I'm having some problems getting my object to gracefully fail out if an invalid parameter is given during instantiation. I have a feeling it's a small syntax thing somewhere that I simply need fresh eyes on. Any help is more than appreciated.
class bib {
protected $bibid;
public function __construct($new_bibid) {
if(!$this->bibid = $this->validate_bibid($new_bibid)) {
echo 'me';
return false;
}
//carry on with other stuff if valid bibid
}
private static function validate_bibid($test_bibid) {
//call this method every time you get a bibid from the user
if(!is_int($test_bibid)) {
return false;
}
return (int)$test_bibid;
}
}
Note that I have an 'echo me' line in there to demonstrate that it is in fact, returning false. The way that I'm calling this in my PHP is as follows:
if(!$bib=new bib('612e436')) {
echo 'Error Message';
} else {
//do what I intend to do with the object
}
This outputs the me from above, but then continues on into the else block, doing what I intend to do with a valid object.
Can anyone spot anything I'm doing wrong in there?
Thanks!
I see several problems in this code.
First of all, I think you want to do something like this:
$myBib=new bib();
if($myBib->validate_bibid('612e436'))
{ ..do stuff.. }
(or something similar)
remember that __construct is not a normal method. It's a constructor, and it shouldn't return anything. It already implicitly returns a reference to the new instance that you've made.
Second, your validate_bibid returns either a boolean or an integer. You won't get immediate problems with that, but I personally don't like the style.
Third, you've declared a protected member $bibid, but you don't set or reference it anywhere. I'd expect it to be set in the constructor for example. After that you can just call validate_bibid without any argument.
This piece of code is obviously confusing you because it has some weird constructs and therefore doesn't behave in a normal way. I'd suggest to rethink and rewrite this piece from scratch.
Update:
Another issue:
I don't think this line does what you think it does:
if(!$this->bibid = $this->validate_bibid($new_bibid)) {
You probably mean this:
if(!$this->bibid == $this->validate_bibid($new_bibid)) {
// Or even better:
if($this->bibid <> $this->validate_bibid($new_bibid)) {
You can't return in a constructor in PHP - the object is still created as normal.
You could use a factory or something similar;
if(!$bib = Bib_Factory::loadValidBib('612e436')){
echo 'Error Message';
} else {
//do what I intend to do with the object
}
Or throw an exception in the constructor and use a try catch instead of the if statement.
definitely, you need to have comparison with == instead of = which is an assign operation.
Related
I am confused about return statement , why we need to use return in end of function , for example
function test($a){blah;
blahh ;
blahhh;
blahhhhh;
return;}
What the use of return here? Function automatically terminates when all the statements executed , I think there is no use of return here , but this picture from http://www.w3resource.com/php/statement/return.php make me confused
So
Can someone please explain the use of return (when we not returning any value).
It depends on what you're trying to achieve.
If you write echo in several places, your code will get confusing. In general, a function that returns a value is also more versatile, since the caller can decide whether to further manipulate that value or immediately print it.
I'd recommend to stick to the convention and use return for a function.
You should check GuardClause.
Example:
function test() {
return 10;
}
$a = test(); // $a stores the value 10
echo $a; // Prints 10
echo $a + 5; // We may want to manipulate the value returned by the function. So, it prints 15.
For further reference, check What is the difference between PHP echo and PHP return in plain English?
In that context: You don't.
return breaks out of the function, but since it is the last statement in that function, you would break out of it anyway without the statement.
return passes its argument back to the caller, but since it doesn't have an argument, there is nothing to pass.
So it does nothing.
Don't assume that every piece of code you stumble across has a purpuse. It might be left over from an earlier version of the code where something else (which gave it meaning) has been removed. It might be written by someone cargo culting. It might be placeholder for future development.
"return" is important when you plan to call this function from other codes, it helps you to:
Know if the function works correctly, or not.
Obtain values from a function.
Make sure other codes are not executed after return.
