sorry for asking this dumb question i will try to explain as good as i can, i have never done a php function before, i could solve this with IF but i want to do my first function
What i want to do is to see if the variable is null and if it is i want to set the variable to be "N/A"
$carplate=mysql_result($result,0,"plate");
$carbrand=mysql_result($result,0,"brand");
function set_null(variablegoeshere?){
if($WHATiWANTtonull==null){
$WHATiWANTtonull="N/A";
}else{ do nothing }
}
set_null($carplate);
set_null($carbrand);
echo "carplate = $carplate | carbrand = $carbrand";
i dont understand how i will get the function to set a variable to "N/A" or do nothing, i don't want it to echo something, just set a variable and i can echo it later whenever i want
[EDIT]: Use this function:
function set_null(&$var)
{
if($var==null)
{
$var="N/A";
}
}
Related
I am confused about return statement , why we need to use return in end of function , for example
function test($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 = $test('foo')), then you can use it later on.
This question already has answers here:
PHP take string and check if that string exists as a variable
(3 answers)
Closed 8 years ago.
I have a field that depends has custom text if a certain condition is met...and if it isn't the field is blank.
I have written a custom function test if a variable is set
function AmISet($fieldName) {
if (isset($fieldName)) {
echo "I am set";
}else{
echo "I am not set";
}
};
but when I attach it the field I get an error that the variable is undefined. But when I do a regular isset($fieldName);
I don't have a problem. Is there any way to get around this and have my function do this instead of the isset()?
I want to add other logic in the function but I only want it to work if the variable is set...but I don't want the undefined error if it is not.
I am new to php and really appreciate any help or direction you can give me.
Thank you for the help!
You need to pass the variable by reference:
function AmISet(&$fieldName) {
if (isset($fieldName)) {
echo "I am set\n";
} else {
echo "I am not set\n";
}
}
Test cases:
$fieldName = 'foo';
AmISet($fieldName); // I am set
unset($fieldName);
AmISet($fieldName); // I am not set
However, this function is not useful as it is, because it will only output a string. You can create a function that accepts a variable and return if it exists (from this post):
function issetor(&$var, $default = false) {
return isset($var) ? $var : $default;
}
Now it can be used like so:
echo issetor($fieldName); // If $fieldName exists, it will be printed
the $fieldName comes from a query that is performed when a checkbox
is checked. If the box is checked the query is made and the variable
is set from the query, if not it doesn't exist and the field is blank.
Filter functions are designed for this kind of tasks:
<input type="foo" value="1">
$foo = filter_input(INPUT_POST, 'foo')==='1';
(There's also a specific FILTER_VALIDATE_BOOLEAN filter you can pass as second argument.)
And now you have a pure PHP boolean that always exists.
I've been at this for two hours, perhaps I could ask some help.
Alright, so I have a basic $_POST variable that a user submits. As you can see below, the code first checks if a value was submitted at all, and if not sets it to a default value. If the user did submit a value, it sets a variable (later used) to the value submitted. You can see my code below.
if(!isset($_POST['pSize'])) {
$pSize = "16";
}
else {
$pSize = ($_POST['pSize']);
};
echo $pSize;
The problem with the above is that I might have 50 or so different areas that the user will submit, and it'd be much nicer to just have myFunction('name','default','value'); than to write out the above for each area. However, I've been running into a problem. Here's a few examples of things I've tried. (excuse any minor errors; I don't have the actual code I tried, just the jist of what I was getting at. The problems of the code I had were not syntax errors).
function newFunction($title, $default, $value)
if(!isset($_POST[$value])) {
$title = $default;
}
else {
$title = ($_POST[$value]);
};
newFunction('pSize','16,'24');
echo $pSize;
I soon learned that the above won't work due to the fact that variables in a function won't be able to be used outside of that function unless they're global. That makes sense, since if the variable $title could be used, it would be set to many different things depending on how many times I called the function. These things in mind, I tried to set global variables.
function newFunction($title, $default, $value)
if(!isset($_POST[$value])) {
global $title . 'Title' = $title;
...
newFunction('pSize','16,'24');
echo $pSizeTitle;
I lastly tried to set the global variable to the name of the $title I supplied for the function with the string 'Title', creating the new global variable pSizeTitle, so I could echo that variable out. And this probably would have actually worked, except for the fact that I cannot define a global variable with something appended to the end, only with a simple name, and that will not work for me since I need a new global variable for every title item I have.
