I was wondering if there was a way that I can declare a variable at the top of my php file that contains some 'array stuff' and have it used in a foreach loop below somewhere?
For example:
$variable = 'Here is some text, and here is the value:'.$items->value.'.';
foreach ( $somearray as $items ) {
echo $variable;
}
*note: I am using ezsql for database actions, hence the $items->value variable....
Make sense? I'm just trying to make my file easy to change for other situations... or is there a better way?
Thanks for looking and for your help.
The nicest way to do this would, I think, be printf. It's not exactly what you asked for, but I think it will make your code cleaner.
$variable = 'Here is some text, and here is the value:%s.';
foreach ($somearray as $items) {
printf($variable, $items->value);
}
The value passed as the second argument to printf will be substituted into the code.
You can use eval(), but you should be aware it's risky, and you better read about the consequences of using it.
Other than that, you can use it like you said, but escape the dollar sign so it won't get treated as a variable until the foreach, or use single quotes:
$variable = 'Here is some text, and here is the value: {$items->value}';
foreach ( $somearray as $items ) {
echo eval('return "'.$variable.'";');
}
or is there a better way?
Use a function/method:
function formatItems($items)
{
return 'Here is some text, and here is the value:'.$items->value.'.';
}
foreach ( $somearray as $items ) {
echo formatItems($items);
}
Using a function allows more complex formatting. If it's going to be as simple as your example, you could use printf, as suggested by lonesomeday.
If appropriate, you could have a display/format method in your Items class definition.
Your $items variable will be overwritten by your for loop - $variable won't be though.
$variable is set only once and takes on the value of $items->value at the time (which will probably result in an error since $items doesn't exist at that time and you are referencing a field.
what you want is this:
$variable = 'Here is some text, and here is the value: {$items->value}';
foreach ( $somearray as $items ) {
echo $variable;
}
Related
I'm trying to define three empty variables through a foreach loop to make my code cleaner. This is what I've tried, however I see the error:
Notice: Undefined variable: hi
foreach(['$hi','$bye','$hello'] as $key) {
$key = "";
}
$hi .= "hello";
When I remove the foreach loop and simply define each empty variable one by one, like this, it works:
$hi = "";
$bye = "";
$hello = "";
You're assigning to $key, not to the variable that's named by it. To indirect through a variable, you need to use $$key. But the value of the variable shouldn't include the $, just the variable name.
foreach (['hi', 'bye', 'hello'] as $key) {
$$key = "";
}
$hi .= "hello";
However, if you ever find yourself using variable variables like this, you're almost certainly doing something wrong. You should probably be using an associative array instead.
You have strings which are saved in $key. So the value of $key is a string and you set it to "".
Later you want to append something to a variable you never used.
Try to remove the ' and write
foreach([$hi, $bye, $hello] as $key) {
Generally thats not the best way to initialise multiple variables. Try this
Initializing Multiple PHP Variables Simultaneously
Easier way:
list($hi, $bye, $hello) = "";
foreach creates a new array variable in memory, so you only clear these values inside the array in memory which is useless out of the foreach sentence. the best way is:
$h1=$bye=$hello="";
I didn't think that a foreach process will work more fast than a Simple equal (=), foreach function uses more CPU resources than a simple =. That's because the math CPU exists.
I have $config variable that have arrays inside it. In smarty I assign the variable like this:
$smarty->assign('config', $config);
when I call it, I used this : {$config.wateverarrayyouwant}
now I want to do the same thing with php. I want to define them in the same manner. How can I define all the arrays in $config in just one line?
I only know how to define a variable one at a time by using this :
define('wateverarrayyouwant', $config['wateverarrayyouwant']);
I tried changing wateverarrayyouwant to a variable because it can be any array :
define('$wateverarrayyouwant', $config[$wateverarrayyouwant]);
but the code above does not work. what is a good way to achieve what I want?
If you want to create a define for each key value pair in the array you can use:
<?php
foreach($config as $key => $value) {
define($key, $value);
}
I will note however that you cannot define array values, all define's must be scalar:
The value of the constant; only scalar and null values are allowed. Scalar values are integer, float, string or boolean values.