It might be useless when the code is simple as your sample, let's make it more complex.
function test($a){
if(file_exists($a)){
if(is_file($a)){
return $a." IS A FILE\n";
} else if(is_dir($a)) {
return $a." IS A DIR\n";
} else {
return $a." EXISTS, BUT I DONT KNOW WHAT IT IS\n"
}
} else {
return $a." NOT EXISTS\n";
}
return 0;
}
$filecheck = test("/abc/def.txt");
if($filecheck){
echo $filecheck;
} else {
echo "unknown error";
}
Above shows how to return a value, and do some basic handling.
It is always good to implement return in your functions so you can even specify error code for your functions for reference.
Based on your example, I'll modify slightly like below:
function test($a){
blah;
blahh;
blahhh;
blahhhhh;
return 1;
}
So I know the code is running to last line.
Otherwise the function just finishes silently.
It's useful if you want a function to return a value.
i.e.
Function FavouritePie($who) {
switch($who) {
case 'John':
return 'apple';
case 'Peter':
return 'Rhubarb';
}
}
So, considering the following:
$WhatPie = FavouritePie('John');
echo $WhatPie;
would give
apple
In a really simple form, it's useful if you want a function to return something, i.e. process it and pass something back. Without a return, you'd just be performing a function with a dead end.
Some further reading:
http://php.net/manual/en/function.return.php
http://php.net/manual/en/functions.returning-values.php
To add some further context specific to the answer, if the question boils down to "Do I need to add a return for the sake of it, at the end of a function, then the answer is no. But that doesn't mean the correct answer is always to leave out your return.
it depends what you want to do with $a.
If you echo $a within the function, that function will spit out $a as soon as it is called. If you return $a, assuming you set the calling of test to a variable (i.e. $something = $test('foo')), then you can use it later on.
My current way:
class A {
public function function_b($myint) {
if (!is_numeric($myint)) return false;
// code ...
}
}
I would like to abandon the function is_numeric() like this:
public function function_b(Integer $myint) {
// code ...
}
It works with arrays like this:
public function function_c(Array $arr) {
// only executes following code if $arr is an array / instance of Array!
}
Note: the function has to return false if the value isn't a number (int)! I don't want to cast it.
How would you short my current code? Thanks in advance!
You can't force strict types in function prototypes in PHP inherently, because it's not a strictly typed language. PHP is a weakly typed language and trying to go against the grain will only hurt you in many situations. Also, is_numeric does not guarantee that your value is of type int (for what it's worth).
What you can do is analyze your need for why you think this approach is necessary in the first place and decide on how to best implement this without creating potential for bugs.
For example, take the following scenario where what your method expects is an ID for a database query.
class MyClass {
public function getUser($id) {
if (!is_int($id)) {
throw new Exception("Invalid argument supplied. Expecting (int), but argument is of type (" . gettype($id) . ").");
}
// Otherwise continue
$db = new PDO($dsn);
$stmt = $db->prepare("SELECT username FROM users WHERE user_id = ?");
$stmt->execute(array($id));
$result = $stmt->fetchAll(PDO::FETCH_ASSOC);
return $result;
}
}
$MyObject = new MyClass;
$result = $MyObject->getUser($_POST['id']);
/* The problem here is $_POST will always be of type string. */
What this should tell you is that it makes no sense to force type checking here since PHP will have done the right thing for you had you just let it alone.
The question you need to be asking yourself is not "How do I force strict typing?", but rather "Why would I need to force strict typing at all?".
You should look into typecasting:
http://php.net/manual/en/language.types.type-juggling.php#language.types.typecasting
Just use (int) when accessing the value to typecast it to an integer.
You could just typecast it:
public function function_b($myint) {
$myint = (int) $myint;
}
Or better yet add a public setter to class A which will do it for you every time you set the value:
class A
{
public function setMyInt($myInt)
{
$this->myInt = (int) $myInt;
}
}
-- Update (based on comment) --
class A
{
public function doSomethingWithAnArray(array $array)
{
....