Hopefully this is clear, sorry if this is an absolute noob problem, I just really can't get past this.
I think the major problem is that you are not returning any data in your function, try to do it
function newFunction($title, $default, $value)
{
if(!isset($_POST[$value])) {
$title = $default;
} else {
$title = ($_POST[$value]);
}
return $title;
//return was missed
}
function checkDefault($index, $default)
{
return isset($_POST[$index]) ? $_POST[$index] : $default;
}
$title = checkDefault("title", "my default value");
I'm trying to pass a value between 2 pages through a $_SESSION variable then empty it as follows:
I'm assigning the session variable with a value on one PHP page:
$_SESSION["elementName"]="a_373";
And trying to store it in a variable on another page as follows:
if (!empty($_SESSION["elementName"])) {
$elemName=$_SESSION["elementName"];
$_SESSION["elementName"]="";
} else {
$elemName="";
}
The value of $elemName is always empty when I print it out. However, I get the correct printout when I remove the $_SESSION["elementName"]=""; line from the above code.
Edit: I'm printing $elemName and not $_SESSION["elementName"] - print($elemName);
I'm on a shared hosting account with PHP 5.3.2 and register_globals set to off (as per phpinfo();).
I need to reset/empty the session variable once I get the value it has, but it's not working and this has been baffling me for the last couple of days. Any ideas why? Thanks!
EDIT:
Additional clues: I tested with the session's var_dump before the if statement and set another value for $elemName in the else section as follows:
var_dump($_SESSION["elementName"]);
$elemName="x";
if (isset($_SESSION["elementName"]) && !empty($_SESSION["elementName"])) {
$elemName=$_SESSION["elementName"];
$_SESSION["elementName"]="";
} else {
$elemName="None";
}
print("<br />".$elemName);
I got this result:
string(5) "a_373"
None
Try using isset($_SESSION["elementName"]) and unset($_SESSION["elementName"]) instead.
Check the Below code and test it
if (isset($_SESSION["elementName"]) && $_SESSION["elementName"]!="") {
$elemName=$_SESSION["elementName"];
$_SESSION["elementName"]="";
} else {
$elemName="";
}
Empty only check value is empty or not, But by isset we can check varaible exit and does not content empty value
Are you printing out $elemName? If yes than there shouldn't be any blank output unless and until your else condition returns true, but if you are printing $_SESSION["elementName"] than you'll get no output as you are making it blank
$_SESSION["elementName"]="";
Correct way to remove a session var completely is to use unset($_SESSION["elementName"])
Though if you want to make it empty, == '' is enough
Also be cautious while using unset() because after you unset the session and later use that index to print, it will show you undefined index error.
Update(As #nvanesch Commented)
I guess your else condition is getting satisfied, because you are
using $elemName=""; in your else, thus you are not returned with any
output
Also if (!empty($_SESSION["elementName"])) { will return false if you are not using session_start() at the very top of your page
I once created this class in order to be able to have this ability, and it works wonderfully:
<?php
class Flash {
public function set($key, $message) {
$_SESSION["flash_$key"] = $message;
}
public function get($key) {
$message = $_SESSION["flash_$key"];
unset($_SESSION["flash_$key"]);
return $message;
}
public function has($key) {
return isset($_SESSION["flash_$key"]);
}
}
It's pretty similar to what you're trying to do, so I'm not sure why it's not working for you, but you may want to give it a try. You obviously need to have the session already started.
I'm trying to edit my co-worker's code (he's on vacation and out of reach) and there is this if statement:
if($this->carrier() == 1 and $this->carrier() == 2) {
return 'V';
}
My hope is that he accidentally put "and" instead of "or" which could explain a bug I'm getting, but he knows much more about PHP than I do and I want to be sure that this is wrong instead of just counter intuitive.
Yes, since it's a function with potential side effects, it might be true.
For example:
function carrier() {
return $someValue++;
}
Yes. The carrier() method could increment whatever value it returns each time you call it.
There's a small chance it could, yes.
He's calling a function twice, and you've not included the text of that function. So that could be doing something we can't see, like counting the number of times it's been called by this process.
On the other hand, it's much more likely that it is indeed a typo.
It is possible. Here is a working example you can run.
class program
{
private $i = 1;
function carrier()
{
$this->i=$this->i+1;
return $this->i-1;
}
function run()
{
if ($this->carrier()==1 && $this->carrier()==2)
{
echo "works";
}
else
{
echo "doesnt work" . $i;
}
}
}
$prog = new Program();
$prog->run();