Титла: BASH - Script Публикувано от: blind_fish в Oct 31, 2002, 12:36 Sdravei te, ot 4isto ljubopitstvo reshih da opitam bash-script sa da resha malak problem koito imah,no sa sazelenie intusiasma mi go pomra4i greshka kojato nemoga da rasbera
![]() #!/bin/bash #script "shell <directory>" cd $1 ls -l for i in * do if test -f $i then cat $i | sed '/require.*/s//require(\"\/somthing\/something\/new.phpformat\")\;/g' > tmp.txt && mv tmp.txt $i else . /http/services/equipment/shell $1/$i # Problema e tuk njakade - kogato nameri directory neiska da # vlese v neja i da ja pretarsi ili po skoro vlisa samo v # parvata i blockira fi done # end bash-script Snam 4e e njakva glupost ama njamam nervi ve4e ![]() Blagodra mnogo na vseki koito mi pomogne Титла: BASH - Script Публикувано от: в Oct 31, 2002, 15:17 Подмени "else . /http/services/equipment/shell $1/$i"
с това "else $0 $i" !!!ВАЖНО!!! викаш скрипта с ПЪЛНИЯ път. Успех Титла: BASH - Script Публикувано от: в Nov 04, 2002, 17:58 Sudo, blagodarja ti sa otgovora.
Tova naistina se okasa 4ast ot problema, no sa sazelenie nebeshe vsi4ko ![]() Doide i drug problem ot "ls" sastoto tasi komanda isvikva samo otnositelnite path-dove na failove te i kogato funkciata vlese v njakoja direktorija i ja pretarsi se opitva da nameri sledvastijat sa obrabotka fail/directorija pokasan parvona4alno ot "ls" ot momentnoto si mestopolozenie - v poslednata namerena direktorija => nemoze da nameri namerenoto ot ls ![]() ![]() Best regards blind_fish Титла: BASH - Script Публикувано от: в Nov 04, 2002, 20:20 Vse pak reshih da postna i celijat script:
#!/bin/bash # use: shell <directory_where_to_start> for i in $(find $1 -name \*.php) do if test -f $i then cat $i | sed '/require.*/s//require\("\/bla\/bla\/new.phpformat"\ ![]() fi done #end bash script ################# best regards blind_fish |