I am trying to fopen()
a debugging text file, which I have simply named debug.txt and put it on my desktop. I am using PHP. My code is simply
$debug_file = fopen( "C:\\Users\\joe\\Desktop\\debug.txt", "w" );
I keep getting this error
Parse error: syntax error, unexpected '$debug_file' (T_VARIABLE) ` on line 755, which is the line of code above.
I have checked the code before this line for a missing semicolon, as that often is the source of a syntax error, but the previous code is fine. If I comment out my one line of code, the PHP file no longer gives a syntax error.
I was thinking that there is something wrong with the way I write the string literal file path to open. I have tried to make it ok by escaping the backslashes. I'm using Windows 10. But that hasn't fixed the problem. For the life of me I can't figure out what the syntax error is.
Thanks for any help.
EDIT: As requested, the previous lines of code are:
add_shortcode('hide-it', 'hide_it_func');
function hide_it_func(){
return;
}