}
}
Notice the keyword array in the signature of the doSomethingWithAnArray method, now if you don't pass an array to this function PHP will throw a fatal error and cease code execution. This is known as typehinting, and can be applied to objects as well.
function needsInteger($int) {
if (((int) $int) != $int) return false;
// ...
}
The advantage here is that you can still accept loosely typed parameters, but the non-strict equality check against the cast value will yield an acceptable result.
Is there a way to check if an object has any fields? For example, I have a soap server I am querying using a soap client and if I call a get method, I am either returned an object containing fields defining the soap query I have made otherwise I am returned object(stdClass)#3 (0) { }.
Is there a way to tell if the object has anything?
public function get($id){
try{
$client = new soapclient($this->WSDL,self::getAuthorization());
$result = $client->__soapCall('get', array('get'=> array('sys_id'=>$id)));
if(empty($result)){$result = false; }
}catch(SoapFault $exception){
//echo $exception;
$result = false;
}
return $result;
}//end get()
This method should return either an object or false and I am only receiving an object with no fields or an object with fields.
Updated to reflect current behavior, 5/30/12
empty() used to work for this, but the behavior of empty() has changed several times. As always, the php docs are always the best source for exact behavior and the comments on those pages usually provide a good history of the changes over time. If you want to check for a lack of object properties, a very defensive method at the moment is:
if (is_object($theObject) && (count(get_object_vars($theObject)) > 0)) {
...
One of the user contributed code on the php empty() page which I think addresses your problem of checking if the array is filled but has empty values.
http://www.php.net/manual/en/function.empty.php#97772
To find if an array has nothing but empty (string) values:
<?php
$foo = array('foo'=>'', 'bar'=>'');
$bar = implode('', $foo);
if (empty($bar)) {
echo "EMPTY!";
} else {
echo "NOT EMPTY!";
}
?>
Anyone has an idea if this is at all possible with PHP?
function foo($var) {
// the code here should output the value of the variable
// and the name the variable has when calling this function
}
$hello = "World";
foo($hello);
Would give me this output
varName = $hello
varValue = World
EDIT
Since most people here 'accuse' me of bad practices and global variables stuff i'm going to elaborate a little further on why we are looking for this behaviour.
the reason we are looking at this kind of behaviour is that we want to make assigning variables to our Views easier.
Most of the time we are doing this to assign variables to our view
$this->view->assign('products', $products);
$this->view->assign('members', $members);
While it would be easier and more readable to just be able to do the following and let the view be responsible to determining the variable name the assigned data gets in our views.
$this->view->assign($products);
$this->view->assign($members);
Short answer: impossible.
Long answer: you could dig through apd, bytekit, runkit, the Reflection API and debug_backtrace to see if any obscure combination would allow you to achieve this behavior.
However, the easiest way is to simply pass the variable name along with the actual variable, like you already do. It's short, it's easy to grasp, it's flexible when you need the variable to have a different name and it is way faster than any possible code that might be able to achieve the other desired behavior.
Keep it simple
removed irrelevant parts after OP edited the question
Regardless of my doubt that this is even possible, I think that forcing a programmer on how to name his variables is generally a bad idea. You will have to answer questions like
Why can't I name my variable $arrProducts instead of $products ?
You would also get into serious trouble if you want to put the return value of a function into the view. Imagine the following code in which (for whatever reason) the category needs to be lowercase:
$this->view->assign(strtolower($category));
This would not work with what you're planning.
My answer therefore: Stick to the 'verbose' way you're working, it is a lot easier to read and maintain.
If you can't live with that, you could still add a magic function to the view:
public function __set($name, $value) {
$this->assign($name, $value);
}
Then you can write
$this->view->product = $product;
I don't think there is any language where this is possible. That's simply not how variables work. There is a difference between a variable and the value it holds. Inside the function foo, you have the value, but the variable that held the value is not available. Instead, you have a new variable $var to hold that value.
