Здравствуйте. В связи с тем, что увидел такое сообщение, возник вопрос.
Перед нажатием на кнопку "Установить отладчик" надо закомментировать строчки в php.ini?:
Code:
zend_extension="\usr\local\php5xt\php_xdebug-2.4.0rc4-5.4-vc9.dll"
xdebug.profiler_enable = 0
xdebug.profiler_enable_trigger = 0
xdebug.profiler_output_dir="/tmp"
xdebug.profiler_output_name = "cachegrind.out.%u"
xdebug.trace_output_dir = "/tmp"
xdebug.profiler_enable = 0
xdebug.profiler_enable_trigger = 0
xdebug.profiler_output_dir="/tmp"
xdebug.profiler_output_name = "cachegrind.out.%u"
xdebug.trace_output_dir = "/tmp"
Полный код фрагмента там такой:
Code:
; XDebug settings.
[xdebug]
zend_extension="\usr\local\php5xt\php_xdebug-2.4.0rc4-5.4-vc9.dll"
xdebug.auto_trace = 0
;xdebug.collect_includes = 1
;xdebug.collect_params = 0
;xdebug.collect_return = 0
;xdebug.collect_vars = 0
xdebug.default_enable = 0
;xdebug.dump.SERVER = REMOTE_ADDR,REQUEST_METHOD
;xdebug.dump.SERVER = REMOTE_ADDR,REQUEST_METHOD
;xdebug.dump.COOKIE = ""
;xdebug.dump.FILES = ""
;xdebug.dump.GET = ""
;xdebug.dump.POST = ""
;xdebug.dump.REQUEST = ""
;xdebug.dump.SERVER = ""
;xdebug.dump.SESSION = ""
;xdebug.dump_globals = 1
;xdebug.dump_once = 1
;xdebug.dump_undefined = 0
;xdebug.extended_info = 1
;xdebug.file_link_format = ""
;xdebug.idekey = ""
;xdebug.manual_url = "http://www.php.net"
;xdebug.max_nesting_level = 100
;xdebug.overload_var_dump = 1
;xdebug.profiler_append = 0
xdebug.profiler_enable = 0
xdebug.profiler_enable_trigger = 0
xdebug.profiler_output_dir="/tmp"
xdebug.profiler_output_name = "cachegrind.out.%u"
;xdebug.remote_autostart = 0
;xdebug.remote_enable = 0
;xdebug.remote_handler = "dbgp"
;xdebug.remote_host = "localhost"
;xdebug.remote_log = "none"
;xdebug.remote_mode = "req"
;xdebug.remote_port = 9000
;xdebug.show_exception_trace = 0
;xdebug.show_local_vars = 0
;xdebug.show_mem_delta = 0
;xdebug.trace_format = 0
;xdebug.trace_options = 0
xdebug.trace_output_dir = "/tmp"
;xdebug.trace_output_name = "trace.%H%R"
;xdebug.var_display_max_children = 128
;xdebug.var_display_max_data = 512
;xdebug.var_display_max_depth = 3
; Local Variables:
; tab-width: 4
; End:
[xdebug]
zend_extension="\usr\local\php5xt\php_xdebug-2.4.0rc4-5.4-vc9.dll"
xdebug.auto_trace = 0
;xdebug.collect_includes = 1
;xdebug.collect_params = 0
;xdebug.collect_return = 0
;xdebug.collect_vars = 0
xdebug.default_enable = 0
;xdebug.dump.SERVER = REMOTE_ADDR,REQUEST_METHOD
;xdebug.dump.SERVER = REMOTE_ADDR,REQUEST_METHOD
;xdebug.dump.COOKIE = ""
;xdebug.dump.FILES = ""
;xdebug.dump.GET = ""
;xdebug.dump.POST = ""
;xdebug.dump.REQUEST = ""
;xdebug.dump.SERVER = ""
;xdebug.dump.SESSION = ""
;xdebug.dump_globals = 1
;xdebug.dump_once = 1
;xdebug.dump_undefined = 0
;xdebug.extended_info = 1
;xdebug.file_link_format = ""
;xdebug.idekey = ""
;xdebug.manual_url = "http://www.php.net"
;xdebug.max_nesting_level = 100
;xdebug.overload_var_dump = 1
;xdebug.profiler_append = 0
xdebug.profiler_enable = 0
xdebug.profiler_enable_trigger = 0
xdebug.profiler_output_dir="/tmp"
xdebug.profiler_output_name = "cachegrind.out.%u"
;xdebug.remote_autostart = 0
;xdebug.remote_enable = 0
;xdebug.remote_handler = "dbgp"
;xdebug.remote_host = "localhost"
;xdebug.remote_log = "none"
;xdebug.remote_mode = "req"
;xdebug.remote_port = 9000
;xdebug.show_exception_trace = 0
;xdebug.show_local_vars = 0
;xdebug.show_mem_delta = 0
;xdebug.trace_format = 0
;xdebug.trace_options = 0
xdebug.trace_output_dir = "/tmp"
;xdebug.trace_output_name = "trace.%H%R"
;xdebug.var_display_max_children = 128
;xdebug.var_display_max_data = 512
;xdebug.var_display_max_depth = 3
; Local Variables:
; tab-width: 4
; End:
В поле "Веб сервер" что вписывать? Какую строку в точности? Установлен Аpache 2.2.31 по пути "C:\WebServers\usr\local\apache" в Windows XP.
Или, если ориентироваться на виртуальный путь, как на скриншоте, где прописан PHP
"Z:\usr\local\apache\"
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