PHP代码中如何让echo $test输出6而非$j7？

Got it, let's figure out how to fix this for you!

First, let's understand why your current code outputs $j7 instead of 6. In PHP, single quotes (') treat everything inside them as literal text—so '$j' doesn't get parsed as the variable $j, it's just the string $j. When you concatenate that with $b (which is 7), you end up with the string $j7, not the value of the actual variable $j7.

Now, here are a few solid ways to get the output 6:

1. Use Double Quotes with Curly Braces #

PHP parses variables inside double quotes, and curly braces let you clearly define the full variable name when it's followed by other characters. This is clean and readable:

<?php 
$j7 = 6; 
$b = 7; 
$test = "${j$b}"; // This resolves to the variable $j7
echo $test; 
?>

The ${j$b} tells PHP to combine the string j with the value of $b (7) to make the variable name j7, then grab its value (6).

2. Use Variable Variables #

This is another straightforward approach. First build the variable name as a string, then use two dollar signs to access its value:

<?php 
$j7 = 6; 
$b = 7; 
$varName = 'j' . $b; // Creates the string 'j7'
$test = $$varName; // $$varName is the same as $j7
echo $test; 
?>

This is great if you need to dynamically build variable names often—it's explicit and easy to follow.

3. Avoid eval() (Unless Absolutely Necessary) #

While this works, eval() is risky if you're dealing with any user-provided input, since it executes arbitrary code. But for completeness, here's how it would look:

<?php 
$j7 = 6; 
$b = 7; 
eval("\$test = \$j$b;"); 
echo $test; 
?>

The backslashes escape the first $ so PHP correctly parses $j$b as $j7.

My go-to recommendations are Method 1 or 2—they're safe, readable, and fit most use cases. Either one will make your echo $test output 6 like you want.

内容的提问来源于stack exchange，提问作者Muhammad Afdhal Bin Hassan