How do I set a piece of code to a function in WordPress and then call that function...this was my first guess, but of course this doesn't work...
Could someone suggest how I can get this to work so I can avoid redundant code. Also if I define code in a function can I call it in different .php files, or can I only call it within that file?
<?php
// This code will never change.
function $test {
echo('test this will be a long string and repeated many times.')
}
?>
<?php
echo $test;
?>
PHP function names (unlike variables) are declared without dollar signs:
function test (
To call a function, use this format:
test();
Your code example will look like this:
<?php
// This code will never change.
function test {
echo('test this will be a long string and repeated many times.')
}
?>
<?php
test();
?>
Related
i code the following
<?php
if ($id = mysql_real_escape_string(#$_GET['pid'])
&& $uid = mysql_real_escape_string(#$_GET['file']))
echo include "foo.php";
else
echo include "bar.php";
?>
When I use the include function in conjunction with a function that's designed to output to the page (e.g., or echo include 'foo.php'), it returns the include but with a "1" after the content that has been included.
echo include "foo.php"
should be
include 'foo.php';
Note that this can also happen when using include with shorthand echo:
<?= include 'foo.php'; ?>
This will also print out the return value of 1 when used inside a script. To get rid of this you need to use the regular PHP opening tag like so:
<?php include 'foo.php'; ?>
PHP will now include the contents of the file without printing the return value.
Okey so the answers here are actually not entirely correct; in some sense even misleading.
include takes the contents of the file and places them in context. One of the more common uses is to pass variable scope around, ie. passing scoped variables in your view by including them in the handler and using include on the view. Common, but there are also other uses; you can also return inside a included file.
Say you have a file like this:
<?php return array
(
'some',
'php'
'based'
'configuration',
'file'
); # config
Doing $config = include 'example-config-above.php'; is perfectly fine and you will get the array above in the $config variable.
If you try to include a file that doesn't have a return statement then you will get 1.
Gotcha Time
You might think that include 'example-config-above.php'; is actually searching for the file in the directory where the file calling the include is located, well it is, but it's also searching for the file in various other paths and those other paths have precedence over the local path!
So if you know you had a file like the above with a return inside it, but are getting 1 and potentially something like weird PEAR errors or such, then you've likely done something like this:
// on a lot of server setups this will load a random pear class
include 'system.php'
Since it's loading a file with out a return you will get 1 instead of (in the case of our example) the configuration array we would be expecting.
Easy fix is of course:
include __DIR__.'/system.php'
That is because the include function returns 1 on success. You get, as you say, 'my name is earl1' because the code inside the included file runs first, printing 'my name is earl' and then you local echo runs printing the return value of include() which is 1.
Let the file.txt contain xxx
echo include("file.txt");
This returns,
xxx1
The '1' is the return value of the include function, denoting the success that the file is accessed. Otherwise it returns nothing, in this case parse error is thrown.
echo print "hello";
This too returns,
hello1
Same as above; '1' denoting the success,it's printed.
echo echo "hello";
print echo "hello";
Both the above cases produces an error.
Since the echo function has no return value, hence undefined.
echo echo "hello"; print echo "hello";
(1st) (2nd)
Now the second 'echo' in the both cases produces an undefined.The first 'echo' or 'print' can't take in the hello output with the undefined (produced by the second echo).
A verification:
if((print "hello")==1)
echo "hey!";
output: hellohey! ('echo' in the 2nd line can be a print, it doesn't matter)
Similarly,
if((include ("file.txt"))==1)
echo "hey!";
output: xxxhey!
Other hand,
if((echo "hello")==1)
echo "hey!";
output: an error
In the first two cases the functions (print and include) returned 1, in the third case 'echo' produces no return value (undefined) hence the third case produces an error.
Well... I am using Codeigniter(php frame work). And I encountered the same problem. What I concluded is that when we try to print/echo the include method then it prints 1 on screen and when we just simply write the include command(example given below) it will only do what it is supposed to do.
