我一直在玩Elm导航包,目前我正在使用Hop框架,我想知道是否有办法防止cannot GET /<url>
消息没有前面的英镑符号网址(示例:#<url>
)。
答案 0 :(得分:2)
我觉得这可能是服务器配置问题。要让#-less网址正常工作,您需要将服务器配置为提供相同的.html
,而不管请求网址是什么。提供后,您的Elm应用程序可以加载并从那里获取。
答案 1 :(得分:2)
为了补充Peter Szerzo的答案,我发现an issue at the Github page of browser-sync页面与我的问题大致相同。解决方案如下:
StringBuilder spawn_async_with_pipes_output;
private static bool process_line (IOChannel channel, IOCondition condition, string stream_name) {
if (condition == IOCondition.HUP) {
stdout.printf ("%s: The fd has been closed.\n", stream_name);
return false;
}
try {
string line;
channel.read_line (out line, null, null);
stdout.printf ("%s: %s", stream_name, line);
} catch (IOChannelError e) {
stdout.printf ("%s: IOChannelError: %s\n", stream_name, e.message);
return false;
} catch (ConvertError e) {
stdout.printf ("%s: ConvertError: %s\n", stream_name, e.message);
return false;
}
return true;
}
public int execute_sync_multiarg_command_pipes(string[] spawn_args) {
debug("Starting to execute async command: "+string.joinv(" ", spawn_args));
spawn_async_with_pipes_output.erase(0, -1); //clear the output buffer
MainLoop loop = new MainLoop ();
try {
string[] spawn_env = Environ.get();
Pid child_pid;
int standard_input;
int standard_output;
int standard_error;
Process.spawn_async_with_pipes ("/",
spawn_args,
spawn_env,
SpawnFlags.SEARCH_PATH | SpawnFlags.DO_NOT_REAP_CHILD,
null,
out child_pid,
out standard_input,
out standard_output,
out standard_error);
// capture stdout:
IOChannel output = new IOChannel.unix_new (standard_output);
output.add_watch (IOCondition.IN | IOCondition.HUP, (channel, condition) => {
return process_line (channel, condition, "stdout");
});
// capture stderr:
IOChannel error = new IOChannel.unix_new (standard_error);
error.add_watch (IOCondition.IN | IOCondition.HUP, (channel, condition) => {
return process_line (channel, condition, "stderr");
});
ChildWatch.add (child_pid, (pid, status) => {
// Triggered when the child indicated by child_pid exits
Process.close_pid (pid);
loop.quit ();
});
loop.run ();
} catch(SpawnError e) {
warning("Failure in executing async command ["+string.joinv(" ", spawn_args)+"] : "+e.message);
spawn_async_with_pipes_output.append(e.message);
}
debug("Completed executing async command["+string.joinv(" ", spawn_args)+"]...");
return 0;
}
public static int main (string[] args) {
spawn_async_with_pipes_output = new StringBuilder ();
execute_sync_multiarg_command_pipes ({"pkexec", "ls", "-l", "-h"});
return 0;
}
如果您要使用Apache或Nginx,当然需要使用不同的解决方案。这是特定于浏览器同步的。
<强>更新强>
我遇到了另一个问题,上面的解决方案不起作用,因为它会从不同的位置再次获取项目。例如:手动导航到const modRewrite = require("connect-modrewrite");
gulp.task("serve", () => {
browserSync.init(null, {
middleware: [
modRewrite([
"!\\.\\w+$ /index.html [L]"
]),
]
}
}
可以很好地工作,因为在更改到该位置时没有任何HTTP请求,但是当重新加载页面时,页面将从/blog/1
获取文件,上面的解决方案没有处理。这就是我解决后一个问题的方法:
/blog