variables

问题 I have one file named style.scss with the following code: @import 'variables'; body { color: $text-color; background: $background-color; } And one partial named _variables.scss: $colorscheme: white; @if $colorscheme == white { $text-color: #333; $background-color: #fff; } @else { $text-color: #ccc; $background-color: #333; } The if-statement works properly, but the variables defined inside, do not work. When I try to compile it, I keep getting: Syntax error: Undefined variable: “$text-color”.

问题 I have one file named style.scss with the following code: @import 'variables'; body { color: $text-color; background: $background-color; } And one partial named _variables.scss: $colorscheme: white; @if $colorscheme == white { $text-color: #333; $background-color: #fff; } @else { $text-color: #ccc; $background-color: #333; } The if-statement works properly, but the variables defined inside, do not work. When I try to compile it, I keep getting: Syntax error: Undefined variable: “$text-color”.

问题 I have one file named style.scss with the following code: @import 'variables'; body { color: $text-color; background: $background-color; } And one partial named _variables.scss: $colorscheme: white; @if $colorscheme == white { $text-color: #333; $background-color: #fff; } @else { $text-color: #ccc; $background-color: #333; } The if-statement works properly, but the variables defined inside, do not work. When I try to compile it, I keep getting: Syntax error: Undefined variable: “$text-color”.

问题 I'm not entirely sure if this is possible in Ruby, but hopefully there's an easy way to do this. I want to declare a variable and later find out the name of the variable. That is, for this simple snippet: foo = ["goo", "baz"] How can I get the name of the array (here, "foo") back? If it is indeed possible, does this work on any variable (e.g., scalars, hashes, etc.)? Edit: Here's what I'm basically trying to do. I'm writing a SOAP server that wraps around a class with three important

问题 I know this is a frequent question on Stack Overflow but I couldn't find scenario where the PHP code is just above JS script in the same file. I guess ajax POST/GET won't do the job in this case? (...) var ww = $(window).width(); </script> <?php $ww = [data here] ?> 回答1: You will have to use POST/GET to send a JS variable to PHP, because PHP is processed by the server and JS is processed by the client. I know this isn't what you want to hear. 回答2: This is the wrong approach. The PHP is all

问题 I know this is a frequent question on Stack Overflow but I couldn't find scenario where the PHP code is just above JS script in the same file. I guess ajax POST/GET won't do the job in this case? (...) var ww = $(window).width(); </script> <?php $ww = [data here] ?> 回答1: You will have to use POST/GET to send a JS variable to PHP, because PHP is processed by the server and JS is processed by the client. I know this isn't what you want to hear. 回答2: This is the wrong approach. The PHP is all

问题 I know this is a frequent question on Stack Overflow but I couldn't find scenario where the PHP code is just above JS script in the same file. I guess ajax POST/GET won't do the job in this case? (...) var ww = $(window).width(); </script> <?php $ww = [data here] ?> 回答1: You will have to use POST/GET to send a JS variable to PHP, because PHP is processed by the server and JS is processed by the client. I know this isn't what you want to hear. 回答2: This is the wrong approach. The PHP is all

问题 This question already has answers here : PHP Pass variable to next page (8 answers) Closed 5 years ago . I need to pass a variable from page one to third page how do I make it? I can pass variable from 1 page to next how can i to pass variable from 1st to second page and same value to 3rd page here is the example <a href="page2.php?myname=Suraj Mittal&age=22&fav_color=White&fav_fruit=Grapes">Click here to send variables in the URL</a> second page <?php $name = $_GET['myname']; $age = $_GET[

问题 Alright, PHP is throwing this error (in the log) and it's quite detrimental to the application. PHP Notice: Undefined index: sessid in methods.php on line 7 Yes, I'm aware of what this error means, what I cannot figure out is why it's obviously defined and saying it's undefined. Here is the relevant code in methods.php $sessid = mysql_real_escape_string($_REQUEST['sessid']); Now before you go off and say that "NO IT'S UNDEFINED!!!", here is the POST request to the methods.php (and yes, I'm

问题 Hello I want to make program wich will say how many big and short letters is in the word and such but run in to the problem I can't declare content of array dynamically. This is all C code. I tried this char something; scanf("%c",somethnig); char somethingmore[]=something; printf("%c",something[0]) but it wasn't possible to compile I also tried something like this char *something; scanf("%c",something); printf("%c",something[0]); wich was possible to compile but crushed when called array