How to solve PHP error: unexpected "$" symbol?
PHP is a widely used programming language that is often used to develop websites and web applications. However, when writing PHP code, we often encounter various errors and errors. This article will focus on a common error, the "unexpected "$" sign", and provide ways to resolve it.
What is the "unexpected "$" sign" error?
In PHP, variables start with the $ symbol, such as $variable. If an invalid $ sign appears in the code, the PHP interpreter will report an error and prompt "Unexpected "$" sign". This error is usually caused by the following reasons:
Solution:
Example:
$var1 = "Hello"; $var2 = 123; echo $var1; // 输出"Hello" echo $var2; // 输出123
In this example, $var1 and $var2 are valid variable names and they are both quoted and used correctly.
Example:
if ($condition) { // 缺少了右花括号 } echo "Condition is true"; }
In this example, due to the missing right curly brace, an "unexpected "$" symbol" error will be reported. We can fix this error by simply adding a closing curly brace.
Example:
$name = "John; // 缺少了右引号 " $age = 25 echo "My name is ".$name." and I'm ".$age." years old";
In this example, due to the missing closing quotation mark, an "unexpected "$" symbol" error will be reported. We simply add a closing quote after the closing quote on the definition line of the $name variable and a semicolon after the definition line of the $age variable to resolve this error.
Summary:
It is not uncommon to encounter "unexpected "$" sign" errors when writing PHP code, but we can do this by carefully checking the use of variable names, syntax, and punctuation. to resolve this error. I hope the solutions provided in this article can help you solve this problem and improve your PHP programming skills.
The above is the detailed content of How to solve PHP error: unexpected '$' symbol?. For more information, please follow other related articles on the PHP Chinese website!