<?php include('file/path'); ?> // this works fine for me
<?= include('file/path'); ?> // this works fine but prints "1" on screen
Hope my explaination will be helful to someone
= is assigning operator
== is for checking equal to
check for php operators
I have solved it returning nothing at the end of the included file:
$data = include "data-row.php";
return $data;
Inside data-row.php:
<div>etc</div>
...
<?php return; //End of file
I found the selected answer from this thread very helpful.
Solution 1
ob_start();
include dirname( __FILE__ ) . '/my-file.php';
$my_file = ob_get_clean();
You might also find this thread about the ob_start() function very insightful.
Solution 2
Add a return statement in the file that is being included.
E.g my-file.php
<?php
echo "<p>Foo.</p>";
return;
Drawn from the answer provided by #gtamborero.
To help you understand it the way I do now, just take this oversimplification:
There should always be a return statement otherwise include will return 1 on success.
Happy coding!
M5
Use return null in foo.php and bar.php
echo substr(include("foo.php"),1,-1);
I have 1 html, 1 javascript file and 3 php files, purposed as below:
script.js : used to pass user-entered variables from html to calculate.php without refreshing the html page. then get the result from calculate.php and pass it to certain div in my html.
calculate.php : include parameters.php and functions.php, after that, do the calculation.
<?php
include 'parameters.php';
include 'functions.php';
call_user_func("do_something()");
?>
parameters.php: where I define variables.
<?php
$var = $_POST['data_from_js']
?>
functions.php: where I define functions and their "if" conditions, and these "if" conditions uses variables defined in parameters.php
<?php
function do_something()
{
if ($var == 100)
{
echo 200;
}
}
?>
everything works fine, except that when calculate.php called the user function "do_something", the "if" condition doesn't work, because the $var is always 0
and if I move the "if" condition from functions.php to calculate.php, like below:
functions.php:
<?php
function do_something()
{
echo 200;
}
?>
Calculate.php:
<?php
include 'parameters.php';
include 'functions.php';
if($var == 100)
{
call_user_func("do_something()");
}
?>
Then it works, the value of $var is just exactly what user entered in the html, and it returns 200 into somewhere in the html if the user enters 100 in the field.
so, variables defined in parameters.php,
can be read and get the value (by $_post method) inside calculate.php, where my javascript sends data to.
But it cannot be read inside the function which I defined in functions.php,
when included and called by calculate.php.
It confuses me a little bit, I guess it's something about synchronization, and how php works on its sequence...
my question is:
Is there a simple way I can make it work without moving the "if" condition from functions.php to calculate.php, or make my javascript send data directly into functions.php?
The reason I'd like to do this is to keep the code organized, categorized and explicit so I can know what to modify if needed. The real calculation.php is actually a little bit complicated that I want to separate functions from it.
It's my first post here so if there's something confusing please let me know, thank you guys !
This is an issue with the variable scope. Check this article from the official PHP site: http://php.net/manual/en/language.variables.scope.php
Try using global $val; at the beginning of the do_something() function, so the function knows that $var is a global version of the variable, instead of the local one to the function:
<?php
function do_something()
{
global $var;
if ($var == 100)
{
echo 200;
}
}
?>
Alternatively, and as suggested by other user, you may want to pass the value as a parameter to the function, and avoid the use of global variables (and their potential issues).
you have to pass the variable
$var = $_POST['data_from_js'];
do_something($var);
function do_something($var){
...
Is it bad practice to echo out a bunch of HTML using a function in php and having it something like this:
function my_function() {
global $post;
$custom_fields = get_post_custom();
$some_field = $custom_fields ['some_field'][0];
?>
<div class="something <?php if ($some_field) { echo $special-clas;} ?>">
<div class="something-else">
/* bunch more of html code */
</div>
</div>
}
And then in the page where you want to use that to echo it?
<html>
<body>
.....
....
<?php echo my_function(); ?>
....
I'm unsure how "accepted" it is to echo out functions?
