尝试使用OSX使用schemacrawler连接到sqlite db - 为什么要求用户?

Ste*_*erg 5 sqlite macos jdbc schemacrawler

见底部得出结论:

我正在尝试使用schemacrawler来绘制sqlite数据库.我的设置:

  • OSX 10.8
  • SchemaCrawler 10.5从Here下载
  • 从Oracle下载的Java版本1.7.0_45
  • sqlite版本:3.7.12 2012-04-03 19:43:07 86b8481be7e76cccc92d14ce762d21bfb69504af

我在安装schemacrawler的目录中,并使用以下命令行:

stebro$ java -classpath lib/*:. schemacrawler.tools.sqlite.Main /Library/Application\ Support/MyApp/Data/MyApp.db -command="select count(*) from myTable" -infolevel=maximum
SchemaCrawler 10.5
Copyright (c) 2000-2013, Sualeh Fatehi.

SchemaCrawler is a database schema discovery and comprehension tool. 
You can search for database schema objects using regular expressions, 
and output the schema and data in a readable text format. You can find 
potential schema design issues with lint. The output serves for 
database documentation is designed to be diff-ed against other database 
schemas. SchemaCrawler also generates schema diagrams.
password: 
java.sql.SQLException: Could not connect to database, for user null
    at schemacrawler.schemacrawler.BaseDatabaseConnectionOptions.getConnection(BaseDatabaseConnectionOptions.java:93)
    at schemacrawler.schemacrawler.BaseDatabaseConnectionOptions.getConnection(BaseDatabaseConnectionOptions.java:70)
    at schemacrawler.tools.commandline.SchemaCrawlerCommandLine.execute(SchemaCrawlerCommandLine.java:173)
    at schemacrawler.tools.commandline.SchemaCrawlerMain.main(SchemaCrawlerMain.java:93)
    at schemacrawler.tools.commandline.SchemaCrawlerMain.main(SchemaCrawlerMain.java:52)
    at schemacrawler.tools.sqlite.Main.main(Main.java:43)
Caused by: java.lang.IllegalArgumentException: Insufficient parameters for database connection URL: missing [database]
    at schemacrawler.schemacrawler.DatabaseConfigConnectionOptions.getConnectionUrl(DatabaseConfigConnectionOptions.java:73)
    at schemacrawler.schemacrawler.BaseDatabaseConnectionOptions.getConnection(BaseDatabaseConnectionOptions.java:89)
    ... 5 more
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如果我指定-password,我会得到相同的错误,并为-user指定各种值也是一样的.Sqlite不需要用户/密码 - 为什么jdbc或schemacrawler会问我一个?

附录:

这是一系列特定的命令,它们创建一个简单的数据库,然后尝试使用schemacrawler绘制它:

bash-3.2$ sqlite3 MyApp.db
sqlite3 MyApp.db
SQLite version 3.7.12 2012-04-03 19:43:07
Enter ".help" for instructions
Enter SQL statements terminated with a ";"
sqlite> create table smbtest(col1 integer, col2 integer);
create table smbtest(col1 integer, col2 integer);
sqlite> ^D 
bash-3.2$ ~/bin/sunjava -classpath lib/*:. schemacrawler.tools.sqlite.Main -database=MyApp.db -infolevel=maximum -command=graph -outputformat=pdf -outputfile=myapp.pdf
SchemaCrawler 10.5
Copyright (c) 2000-2013, Sualeh Fatehi.

SchemaCrawler is a database schema discovery and comprehension tool. 
You can search for database schema objects using regular expressions, 
and output the schema and data in a readable text format. You can find 
potential schema design issues with lint. The output serves for 
database documentation is designed to be diff-ed against other database 
schemas. SchemaCrawler also generates schema diagrams.
password: 

Graphviz was not available to create the requested graph. Please reinstall 
Graphviz, and make it available on the system PATH. Meanwhile, a .dot text file 
has been created instead. This .dot file can be opened in any Graphviz file 
viewer.
java.io.IOException: Cannot run program "dot": error=2, No such file or directory
at java.lang.ProcessBuilder.start(ProcessBuilder.java:1041)
at schemacrawler.tools.integration.graph.ProcessExecutor.execute(ProcessExecutor.java:124)
at schemacrawler.tools.integration.graph.GraphGenerator.generateDiagram(GraphGenerator.java:87)
at schemacrawler.tools.integration.graph.GraphExecutable.executeOn(GraphExecutable.java:123)
at schemacrawler.tools.executable.SchemaCrawlerExecutable.executeOn(SchemaCrawlerExecutable.java:87)
at schemacrawler.tools.executable.BaseExecutable.execute(BaseExecutable.java:77)
at schemacrawler.tools.commandline.SchemaCrawlerCommandLine.execute(SchemaCrawlerCommandLine.java:176)
at schemacrawler.tools.commandline.SchemaCrawlerMain.main(SchemaCrawlerMain.java:93)
at schemacrawler.tools.commandline.SchemaCrawlerMain.main(SchemaCrawlerMain.java:52)
at schemacrawler.tools.sqlite.Main.main(Main.java:43)
Caused by: java.io.IOException: error=2, No such file or directory
at java.lang.UNIXProcess.forkAndExec(Native Method)
at java.lang.UNIXProcess.<init>(UNIXProcess.java:135)
at java.lang.ProcessImpl.start(ProcessImpl.java:130)
at java.lang.ProcessBuilder.start(ProcessBuilder.java:1022)
... 9 more
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最后的结论:问题在于我的数据库的路径中有一个空格; 没有反斜杠,双引号或单引号的组合,或两者都解决了问题.从与我的db文件相同的目录运行命令就可以了.空间问题可能是由于我不理解在bash中使用的正确转义序列 - 如果schemacrawler使用我的名字作为命令行的一部分产生其他进程,我可能必须双重转义空格.

但是,应用程序确实继续询问我的密码,但只需按Enter即可继续成功.

Sua*_*ehi 1

史蒂夫,

您需要像这样指定数据库:

"-database=/Library/Application Support/MyApp/Data/MyApp.db"
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请注意双引号以允许数据库文件路径中的空格以及 -database 命令行开关。我还删除了反斜杠,因为整个开关都用双引号引起来。

如需命令行帮助,只需运行:

java -classpath lib/*:. schemacrawler.tools.sqlite.Main
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Sualeh Fatehi,SchemaCrawler