If you check the OP's answer for further explanation of what he's trying to achieve, it can be done with:
<?php
foreach($config as $key => $value){
$$key = $value;
}
?>
This question cannot be done. because I am trying to define a variable as a constant. I was just thinking about how can I reduce the letters for variables and never though that I better leave them alone. Logically, why do somebody need to change $config[$wateverarrayyouwant] to wateverarrayyouwant. I was only thinking about maintaining a neat code. but now I am thinking about it.. it is better to leave it as it is : $config[$wateverarrayyouwant]
This can be done with:
foreach($config as $key => $value){
$$key = $value;
}
You may not even want to use define here. define is used to create constants not plain variables and that carries with it certain connotations:
they are immutable for the life of the script
they must be scalar
If you just want an array variable then define it like normal with:
$whatever = array(
'key1' => 'value1'
);
ok, So I have this array:
$choices = array($_POST['choices']);
and this outputs, when using var_dump():
array(1) { [0]=> string(5) "apple,pear,banana" }
What I need is the value of those to become variables as well as adding in value as the string.
so, I need the output to be:
$apple = "apple";
$pear = "pear";
$banana = "banana";
The value of the array could change so the variables have to be created depending on what is in that array.
I would appreciate all help. Cheers
Mark
How about
$choices = explode(',', $_POST['choices']);
foreach ($choices as $choice){
$$choice = $choice;
}
$str = "apple,pear,pineapple";
$strArr = explode(',' , $str);
foreach ($strArr as $val) {
$$val = $val;
}
var_dump($apple);
This would satisfy your requirement. However, here comes the problem, since you could not predefine how many variables are there and what are they, it's hard for you to use them correctly. Test "isset($VAR)" before using $VAR seems to be the only safe way.
You'd better just split the source string in just one array and just operate the elements of the specific array.
I have to concur with all the other answers that this is a very bad idea, but each of the existing answers uses a somewhat roundabout method to achieve it.
PHP provides a function, extract, to extract variables from an array into the current scope. You can use that in this case like so (using explode and array_combine to turn your input into an associative array first):
$choices = $_POST['choices'] ?: ""; // The ?: "" makes this safe even if there's no input
$choiceArr = explode(',', $choices); // Break the string down to a simple array
$choiceAssoc = array_combine($choiceArr, $choiceArr); // Then convert that to an associative array, with the keys being the same as the values
extract($choiceAssoc, EXTR_SKIP); // Extract the variables to the current scope - using EXTR_SKIP tells the function *not* to overwrite any variables that already exist, as a security measure
echo $banana; // You now have direct access to those variables
For more information on why this is a bad approach to take, see the discussion on the now deprecated register_globals setting. In short though, it makes it much, much easier to write insecure code.
Often called "split" in other langauges, in PHP, you'd want to use explode.
EDIT: ACTUALLY, what you want to do sounds... dangerous. It's possible (and was an old "feature" of PHP) but it's strongly discourage. I'd suggest just exploding them and making their values the keys of an associative array instead:
$choices_assoc = explode(',', $_POST['choices']);
foreach ($choices as $choice) {
$choices_assoc[$choice] = $choice;
}
The title may be a little confusing. This is my problem:
I know you can hold a variable name in another variable and then read the content of the first variable. This is what I mean:
$variable = "hello"
$variableholder = 'variable'
echo $$variableholder;
That would print: "hello". Now, I've got a problem with this:
$somearray = array("name"=>"hello");
$variableholder = "somearray['name']"; //or $variableholder = 'somearray[\'name\']';
echo $$variableholder;
That gives me a PHP error (it says $somearray['name'] is an undefined variable). Can you tell me if this is possible and I'm doing something wrong; or this if this is plain impossible, can you give me another solution to do something similar?
Thanks in advance.
For the moment, I could only think of something like this:
<?php
// literal are simple
$literal = "Hello";
$vv = "literal";
echo $$vv . "\n";
// prints "Hello"
// for containers it's not so simple anymore
$container = array("Hello" => "World");
$vv = "container";
$reniatnoc = $$vv;
echo $reniatnoc["Hello"] . "\n";
// prints "World"
?>
The problem here is that (quoting from php: access array value on the fly):
the Grammar of the PHP language only allows subscript notation on the end of variable expressions and not expressions in general, which is how it works in most other languages.
Would PHP allow the subscript notation anywhere, one could write this more dense as
echo $$vv["Hello"]
Side note: I guess using variable variables isn't that sane to use in production.
How about this? (NOTE: variable variables are as bad as goto)
$variablename = 'array';
$key = 'index';
echo $$variablename[$key];
I know this is not exactly reflection, but kind of.
I want to make a debug function that gets a variable and prints a var_dump and the variable name.
Of course, when the programmer writes a call to the function, they already know the variable's name, so they could write something like:
debug( $myvar, 'myvar' );
But I want it to be quick and easy to write, just the function name, the variable, and voilà !
debug( $myvar ); // quicker and easier :)
You can do it by converting the variable to a key/value set before passing it to the function.
function varName($theVar) {
$variableName = key($theVar);
$variableValue = $theVar[$variableName];
echo ('The name of the variable used in the function call was '.$variableName.'<br />');
echo ('The value of the variable used in the function call was '.$variableValue.'<br />');
}
$myVar = 'abc';
varName(compact('myVar'));
Though I don't recommend creating a reference to a nameless variable, function varName(&$theVar) works too.
