I'm confused in how to use $$ to use a string as a variable, mainly when it comes to use a string to refer an array index.
Consider the following case:
$colors = array(
'r'=>"red",
'b'=>"blue"
);
$vr = "colors[r]"; //I tried even this "color['r']"
echo $$vr; // I tried even this ${$vr}
Can anyone tell if it is possible to do the above.
expected o/p is red using "color[r]" as string and then using it as variable.
You can't do that directly. Consider the following:
$varName = array_shift(explode('[', $vr));
foreach($$varName as $key=>$value){
echo $key.": ".$value."<br />";
}
this will print out:
r: red
b: blue
The variable variable is just the first part (colors). You can't include the key in this.
Related
I'm not even sure if what I am trying to do is possible, I have a simple php echo line as below..
<?php echo $T1R[0]['Site']; ?>
This works well but I want to make the "1" in the $T1R to be fluid, is it possible to do something like ..
<?php echo $T + '$row_ColNumC['ColNaumNo']' + R[0]['Site']; ?>
Where the 1 is replaced with the content of ColNaumNo i.e. the returned result might be..
<?php echo $T32R[0]['Site']; ?>
It is possible in PHP. The concept is called "variable variables".
The idea is simple: you generate the variable name you want to use and store it in another variable:
$name = 'T'.$row_ColNumC['ColNaumNo'].'R';
Pay attention to the string concatenation operator. PHP uses a dot (.) for this, not the plus sign (+).
If the value of $row_ColNumc['ColNaumNo'] is 32 then the value stored in variable $name is 'T32R';
You can then prepend the variable $name with an extra $ to use it as the name of another variable (indirection). The code echo($$name); prints the content of variable $T32R (if any).
If the variable $T32R stores an array then the syntax $$name[0] is ambiguous and the parser needs a hint to interpret it. It is well explained in the documentation page (of the variable variables):
In order to use variable variables with arrays, you have to resolve an ambiguity problem. That is, if you write $$a[1] then the parser needs to know if you meant to use $a[1] as a variable, or if you wanted $$a as the variable and then the [1] index from that variable. The syntax for resolving this ambiguity is: ${$a[1]} for the first case and ${$a}[1] for the second.
You can do like this
$T1R[0]['Site'] = "test";
$c = 1;
$a = "T".$c."R";
$b = $$a;
echo "<pre>";
print_r($b[0]['Site']);
Or more simpler like this
$T1R[0]['Site'] = "test";
$c = 1;
$a = "T".$c."R";
echo "<pre>";
print_r(${$a}[0]['Site']);
Im trying to get a list of variables into an array (for an error reporting class), but if the variable is NOT set it is not being "compacted".
The below is extracts of the code:
$testVar1 = 123;
$testVar2 = 'ABC';
$ErrorArray = compact('testVar1', 'testVar2', 'notSetVar');
I then walk through the $ErrorArray with :
foreach($ErrorArray as $key => $value) {
$TempErrorMessage .= '$'.$key.' == '.$value.' ---- ';
}
The resulting output is :
$testVar1 == 123 ---- $testVar2 == ABC ----
The problem is, i would like to to output "notSetVar" as ""/NULL, as this is likely to be where my error is....
Any suggestions would be greatly welcomed!
Best Regards
Ford
According to PHP doc
http://php.net/manual/en/function.compact.php
compact creates an array containing variables and their values.
For each of these, compact() looks for a variable with that name in the current symbol table and adds it to the output array such that the variable name becomes the key and the contents of the variable become the value for that key. In short, it does the opposite of extract().
Any strings that are not set will simply be skipped.
So, it is not possible to pass variable via compact unless its set. My suggestion is, check variable before compact().
$testVar1 = 123;
$testVar2 = 'ABC';
if (!isset($notSetVar) {
$notSetVar = null;
}
$ErrorArray = compact('testVar1', 'testVar2', 'notSetVar');
var_dump($ErrorArray);
I have one variable that comes from a database.
I then want to check whether that value is the same of one of the values in an array.
If the variable matches one of the array values, then I want to print nothing, and if the variable does not match one of the array values, then I want to print something.