Look at it like this: a variable is like a bucket with a name on it. The content (value) of the variable is what's inside the bucket. When you call a function, it comes with its own buckets (parameter names), and you pour the content of your bucket into those (well, the metaphor breaks down here because the value is copied and still available outside). Inside the function, there is no way to know about the bucket that used to hold the content.
What you're asking isn't possible. Even if it was, it would likely be considered bad practice as its the sort of thing that could easily get exploited.
If you're determined to achieve something like this, the closest you can get would be to pass the variable name as a string and reference it in the function from the $GLOBALS array.
eg
function this_aint_a_good_idea_really($var) {
print "Variable name: {$var}\n";
print "Variable contents: {$GLOBALS[$var]}\n";
}
$hello="World";
this_aint_a_good_idea_really('hello');
But as I say, that isn't really a good idea, nor is it very useful. (Frankly, almost any time you resort to using global variables, you're probably doing something wrong)
Its not impossible, you can find where a function was invoked from debug_backtrace() then tokenize a copy of the running script to extract the parameter expressions (what if the calling line is foo("hello $user, " . $indirect($user,5))?),
however whatever reason you have for trying to achieve this - its the wrong reason.
C.
Okay, time for some ugly hacks, but this is what I've got so far, I'll try to work on it a little later
<?php
class foo
{
//Public so we can test it later
public $bar;
function foo()
{
//Init the array
$this->bar = array();
}
function assign($__baz)
{
//Try to figure out the context
$context = debug_backtrace();
//assign the local array with the name and the value
//Alternately you can initialize the variable localy
//using $$__baz = $context[1]['object']->$__baz;
$this->bar[$__baz] = $context[1]['object']->$__baz;
}
}
//We need to have a calling context of a class in order for this to work
class a
{
function a()
{
}
function foobar()
{
$s = "testing";
$w = new foo();
//Reassign local variables to the class
foreach(get_defined_vars() as $name => $val)
{
$this->$name = $val;
}
//Assign the variable
$w->assign('s');
//test it
echo $w->bar['s'];
}
}
//Testrun
$a = new a();
$a->foobar();
impossible - the max. ammount of information you can get is what you see when dumping
debug_backtrace();
Maybe what you want to do is the other way around, a hackish solution like this works fine:
<?php
function assign($val)
{
global $$val;
echo $$val;
}
$hello = "Some value";
assign('hello');
Ouputs: Some value
What you wish to do, PHP does not intend for. There is no conventional way to accomplish this. In fact, only quite extravagant solutions are available. One that remains as close to PHP as I can think of is creating a new class.
You could call it NamedVariable, or something, and as its constructor it takes the variable name and the value. You'd initiate it as $products = new NamedVariable('products', $productData); then use it as $this->view->assign($products);. Of course, your declaration line is now quite long, you're involving yet another - and quite obscure - class into your code base, and now the assign method has to know about NamedVariable to extract both the variable name and value.
As most other members have answered, you are better off suffering through this slight lack of syntactic sugar. Mind you, another approach would be to create a script that recognizes instances of assign()'s and rewrites the source code. This would now involve some extra step before you ran your code, though, and for PHP that's silly. You might even configure your IDE to automatically populate the assign()'s. Whatever you choose, PHP natively intends no solution.
This solution uses the GLOBALS variable. To solve scope issues, the variable is passed by reference, and the value modified to be unique.
function get_var_name(&$var, $scope=FALSE) {
if($scope) $vals = $scope;
else $vals = $GLOBALS;
$old = $var;
$var = $new = 'unique'.rand().'value';
$vname = FALSE;
foreach ($vals as $key => $val) {
if($val === $new) $vname = $key;
}
$var = $old;
return $vname;
}
$testvar = "name";
echo get_var_name($testvar); // "testvar"
function testfunction() {
$var_in_function = "variable value";
return get_var_name($var_in_function, get_defined_vars());
}
echo testfunction(); // "var_in_function"
class testclass {
public $testproperty;
public function __constructor() {
$this->testproperty = "property value";
}
}
$testobj = new testclass();
echo get_var_name($testobj->testproperty, $testobj); // "testproperty"
Do you consider this a code smell?
foreach((array)$foo as $bar)
{
$bar->doStuff();
}
Should i use that instead?
if (isset($foo) && is_array($foo))
{
foreach($foo as $bar)
{
$bar->doStuff();
}
}
Any other good practices to cover not set variables and assert an array?
