Dollar ($) Sign in Password String Treated as Variable

Dollar ($) sign in password string treated as variable

$_DB['password'] = 'mypas$word';

Single quote strings are not processed and are taken "as-is". You should always use single quote strings unless you specifically need the $variable or escape sequences (\n, \r, etc) substitutions. It's faster and less error prone.

PHP adding $ signs into strings without it being a variable?

As PHP Document says

If the string is enclosed in double-quotes ("), PHP will interpret the
following escape sequences for special characters: \$ Dollar Sign

http://php.net/manual/en/language.types.string.php

So if you want to use $ in a double-quotes strings use \$ instead of $.

Or simply use single quotes

What is the purpose of the dollar sign in JavaScript?

A '$' in a variable means nothing special to the interpreter, much like an underscore.

From what I've seen, many people using jQuery (which is what your example code looks like to me) tend to prefix variables that contain a jQuery object with a $ so that they are easily identified and not mixed up with, say, integers.

The dollar sign function $() in jQuery is a library function that is frequently used, so a short name is desirable.

How to delete dollar signs($) from a PHP variable only if the dollar signs are surrounding a word

First of all, i will answer your question

Use preg_replace instead, that is uses with a pattern that detects only those dollar signs, that are surrounding a word.

$string = preg_replace('/\$(\w+)\$/', '$1', $string);

Now an improvement for your code

As mentioned in the comments of your question, you would need to surround your string with single quotes (') to prevent PHP from replacing your words, that are prefixed by a dollar sign, with their corresponding variables.

A good source to understand the differences between double- and single-enquoted strings can be found from this question.

Escape $ character inside a MS SQL query in PHP

You can escape the $ and then it won't be read as a variable by PHP.

echo "\$test";

Demo: https://eval.in/605867 Vs. https://eval.in/605866 (which is empty because $test doesnt exist)

...

or since your query doesn't use single quotes just use single quotes for the string encapsulation.

echo '$test';

Use the $ symbol as text, not as a variable in PHP

It's been a while since I wrote PHP, but I believe that if you use single quotes(') instead of double quotes("), PHP won't evaluate it.

You may also be able to escape it with a backslash: \$.

How to echo a $ sign in php

This won't work:

$replacements[1] = '$name';
$replacements[0] = '$company';

Use double-quotes:

$replacements[1] = "$name";
$replacements[0] = "$company";

or no quotes:

$replacements[1] = $name;
$replacements[0] = $company;

Use this:

$body = "<html><head>...</head><body> " . $inserted_text_dynamic . " </body>";

Instead of:

$body = <html><head>...</head><body> $inserted_text_dynamic </body>

And remove firts brackets in

$inserted_text_dynamic = (preg_replace($patterns, $replacements, $inserted_text));

like this:

$inserted_text_dynamic = preg_replace($patterns, $replacements, $inserted_text);

Dollar ($) Sign in Password String Treated as Variable