Bash 脚本 SQLITE - 选项太多
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Bash Script SQLITE - Too Many Options
提问by crankshaft
Can't figure this one out, if I execute the command directly in the shell, then it executes successfully, but running from a script file I get the following error:
想不通，如果我直接在shell中执行命令，那么它执行成功，但是从脚本文件运行我得到以下错误：
sqlite3: Error: too many options: "INTO"
sqlite3：错误：选项太多：“INTO”
#!/bin/bash
sql="\"INSERT INTO keys ('date','chan','key','name','desc','ser','ep','cat') VALUES('xxx','xxx','xxxx','xxxx','xxxxx.','xxx','xxx','xxxx');\""
echo $sql
sqlite3 mydb.db $sql
Any ideas on what the problem might be ?
关于问题可能是什么的任何想法？
采纳答案by higuaro
Double quote the $sqlvariable in the script:
双引号$sql脚本中的变量：
#!/bin/bash
sql="\"INSERT INTO keys ('date','chan','key','name','desc','ser','ep','cat') VALUES('xxx','xxx','xxxx','xxxx','xxxxx.','xxx','xxx','xxxx');\""
echo $sql
sqlite3 mydb.db "$sql"
回答by Drey
In my case the problem was in the binary I execute:
就我而言，问题出在我执行的二进制文件中：
cat /opt/sbin/sqlite3:
#!/bin/sh
export LD_LIBRARY_PATH="/opt/lib:$LD_LIBRARY_PATH"
/opt/sbin/sqlite3.real $@
So if I execute /opt/sbin/sqlite3.real I have no problems.
所以如果我执行 /opt/sbin/sqlite3.real 我没有问题。
