无法在PHP中打印$

This has a lot of answers on other pages of stackoverflow, like this question however none of them seem to work for me.
I've this small piece of code as:

echo $optionss='$'."q_option".$de;

In above code, i have a few variables by the name of "q_option0", "q_option1", "q_option2"... up to "q_option4". $de is a variable that holds count from 0-4.

Now i want to print $optionss= $q_option0 (this is in a for loop to get the increment of last number upto 4), but i'm unable to do so!
Please suggest me what i'm doing wrong!!

I Think you dint process the array $de. you can use foreach or for loop .Try this .

<?php
$de=array(1,2,3,4);
foreach($de as $val){
echo $optionss='$'."q_option".$val."<br>";  
}
?>

or use single quotes as given

<?php
$de=array(1,2,3,4);
foreach($de as $val){
echo $optionss='$q_option'.$val."<br>"; 
}
?>

or try this

<?php
$dollar="$";
$de=array(1,2,3,4);
foreach($de as $val){
echo $optionss=$dollar.'q_option'.$val."<br>";  
}
?>

output

$q_option1
$q_option2
$q_option3
$q_option4

Use single quote to print special characters

'$q_option'.$de;

Also you have forgotten to place $optionss within single quotes if you do not want its value printed out.

echo '$optionss= $q_option'.$de; // print the whole expression

But if you do this..

echo $optionss='$'."q_option".$val."<br>"; // assign right hand side of the expression to the variable '$optionss' and print that out
    // so you will only get output $q_option0, $q_option1... instead of $optionss=$q_option0, $optionss=$q_option1....

Below is solution to a different problem as I misunderstood it:

I assume its to dynamically create variable names and read their values

you can either create a temporary variable with the value as the name of the variable you want to access and then use double $$ to access the value of the variable inside the variable. Make sense?

$tmp = "q_option".$de;
echo $optionss=$$tmp;

If "q_option0", "q_option1", "q_option2" etc are variables, then you reference the value of these varaibles with $$. This will output "abc". Note that the variable $optionss is being assigned the value of the variable $q_option0 in this code at the same time as its value is being echoed.

$de = 0;
$q_option0 = "abc";

$var = "q_option".$de;
echo $optionss = $$var;

Remove the doller symbol '$'and use double $$ to access the value of the variable.

Assume $q_option0 = "400"; Try This :

<?php
$q_option0 = "400"; 
$de = 0;

$optionss="q_option".$de;
echo "Your Output: " . $$optionss;

Output :

Your Output: 400