Since compact() takes the variable name as a string rather than the actual variable, iterating over a list of variable names should be easy.
As to why you would want to do this -- don't ask me but it seems like a lot of people ask the question so here's my solution.
I know I'm answering a 4 year old question but what the hell...
compact() might help you is your friend here!
I made a similar function to quickly dump out info on a few chosen variables into a log for debugging errors and it goes something like this:
function vlog() {
$args = func_get_args();
foreach ($args as $arg) {
global ${$arg};
}
return json_encode(compact($args));
}
I found JSON to be the cleanest and most readable form for these dumps for my logs but you could also use something like print_r() or var_export().
This is how I use it:
$foo = 'Elvis';
$bar = 42;
$obj = new SomeFancyObject();
log('Something went wrong! vars='.vlog('foo', 'bar', 'obj'));
And this would print out like this to the logs:
Something went wrong! vars={"foo":"Elvis","bar":42,"obj":{"nestedProperty1":1, "nestedProperty2":"etc."}}
Word of warning though: This will only work for variables declared in the global scope (so not inside functions or classes. In there you need to evoke compact() directly so it has access to that scope, but that's not really that big of a deal since this vlog() is basically just a shortcut for json_encode(compact('foo', 'bar', 'obj')), saving me 16 keystrokes each time I need it.
Nope, not possible. Sorry.
Not elegantly... BUT YOU COULD FAKE IT!
1) Drink enough to convince yourself this is a good idea (it'll take a lot)
2) Replace all your variables with variable variables:
$a = 10
//becomes
$a = '0a';
$$a = 10;
3) Reference $$a in all your code.
4) When you need to print the variable, print $a and strip out the leading 0.
Addendum: Only do this if you are
Never showing this code to anyone
Never need to change or maintain this code
Are crazy
Not doing this for a job
Look, just never do this, it is a joke
I know this is very very late, but i did it in a different way.
It might honestly be a bit bad for performance, but since it's for debugging it shouldn't be a problem.
I read the file where the function is called, on the line it was called and I cut out the variable name.
function dump($str){
// Get the caller with debug backtrace
$bt = debug_backtrace();
$caller = array_shift($bt);
// Put the file where the function was called in an array, split by lines
$readFileStr = file($caller['file']);
// Read the specific line where the function was called
$lineStr = $readFileStr[$caller['line'] -1];
// Get the variable name (including $) by taking the string between '(' and ')'
$regularOutput = preg_match('/\((.*?)\)/', $lineStr, $output);
$variableName = $output[1];
// echo the var name and in which file and line it was called
echo "var: " . $variableName . " dumped in file: " . $caller['file'] . ' on line: ' . $caller['line'] . '<br>';
// dump the given variable
echo '<pre>' . var_export($str, true) . '</pre>';
}
i've had the same thought before, but if you really think about it, you'll see why this is impossible... presumably your debug function will defined like this: function debug($someVar) { } and there's no way for it to know the original variable was called $myvar.
The absolute best you could do would be to look at something like get_defined_vars() or $_GLOBALS (if it were a global for some reason) and loop through that to find something which matches the value of your variable. This is a very hacky and not very reliable method though. Your original method is the most efficient way.
No, the closer you will get is with get_defined_vars().
EDIT: I was wrong, after reading the user comments on get_defined_vars() it's possible with a little hack:
function ev($variable){
foreach($GLOBALS as $key => $value){
if($variable===$value){
echo '<p>$'.$key.' - '.$value.'</p>';
}
}
}
$lol = 123;
ev($lol); // $lol - 123
Only works for unique variable contents though.
Bit late to the game here, but Mach 13 has an interesting solution: How to get a variable name as a string in PHP
You could use eval:
function debug($variablename)
{
echo ($variablename . ":<br/>");
eval("global $". $variablename . ";");
eval("var_dump($" . $variablename . ");");
}
Usage: debug("myvar") not debug($myvar)
This is late post but I think it is possible now using compact method
so the code would be
$a=1;
$b=2;
$c=3
var_dump(compact('a','b','c'));
the output would be
array (size=3)
'a' => int 1
'b' => int 2
'c' => int 3
where variable name a, b and c are the key
Hope this helps
I believe Alix and nickf are suggesting this:
function debug($variablename)
{
echo ($variablename . ":<br/>");
global $$variablename; // enable scope
var_dump($$variablename);
}
I have tested it and it seems to work just as well as Wagger's code (Thanks Wagger: I have tried so many times to write this and the global variable declaration was my stumbling block)