This is the code I have been trying without luck, I know that contains is not valid code, but that is the bit I cannot find any info for:
<?php
$site = getStuff();
$codes = array('value2', 'value4');
if ($codes contains $site)
{
echo "";
}
else
{
echo "something";
?>
So if the database would return value1 for $site, then the code should print "something" because value1 is not in the array.
The function you are looking for is in_array.
if(in_array($site, array('value2', 'value4')))
if(!in_array($site,$codes)) {
echo "something";
}
To provide another use way to do what the other answers suggest you can use a ternary if
echo in_array($site, $codes)?"":"something";
$arr[0]=123;
$a="arr[0]";
echo $$a;
gives me error
Notice: Undefined variable: arr[0]
on the last line.
What should I do to make it work?
EDIT:
Above is the simplification of what I want to do. If someone wants to know why I want to do this, then here's the explanation:
This is something like what I want to do:
if(condition){
$a=$arr1[0][0];
$b=$arr1[0][1];
$c=$arr1[0][2];
}
else{
$a=$arr2[0];
$b=$arr2[1];
$c=$arr2[2];
}
I can compact it like this:
if(condition)
$arr=$arr1[0];
else
$arr=$arr2;
$a=$arr[0];
$a=$arr[1];
$a=$arr[2];
But I wanted to try doing this using variable variable:
if(condition)
$arr="$arr1[0]";
else
$arr="$arr2";
$a={$$arr}[0];
$b={$$arr}[1];
$c={$$arr}[2];
Sure, we don't need variable variables as we can still code without them. I want to know, for learning PHP, why the code won't work.
Now that you said what you’re actually trying to accomplish: Your code doesn’t work because if you look at $arr1[0][0], only arr is the variable name; the [0] are special accessors for certain types like strings or arrays.
With variable variables you can only specify the name but not any accessor or other operation:
A variable variable takes the value of a variable and treats that as the name of a variable.
Your solution with the additional variable holding the array to access later on would be the best solution to your problem.
What you are trying to do just won't work - the code $arr[0] is referencing a variable called $arr, and then applying the array-access operator ([$key]) to get the element with key 0. There is no variable called $arr[0], so you cannot reference it with variable-variables any more than you could the expression $foo + 1 .
The real question is why you want to do this; variable variables are generally a sign of very messy code, and probably some poor choices of data structure. For instance, if you need to select one of a set of variables based on some input, you probably want a hash, and to look up an item using $hash[$item] or similar. If you need something more complex, a switch statement can often cover the cases you actually need.
If for some reason you really need to allow an arbitrary expression like $arr[0] as input and evaluate it at runtime, you could use eval(), but be very very careful of where the input is coming from, as this can be a very easy way of introducing security holes into your code.
FROM PHP DOC
In order to use variable variables with arrays, you have to resolve an ambiguity problem. That is, if you write $$a[1] then the parser needs to know if you meant to use $a[1] as a variable, or if you wanted $$a as the variable and then the [1] index from that variable. The syntax for resolving this ambiguity is: ${$a[1]} for the first case and ${$a}[1] for the second.
Use
echo ${$a}[0]; // 123
Edit : Based on your edit you can simply have
list($a, $b, $c) = (condition) ? $arr1[0] : $arr2;
Or
$array = (condition) ? $arr1[0] : $arr2;
$a = $array[0];
$b = $array[1];
$c = $array[2];
As pointed out you don't need variable variables. To get a PHP variable variable name containing index (a key) use array_keys() or array_search() or other array parsers. From php's site:
$array = array(0 => 'blue', 1 => 'red', 2 => 'green', 3 => 'red');
$key = array_search('green', $array); // $key = 2;
$key = array_search('red', $array); // $key = 1;
You could also use the following (using $var= instead of echo):
$arr[0]=123;
$arr[1]=456;
foreach ($arr as $key => $value) {
echo "arr[{$key}] = {$value} \r\n";
}
Which outputs:
arr[0] = 123
arr[1] = 456
But I don't see why you'd do that, since the whole point of the array is not doing that kind of stuff.
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];