I have a Zim Database myfile file with the content:
output "hello world"
I want to execute the file from command line. But, when I try to call zimmu myfile from shell, I see the following error:
*** Error *** "myfile" is not a know name.
>
In order to work, 'myfile' must have been created in directory 'zim' and must be located in zim0001.ws (if compiled).
If not compiled, 'myfile' must be located in the database directory.
Related
We are executing SQL loader command from shell script on RH Linux OS. The command passes both control file and data file as command line parameters. The syntax used is given below (masked sensitive data):
sqlldr userid=$connstring control=/local/abc-1.2.3/instances/www.abc.com/apps/int/script/bin/ABC_CONTROL.ctl data=$f log=/local/abc-1.2.3/instances/www.abc.com/apps/int/script/logs/ABC.log bad=/local/abc-1.2.3/instances/www.abc.com/apps/int/script/logs/ABC.bad
The data file name is passed as dynamic variable in a FOR loop to process multiple files. The data file extension is *.app and path is /local/abc-1.2.3/instances/www.abc.com/apps/int/script/input.
We have verified that $f variable is able to correctly point to data file. Also verified file permissions. We tried changing the directory paths as well.
Still script fails with below error: **SQL Loader-500 Unable to open file (ABC_CTL.dat)**, SQL Loader-553 File not found, SQL Loader-509 System error: No such file or directory
The same script runs with exact same syntax on another server. Please suggest any solutions.
With the above I get the error msg below in the console:
Can't open file 'ini=false': the system cannot find the file specified.
It's trying to open the filename 'ini=false'!?
ini=false is the default value. The documentation you referenced provides several examples: ini='file.ini' will create file.ini in the default system-dependent location, ini='./file.ini' will create file.ini in the ZeroBrane Studio folder, and ini='d:/file.ini' will create file.ini in the root directory of the D: drive.
I don't get any error messages in the console when I run it with zbstudio.exe -cfg "ini=false".
I am attempting to move my learn.db file to a new folder, however when I entered the necessary commands :
.output ./iivri.andre/lxdb/teenager.sql
I received this error message
sqlite> .output ./iivri.andre/lxdb/teenager.sql
Error: cannot open "./iivri.andre/lxdb/teenager.sql"
Error: cannot write to "./iivri.andre/lxdb/teenager.sql"
How can I fix this ?
Cheers,
Andre
I want to use a resource file with the name clean_environment.robot on windows. It is in the same folder like my robot file (C:\Users\xxxxx\Desktop\git\src), which is callig the resource file.
Basic file 30.robot:
** Settings **
Resource clean_environment.robot
** Test Cases **
MyTestCase
Clean environment
I get the error:
[ ERROR ] Error in file
'C:\Users\xxxxx\Desktop\git\src\30.robot':
Resource file 'Clean environment' does not exist.
The resource file:
*** Settings ***
*** Test Cases ***
Clean environment
Log to console 111
Edit:
The computer said: Resource file 'Clean environment' does not exist! But 'Clean environment is not the resource file, it is the keyword right?
If the file has a test case in it, it is not considered to be a resource file by robot. Resource files can only have keywords, variables and settings.
If you expect Clean environment to be a keyword, put it in a keyword table:
*** Keywords ***
Clean environment
Log to console 1111
I compiled a C code using make then afterwards I issued this command on linux terminal
./regex --parse /ruleset/snort34.re --debug
where snort34.re is a file which I want to open for parsing but it gives me this error
Unable to open file in /ruleset/snort.re in modeFatal
Can anyone please help me why I am getting this error and what is modeFatal?
You have passed /ruleset/snort34.re to the program, which will look in a folder in your root directory called ruleset. You probably want to use a relative path instead, ruleset/snort34.re.