Cannot use php String variable in place of html plaintext

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cannot
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If the variable $a is replaced with plaintext then the code works, however when it links to a string variable it does not.,Connect and share knowledge within a single location that is structured and easy to search., Why do electricians in some areas choose wire nuts over reusable terminal blocks like Wago offers? , Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers

Replace

< ? php $a; ? >
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$a = '12345';

// This works:
echo "qwe{$a}rty"; // qwe12345rty, using braces
echo "qwe".$a.
"rty"; // qwe12345rty, concatenation used

// Does not work:
echo 'qwe{$a}rty'; // qwe{$a}rty, single quotes are not parsed
echo "qwe$arty"; // qwe, because $a became $arty, which is undefined
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Including expressions in strings,This can work just like a normal string, except you can include variables in it, wrapped inside ${ } characters, and the variable's value will be inserted into the result:,Vue conditional rendering: editing existing todos,To have the equivalent output using a normal string you'd have to include line break characters (\n) in the string:

const string = 'The revolution will not be televised.';
console.log(string);
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Please note that variable variables cannot be used with PHP's Superglobal arrays within functions or class methods. The variable $this is also a special variable that cannot be referenced dynamically. , A variable variable takes the value of a variable and treats that as the name of a variable. In the above example, hello, can be used as the name of a variable by using two dollar signs. i.e. , Variable variables , Sometimes it is convenient to be able to have variable variable names. That is, a variable name which can be set and used dynamically. A normal variable is set with a statement such as:

<
? php$a = 'hello'; ? >
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The goal of the printf templating language is to replace variable expansion. Instead, you put variable placeholders inside a string. printf will then replace them with the arguments that you passed to the function.,Above is an example of the simple syntax. Anytime PHP encounters a $, it’ll try to match the text after to a valid variable name. That said, it can’t expand variables like PHP_VERSION since they don’t use a $. That’s why our example assigned the value of the constant to the version variable.,Here's the full definition of a printf placeholder:,s tells printf that the argument is a string and to leave it as is.

Simple syntax

$version = PHP_VERSION;
echo "Current PHP version:  $version";
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