Consider two functions:
function does_return() {
return 'foo';
}
function does_echo() {
echo 'bar';
}
does_return(); // nothing displayed
echo does_return(); // 'foo' displayed
does_echo(); // 'bar' displayed
echo does_echo(); // 'bar' displayed
In both cases, output CAN be performed, but how it happens differs. Since does_return() does not itself have ANY code to perform output within its definition, output is up to the calling code, e.g. the echo you perform.
With does_echo(), it doesn't matter how you call the function (with or without echo), since the function does the output itself. you'll get bar regardless.
Now consider this:
function this_is_fun();
echo 'foo';
return 'bar';
}
this_is_fun(); // outputs 'foo'
echo this_is_fun(); // outputs 'foobar';
This is bad practice, because it makes your code hard to maintain.
With a function like that you are mixing the logic and presentation. So, when you see something in your output that you don't like you can not be sure where to go first to go and change it. Do you go to the page code or the function code?
Functions are supposed to return data, and then your application deals with it how you wish, whether that’s assigning it to a variable or echoing it out.
I don't see how it's bad practice. As long as you're reusing the function, then it seems like you're using it the right way.
The only thing you shouldn't be doing is using global; rather pass $post to the function. See this answer for why.
Since your function already has an output, you don't need the echo.
my_function( $post );
That's fine. I'd rather see that than the PHP mixed in completely to the HTML.
You can use <?= my_function() ?> instead if you want to write a little less code.
What #DaveRandom said in his comment. Aside from that, no, it's not necessarily bad practice. It can though make for code that's hard to debug. Consider a MVC approach instead where the logic is largely in the Controller and the View simply handles rendering the view based on that logic.
I have a file, lets say it's index.php where the very beginning of the file has an include for "include.php". In include.php I set a variable like this:
<?php $variable = "the value"; ?>
then further down the in index.php I have another include, say "include2.php" that is included like this:
<?php include(get_template_directory_uri() . '/include2.php'); ?>
How can I call the "$variable" that I set in the first include, in "include2.php"?
The exact code that I am using is as follows:
The very first line of the index.php I have this line
<?php include('switcher.php'); ?>
Inside switcher.php I have this
<?php $GLOBALS["demo_color"] = "#fffffe"; ?>
If I use this in index.php, it works
<?php echo $GLOBALS["demo_color"]; ?>
However, If I use the following code to include another php file
<?php include(get_template_directory_uri() . '/demo_color.php'); ?>
then inside demo_color.php I have this code:
<?php echo "demo color:" . $GLOBALS["demo_color"]; ?>
The only thing it outputs is "demo color:"
edited for code-formatting
It simply can be used in include2.php, unless the inclusion of include.php happens inside of a different scope (i.e. inside a function call). see here.
If you want to be completely explicit about the intention of using the variable across the app, use the $GLOBALS["variable"] version of it's name, which will always point to the variable called variable in the global scope.
EDIT: I conducted a test against php 5.3.10 to reconstruct this:
// index.php
<?php
include("define.php");
include("use.php");
// define.php
$foo = "bar";
// use.php
var_dump($foo);
This works exactly as expected, outputting string(3) "bar".
<?PHP
//index.php
$txt='hello world';
include('include.php');
<?PHP
//include.php
echo $txt; //will output hello world
So it does work. Though there seems to be a bigger issue since this is likely to be difficult to maintain in the future. Just putting a variable into global namespace and using it in different files is not a best practice.
To make the code more maintainable it might be an idea to use classes so you can attach the variables you need explicit instead of just using them. Because the code around is not showed it is not clear what is your exact need further but it will be likely the code can be put in classes, functions etc. If it is a template you could think about an explicit set() function to send the variable data to the templates and extract() it there.
edit:
In addition based on the information first set your error_reporting to E_ALL and set the display_errors to 1. So you get all errors since the information you placed in your updated question gives indications that a missing variable is used as a constant which should raise errors all over the place.
i code the following
<?php
if ($id = mysql_real_escape_string(#$_GET['pid'])
&& $uid = mysql_real_escape_string(#$_GET['file']))
echo include "foo.php";
else
echo include "bar.php";
?>
When I use the include function in conjunction with a function that's designed to output to the page (e.g., or echo include 'foo.php'), it returns the include but with a "1" after the content that has been included.
echo include "foo.php"
should be
include 'foo.php';
Note that this can also happen when using include with shorthand echo:
<?= include 'foo.php'; ?>
This will also print out the return value of 1 when used inside a script. To get rid of this you need to use the regular PHP opening tag like so:
<?php include 'foo.php'; ?>
PHP will now include the contents of the file without printing the return value.
