我为sqlite3操作编写了两个脚本。在我的第一个脚本中,我创建了数据库和表名。在我的第二个脚本中,我创建了表记录。在执行第二个脚本时,我想获取已经创建的数据库和选择表名。请任何人帮助我..!
script1.sh
#!/bin/bash
echo " --- Enter the Database name ---" #name of the database
read databasename
echo " --- enter the table name --- " #name of the table
read table_name
sqlite3 $databasename.db "DROP TABLE IF EXISTS $table_name;"
sqlite3 $databasename.db "CREATE TABLE IF NOT EXISTS $table_name(cus_id INTEGER PRIMARY KEY AUTOINCREMENT UNIQUE,cus_name TEXT NOT NULL UNIQUE ,cus_domain TEXT UNIQUE, cus_status TEXT NOT NULL,Port INTEGER NOT NULL);" #table creation
Script2.sh
echo "choose the database"
#Here i want to choose the already create database.
echo "choose the table"
#Here i want to choose the table in the alreadt created database.
echo " --- Enter the total number of customer records do you want ---"
read cus_count # number of rows value
echo "--- Enter the following details one by one---"
RED='\033[0;31m'
NC='\033[0m'
port_num=8080
declare -a customer
for((i=1;i<=cus_count;i++))
do
echo "enter the $i customer details"
echo "---Enter the customer name---"
read c_name
d_name=${c_name,,}
#echo $d_name
customer=$(sqlite3 $databasename.db "select cus_domain from $table_name where cus_domain like '$d_name';")
for cus in "${customer[@]}"
do
if [[ $d_name != $customer ]];
then
echo "---Enter the Status(Active/Inactive)---"
read c_status
if [[ "$port_num" == "$port_num" ]]; then
port_num=$(($port_num + 1))
c_domain="$c_name"
sqlite3 $databasename.db "INSERT OR IGNORE INTO $table_name (cus_name,cus_domain,cus_status, Port) VALUES(\"$c_name\",\"${c_domain,,}\",\"${c_status,,}\",\"$port_num\") ;"
fi
else
echo -e "${RED}!!!OOPS for you entered customer name already domain name assigned!!!${NC}"
echo -e "${RED}Please enter new customer name${NC}"
i=$(($i - 1))
fi
done
done
echo " --- Records from the $table_name ---"
sqlite3 $databasename.db "select * from $table_name;"
请任何人帮助我......谢谢...
答案 0 :(得分:0)
您可以将数据库和表名写入文件。
echo $databasname >> /tmp/new_database
echo $table >> /tmp/new_database
并配置第二个脚本以读取此文件
或者您可以从脚本1调用脚本2,将数据库名称和表名称作为args
script1
#!/bin/bash
...
/bin/bash script2.sh $databasename $tablename
script2.sh
#!/bin/bash
databasename = $1
tablename = $2