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linux shell substitute variable

Posted on 2013-05-14
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Last Modified: 2013-05-14
I have list of tables in t.txt file, I need to substitute all the ${table} instance with the table name in t.txt file. the challenge here is after the single quote, the ${table} variable will not be substituted, like the following code, only the first ${table} will be substituted, the last two does not. How can I do this? Thanks.

for table in `cat t.txt`
do

curl -XPUT "http://localhost:9200/_river/${table}_river/_meta" -d '{
    "type" : "jdbc",
    "jdbc" : {
        "driver" : "oracle.jdbc.driver.OracleDriver",
        "url" : "jdbc:oracle:oci:@test",
        "user" : "test",
        "password" : "test",
        "sql" : "select * from ${table} where rownum<3"
    },
    "index" : {
        "index" : "database",
        "type" : "${table}"
    }
}' 

done

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Question by:xiaoyunwu
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ozo earned 500 total points
ID: 39166512
curl -XPUT "http://localhost:9200/_river/${table}_river/_meta" -d '{
    "type" : "jdbc",
    "jdbc" : {
        "driver" : "oracle.jdbc.driver.OracleDriver",
        "url" : "jdbc:oracle:oci:@test",
        "user" : "test",
        "password" : "test",
        "sql" : "select * from '${table}' where rownum<3"
    },
    "index" : {
        "index" : "database",
        "type" : "'${table}'"
    }
}'
0
 

Author Closing Comment

by:xiaoyunwu
ID: 39166537
Thank you, ozo!
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