Okey so the answers here are actually not entirely correct; in some sense even misleading.
include takes the contents of the file and places them in context. One of the more common uses is to pass variable scope around, ie. passing scoped variables in your view by including them in the handler and using include on the view. Common, but there are also other uses; you can also return inside a included file.
Say you have a file like this:
<?php return array
(
'some',
'php'
'based'
'configuration',
'file'
); # config
Doing $config = include 'example-config-above.php'; is perfectly fine and you will get the array above in the $config variable.
If you try to include a file that doesn't have a return statement then you will get 1.
Gotcha Time
You might think that include 'example-config-above.php'; is actually searching for the file in the directory where the file calling the include is located, well it is, but it's also searching for the file in various other paths and those other paths have precedence over the local path!
So if you know you had a file like the above with a return inside it, but are getting 1 and potentially something like weird PEAR errors or such, then you've likely done something like this:
// on a lot of server setups this will load a random pear class
include 'system.php'
Since it's loading a file with out a return you will get 1 instead of (in the case of our example) the configuration array we would be expecting.
Easy fix is of course:
include __DIR__.'/system.php'
That is because the include function returns 1 on success. You get, as you say, 'my name is earl1' because the code inside the included file runs first, printing 'my name is earl' and then you local echo runs printing the return value of include() which is 1.
Let the file.txt contain xxx
echo include("file.txt");
This returns,
xxx1
The '1' is the return value of the include function, denoting the success that the file is accessed. Otherwise it returns nothing, in this case parse error is thrown.
echo print "hello";
This too returns,
hello1
Same as above; '1' denoting the success,it's printed.
echo echo "hello";
print echo "hello";
Both the above cases produces an error.
Since the echo function has no return value, hence undefined.
echo echo "hello"; print echo "hello";
(1st) (2nd)
Now the second 'echo' in the both cases produces an undefined.The first 'echo' or 'print' can't take in the hello output with the undefined (produced by the second echo).
A verification:
if((print "hello")==1)
echo "hey!";
output: hellohey! ('echo' in the 2nd line can be a print, it doesn't matter)
Similarly,
if((include ("file.txt"))==1)
echo "hey!";
output: xxxhey!
Other hand,
if((echo "hello")==1)
echo "hey!";
output: an error
In the first two cases the functions (print and include) returned 1, in the third case 'echo' produces no return value (undefined) hence the third case produces an error.
Well... I am using Codeigniter(php frame work). And I encountered the same problem. What I concluded is that when we try to print/echo the include method then it prints 1 on screen and when we just simply write the include command(example given below) it will only do what it is supposed to do.
<?php include('file/path'); ?> // this works fine for me
<?= include('file/path'); ?> // this works fine but prints "1" on screen
Hope my explaination will be helful to someone
= is assigning operator
== is for checking equal to
check for php operators
I have solved it returning nothing at the end of the included file:
$data = include "data-row.php";
return $data;
Inside data-row.php:
<div>etc</div>
...
<?php return; //End of file
I found the selected answer from this thread very helpful.
Solution 1
ob_start();
include dirname( __FILE__ ) . '/my-file.php';
$my_file = ob_get_clean();
You might also find this thread about the ob_start() function very insightful.
Solution 2
Add a return statement in the file that is being included.
E.g my-file.php
<?php
echo "<p>Foo.</p>";
return;
Drawn from the answer provided by #gtamborero.
To help you understand it the way I do now, just take this oversimplification:
There should always be a return statement otherwise include will return 1 on success.
Happy coding!
M5
Use return null in foo.php and bar.php
echo substr(include("foo.php"),1,-1);