They're both code smells. The second one is just evading all the error messages, kind of like turning off the fire alarm before you set your kitchen on fire. Both of those tell you that you have no idea what's in the variable $foo or if it's even been defined in the code above. You need to go back up through the code and find out exactly what's going on with $foo.
If it was my code, $foo would probably be always defined either as an array, or else false to indicate the array isn't needed:
if(do_we_need_an_array())
$foo = function_returning_an_array();
else
$foo = false;
[...snip...]
if($foo)
foreach($foo as $f) { ... }
If you are testing if variables are set, you can initialize them:
if (! $foo or !is_array($foo))
$foo = array();
foreach($foo as $bar)
{
$bar->doStuff();
}
Personally, I would never do the first method and always opt for the second.
If $foo should always be an array, then the second form would be much better if you did some kind of handling for the error case, e.g.:
if (isset($foo) && is_array($foo))
{
foreach($foo as $bar)
{
$bar->doStuff();
}
}
else
{
// This should not happen, exit angrily.
exit("Oh crap, foo isn't an array!");
}
Of course you don't have to just exit the application, but do whatever is appropriate in that case, maybe logging or some alternate logic.
(array)$foo != if (isset($foo) && is_array($foo))
The (array) cast can be useful for casting objects to arrays or scalars to arrays so you can create consistent interfaces to variables that may contain single values or arrays.
(array)$foo == array($foo)
As defined in the PHP Manual for Array Types.
So if you need to always use an array then the first code snippet you presented would be the answer. However type casting rules still apply so you may not get what you want, so look to the manual for more info. Otherwise the second option would prevent accessing unset variables that are not arrays.
As far as a code smell, I would say that checking for unset variables can certainly be avoided, however always knowing that a variable is going to have an array is more often than not, going to creep up. So I would aim to keep code wrapped in is_array($foo) if-then statements to a minimum.
I usually do this to make sure a foreach can handle both scalars and collections:
<?php
foreach (makeSureTraversable($scalarOrCollection) as $val)
{
// Do something.
}
function
makeSureTraversable($anything)
{
if (is_array($anything) || ($anything instanceof Traversable))
{
return $anything;
}
else
{
return array($anything);
}
}
This way I also handle classes that implement Traversable (from the SPL), which means allowing them to be used in foreaches.
if (!isset($foo) && !is_array($foo)) {
throw new InvalidArgumentException('Wrong array passed');
// Or do something to recover lost array
}
foreach($foo as $bar) {
$bar->doStuff();
}
There's quite a few times that you'd like to write a function to take one or more values for a parameter:
function getNamesById($id) { }
In this case, it would make sense that if this function was called with an array of ids, it should probably return an array of names. Similarly, to save the calling code from having to wrap the input in an array and then unwrap the output, if you just pass a scalar, then a scalar should be returned. Consider the likely contents of the function designed to handle both scalar and array parameters:
function getNamesById($id) {
$returnAnArray = is_array($id);
$output = array();
foreach ((array)$id as $theId) {
// perform some logic
$output[] = someFunction($theId);
}
return $returnAnArray ? $output : $output[0];
}
You can see that in this case, casting to an array definitely makes things a lot easier for everyone. As they say, be liberal in what you accept... As long as it is documented that it is expected that a variable could be either, then I see no problem. PHP is a duck-typed language, which has both benefits and drawbacks, but this is one of the benefits, so enjoy it!