmyprotected () {} - the //declaring a private method - Private functionmyprivate () {} - - //This method is public + functionFoo () - { + $this-mypublic (); a $this-myprotected (); at $this-myprivate (); - } - } - - $myclass=NewMyClass; - $myclass->mypublic ();//This line can be executed normally, and it can call its public method Directly. in $myclass->myprotected ();//This guild produces a fatal error and cannot call its protected method directly - $myclass->myprivate ();//This guild produces a fatal error that cannot be called directly to its private method to $myclass->foo ();//public, protected, Private can be executed, can indirectly invoke private and protected methods through a public method</pre>What if we now define a new class MyClass 2, which inherits the above MyClass, and then calls the public and private methods?<pre>1 /**2 * Define MyClass23 */4 classMyClass2extendsMyClass5 {6 //This method is public7 functionFoo2 ()8 {9 $this-mypublic ();Ten $this-myprotected (); one $this->myprivate ();//This guild produces a fatal error . a } - } - the $myclass 2=NewMyClass2; - $myclass 2->mypublic ();//This line can be executed normally. - $myclass 2->foo2 ();//both public and protected can be enforced, but not Private.</pre>The 16th line, after the call, first in MyClass 2 to find the Mypublic method, did not find, so find his father set, also in its inheritance MyClass found the Mypublic method, because it is the public method, so can call, execute normal.Line 17th, After calling the public method Foo2, contains 3 calls, for the public, and the protected method is the same as the 16th line, first from their own inside, not found, then go to his father set MyClass find, find the normal method of invocation, and for the third row of the private method call, It can only find its own internal there is no this method, no return error, cannot go to the parent set directly call its private method.Shrimp noodles, Look at another example to see how to invoke the private method in the parent set, and compare the example above:<pre>1 classBar2 {3 public functionTest () {4 $this-testprivate ();5 $this-testpublic ();6 }7 8 public functiontestpublic () {9 Echo"bar::testpublic\n";Ten } one a Private functiontestprivate () { - Echo"bar::testprivate\n"; - } the } - - classFooextendsBar - { + public functiontestpublic () { - Echo"foo::testpublic\n"; + } a at Private functiontestprivate () { - Echo"foo::testprivate\n"; - } - } - - $myFoo=NewFoo (); in $myFoo->test ();//bar::testprivate - //foo::testpublic</pre>The 29th line, after calling the test method, finds it in itself, does not find it, so find the parent set Bar, finds the test method on line 3rd, and Executes. Test method, There are two calls, the first, call testprivate private method, so in their own inside, this refers to the inside of bar, because this call is in the parent set, and is through a public test method, call its own private method, and then return the Is the Testprivate private method of bar. The second call testpublic, because Foo has redefined a public method, so the testpublic in Foo will overwrite the Testpublic method of the parent set, first call the Testpublic method in foo, so the return is Foo The Testpublic.Just started to shoot the yellow film, understand as wrong, please guide ~One last word: learning makes me happy:)I want to learn PHP well for the fun of making the yellow FILM.

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