Starting the database server with -c %%

1) If I run dbsrv12.exe -c 30%% then

a) For 12.0.1.3537 - OK

    2: -c

    3: 30%%

    ...

    SQL Anywhere Network Server Version 12.0.1.3537

    ...

    Running Windows 7 Build 7601 Service Pack 1 on X86_64

    Server built for X86_64 processor architecture

    2511152K of memory used for caching

    Minimum cache size: 2511152K, maximum cache size: 7529760K

b) For 12.0.1.4178: Invalid database server command line error

2) If I run dbsrv12.exe -c 30% then
a) For 12.0.1.3537 - OK
b) For 12.0.1.4178 - OK

    2: -c

    3: 30%

    ...

    SQL Anywhere Network Server Version 12.0.1.4178

    ...

    Running Windows 7 Build 7601 Service Pack 1 on X86_64

    Server built for X86_64 processor architecture

    2511152K of memory used for caching

    Minimum cache size: 2511152K, maximum cache size: 7529760K

Is it bug? For 12.0.1.3537 or for 12.0.1.4178 ?

sa-12 command-line

asked 30 Mar '15, 07:56

Ilia63
1.2k●51●55●78
accept rate: 44%

edited 30 Mar '15, 11:18

Mark Culp
24.9k●10●139●297

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One Answer:

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You want one "%" sign to make it to the server. If the server is invoked from a shell (cmd.exe on Windows), "%%" gets converted to "%" by the shell because "%" is used for environment variable substitutions. You appear to be invoking the server without using a shell so you only want -c 30%.

To avoid issues with shells and %, you could just use -c 30p.

-john.

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answered 30 Mar '15, 08:05

John Smirnios
11.7k●3●94●161
accept rate: 37%

Fortunately, "p" in this case doesn't stand for "petabytes." That would be a big cache.

(02 Apr '15, 14:41) Phil Mitchell

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question asked: 30 Mar '15, 07:56

question was seen: 1,556 times

last updated: 02 Apr '15, 14:41

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Starting the database server with -c %%

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