Escape a star (*) character in bash

Always put double quotes around variable substitutions, otherwise characters like spaces and * appearing in the value are interpreted by the shell. I.e., write "$c", not $c.
The syntax mysql <"$c" makes mysql execute commands from a file whose name is the value of $c. What you're looking for is
```
printf '%s\n' "$c" | mysql
```
or simpler, as long as you remember the restrictions ($c must not start with a -, and if it contains \ that's ok in bash but not in some other variants of sh)
```
echo "$c" | mysql
```
There's another alternative that's more comfortable if the command is multiline. It's called a “here-document”. The string EOF isn't special (though it's traditional), any sequence of letters and digits will do. The terminating EOF may not be preceded by whitespace. You need to put a \ before every $, \ and ` unless you want them interpreted by the shell.
```
mysql <<EOF
GRANT ALL ON *.* TO '$1'@'localhost';
EOF
```
Beware that if the argument to the shell contains a single quote, you have an injection vector. The following snippet adds a \ before every \ and '.
```
set -- "${1//\\/\\\\}"
set -- "${1//\'/\'}"
```
This is fairly ugly, which is why if you're going to do anything complicated, forget about using a shell and use a language with actual SQL bindings (perl, python, whatever) where the library handles all the quoting and procedure building for you.

Use single quotes for your string:

c='GRANT ALL ON *.* TO';
c="${c} '$1'@'localhost';";

There is probably a better way to do that but including $1 in the string made it weird

This will work in bash, no escaping necessary

#!/bin/bash
mysql -u root -e "GRANT ALL ON *.* TO '$1'@'localhost'"
exit 0;

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