无法让Zorba使用PHP和Nginx在Windows 7上运行(Can't get Zorba working on Windows 7 with PHP and Nginx)
我试图在这些说明的帮助下在Windows 7机器上安装Zorba。 我已经完成了“验证Zorba”部分，但我无法完成“在PHP中启用Zorba扩展”部分。 当我尝试重启PHP时，弹出一个Windows对话框说;
我需要从源代码编译吗？ 说明说从源代码编译Zorba，而是从Zorba下载页面下载了Windows安装包。 我还必须从源代码编译吗？ 当然不是吗？
缺少zorba_api_wrapper.php说明说“找到文件zorba_api_wrapper.php”，但我找不到该名称的文件。 有一个名为zorba_api.php的文件，所以我用它来代替。 这是正确的文件吗？
php-cgi.exe我正在运行PHP的CGI版本。 我从命令提示符开始做;
php-cgi -b 127.0.0.1:9000这可能是导致错误对话的原因吗？ 我想让Apache启动PHP更常见。 （我使用的是Nginx而不是Apache。）
更新
正如Rodolfo所建议的那样，我已将
C:\Program Files\Zorba XQuery Processor 2.0.2\bin到PATH环境变量中并卸载了旧版本的Zorba。 现在，当我尝试通过执行来启动PHP时;php-cgi -b 127.0.0.1:9000我得到一个不同的Windows对话框;
问题详细信息中的信息是;
Problem signature: Problem Event Name: APPCRASH Application Name: php-cgi.exe Application Version: 5.3.2.0 Application Timestamp: 4b8ec866 Fault Module Name: php5ts.dll Fault Module Version: 5.3.2.0 Fault Module Timestamp: 4b8ec7e7 Exception Code: c0000005 Exception Offset: 000f56c0 OS Version: 6.1.7601.2.1.0.768.3 Locale ID: 2057 Additional Information 1: 0a9e Additional Information 2: 0a9e372d3b4ad19135b953a78882e789 Additional Information 3: 0a9e Additional Information 4: 0a9e372d3b4ad19135b953a78882e789如果我从
php.ini删除行extension=zorba_api.dll，PHP启动正常。I'm trying to install Zorba on a Windows 7 machine with the help of these instructions. I've completed the "Verify Zorba" section ok, but I can't complete the section "Enable Zorba extension in PHP". When I attempt to restart PHP, a Windows dialog box pops up saying;
Do I need to compile from source? The instructions say to compile Zorba from source but instead have downloaded the Windows installation package from the Zorba download page. Do I also have to compile from source? Surely not?
Missing zorba_api_wrapper.php The instructions say "locate the file zorba_api_wrapper.php" but I can't find a file of that name. There is a file called zorba_api.php so I've used that instead. Is that the correct file?
php-cgi.exe I'm running the CGI version of PHP. I start it from the command prompt by doing;
php-cgi -b 127.0.0.1:9000Could that be what's causing the error dialog? I guess it's more common to have Apache start PHP. (I'm using Nginx not Apache.)
Update
As suggested by Rodolfo, I've added
C:\Program Files\Zorba XQuery Processor 2.0.2\binto thePATHenvironment variable and uninstalled an older version of Zorba. Now when I try to start PHP by doing;php-cgi -b 127.0.0.1:9000I get a different Windows dialog;
The info in the Problem Details is;
Problem signature: Problem Event Name: APPCRASH Application Name: php-cgi.exe Application Version: 5.3.2.0 Application Timestamp: 4b8ec866 Fault Module Name: php5ts.dll Fault Module Version: 5.3.2.0 Fault Module Timestamp: 4b8ec7e7 Exception Code: c0000005 Exception Offset: 000f56c0 OS Version: 6.1.7601.2.1.0.768.3 Locale ID: 2057 Additional Information 1: 0a9e Additional Information 2: 0a9e372d3b4ad19135b953a78882e789 Additional Information 3: 0a9e Additional Information 4: 0a9e372d3b4ad19135b953a78882e789If I remove the line
extension=zorba_api.dllfromphp.ini, PHP starts ok.
原文：https://stackoverflow.com/questions/7650585
最满意答案
那么，考虑到你的用例看起来是什么，你就会错误地解决它。 你真的应该做更多的事情
version(A) { immutable int var = 1; } else version(B) { immutable int var = 2; } else { immutable int var = 3; }但在一般情况下，如果您正在专门测试符号是否存在，请使用
is(typeof(symbol))，其中symbol是要测试的符号的名称。 所以，如果你想测试变量var是否存在，你会做类似的事情static if(is(typeof(var))) { //var exists }当然要测试它不存在，你只是否定了这个条件：
static if(!is(typeof(var))) { //var does not exist }
typeof(exp)获取表达式的类型，如果表达式无效（由于变量不存在或者表达式中的函数不适用于这些参数或其他），则结果为void。is(type)检查类型是否void。 因此，is(typeof(exp))测试exp是否是一个有效的表达式，并且在它只是一个符号名称的情况下，这意味着它正在测试它是否是有效的符号。Well, given what your use case appears to be, you're going about it incorrectly. You really should do something more like
version(A) { immutable int var = 1; } else version(B) { immutable int var = 2; } else { immutable int var = 3; }But in the general case, if you're looking specifically to test whether a symbol exists, use
is(typeof(symbol))wheresymbolis the name of the symbol that you're testing for. So, if you wanted to test whether the variablevarexisted, you would do something likestatic if(is(typeof(var))) { //var exists }and of course to test that it doesn't exist, you just negate the condition:
static if(!is(typeof(var))) { //var does not exist }
typeof(exp)gets the type of an expression, and if the expression is invalid (because of a variable which doesn't exist or a function in the expression doesn't work with those arguments or whatever), then the result isvoid.is(type)checks whether the type is non-void. So,is(typeof(exp))tests whether exp is a valid expression, and in the case where it's just a symbol name, that means that it's testing whether it's a valid symbol or not.
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那么，考虑到你的用例看起来是什么，你就会错误地解决它。 你真的应该做更多的事情 version(A) { immutable int var = 1; } else version(B) { immutable int var = 2; } else { immutable int var = 3; } 但在一般情况下，如果您正在专门测试符号是否存在，请使用is(typeof(symbol)) ，其中symbol是要测试的符号的名称。 所以，如果你想测试变量var是否存在，你会做类 ...
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