path

问题 When I try to run python or any language through Sublime Text 2, all it says is [Error 2] The system cannot find the file specified [cmd: [u'python', u'-u', u'C:\\Users\\Owner\\Desktop\\RandomStuff\\python\\java\\bob.java']] [dir: C:\Users\Owner\Desktop\RandomStuff\python\java] [path: C:\Windows\system32;C:\Windows;C:\Windows\System32\Wbem;C:\Windows\System32\WindowsPowerShell\v1.0\;C:\Program Files\QuickTime\QTSystem\;C:\Program Files\Java\jdk1.7.0_03\bin] [Finished] I was told to change my

问题 I know this has been asked many times and I have searched far and wide for a solution to this probably simple problem. I'm trying to follow the simple javaFX component tutorials on Oracle's website. I can define an image this way: Image img = new Image("images/portal.png", 50, 50, true, true); This works when the image is in a folder inside the "src" folder, but I'm trying to get it to find the image when I have the image folder outside of the "src" folder, like this: http://puu.sh/drF7K

问题 Normally, I launch jupyter from a new bash terminal: ~$ jupyter notebook and then navigate down the tree to the files I want to work at. I just did it from a terminal that was already running at a different path: ~/blah/blah/blah$ jupyter notebook and realized that that directory becomes the home for jupyter. I wanted to work at a file located at, say ~/blah/thingy/luis.ipynb , but couldn't find my way up . How do I navigate the directory tree upwards in jupyter notebook? 回答1: As explained in

问题 public/images/portfolio/i-vis/1.jpg How could i remove all the path regardless of what the filename is using php? 回答1: Have a look at basename() $path = 'public/images/portfolio/i-vis/1.jpg' $name = basename($path); // $name == '1.jpg' Also, dirname() fetches the other part $dir = dirname($path); // $dir == 'public/images/portfolio/i-vis' If you need even more information - there is pathinfo() $info = pathinfo($path); var_dump($info); produces array(4) { ["dirname"]=> string(29) "public

问题 public/images/portfolio/i-vis/1.jpg How could i remove all the path regardless of what the filename is using php? 回答1: Have a look at basename() $path = 'public/images/portfolio/i-vis/1.jpg' $name = basename($path); // $name == '1.jpg' Also, dirname() fetches the other part $dir = dirname($path); // $dir == 'public/images/portfolio/i-vis' If you need even more information - there is pathinfo() $info = pathinfo($path); var_dump($info); produces array(4) { ["dirname"]=> string(29) "public

问题 I want to use functions of Path class ( GetDirectoryName , GetFileName , Combine ,etc.) with paths in URL format with slash (/). Example of my path: "xxx://server/folder1/folder2/file" I tried to do the job with Path functions and in the end just replaced the separator. I've found that the GetDirectoryName function does not correctly replace the slashes: Path.GetDirectoryName(@"xxx://server/folder/file") -> @"xxx:\server\folder" Like you see one slash is lost. How can I cause the Path

问题 I'm trying to use the include.path config described here, but I can't seem to find the correct path syntax on Windows. My current .gitconfig: [include] path = 'D:\Scott\Shared\config\common.gitconfig' But git complains: fatal: bad config file line 2 in C:\Users\Scott/.gitconfig What's the proper way to escape paths for Windows? Note: I'm using git from Powershell and not git bash. 回答1: Ok, figured this out. The trick is: Surround in double-quotes Convert backslashes to forward slashes. Begin

问题 I’ve got several Perl modules installed on my hosting machine in ~/perl , how do I add them to Perl module path? Setting PERL5LIB or unshift ing the paths to @INC surely works, but the environment variable does not help when running as a CGI script and the @INC way is not very portable. Is there a better way? This has to be a common problem, am I missing something? 回答1: PERL5LIB works just fine for CGI scripts. You just have to set the variable in the right place, such as the server

问题 i need to create a function that will accept a directory path. But in order for the compiler to read backslash in i need to create a function that will make a one backslash into 2 backslash.. so far this are my codes: string stripPath(string path) { char newpath[99999]; //char *pathlong; char temp; strcpy_s(newpath, path.c_str()); //pathlong = newpath; int arrlength = sizeof(newpath); for (int i = 0; i <= arrlength ;i++) { if(newpath[i] == '\\') { newpath[i] += '\\'; i++; } } path = newpath;

问题 i have the code below and wonder, why the textpath isn't displayed? Can anyone help? var areas = svg.append("g").selectAll("path") .data(chord.groups) .enter().append("path") .style("fill", function(d) { return fill(d.index); }) .style("stroke", function(d) { return fill(d.index); }) .attr("class", "bar") .attr("id", function(d, i) { return "group" + i; }) .attr("d", d3.svg.arc().innerRadius(innerRadius).outerRadius(outerRadius)) .on("mouseover", fade(.1)) .on("mouseout", fade(1)); svg