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How to pass an argument to a script?
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on “sed” command EmpData(Sample Database) 1122|j.b. saxena |g.m. |account |12/12/52|6000 2233|n.k. gupta |d.g.m |sales |31/12/40|9000 4545|anil agarwal |director |account |06/07/47|7500 5656|lalit choudhury |executive|marketing|07/09/50|5000 1265|chanchal singhvi|g.m. |admin |12/09/63|6000 0110|shyam saksena |chairman |marketing|12/12/43|8000 5566|jai sharma |director |account |23/12/89|7000 7733|jayant |d.g.m |sales |29/02/70|6000 1. From the above database substitute the delimiter of first 3 lines with “ : “ 2. From the above database substitute the delimiter with “ : ” 3. Insert the string “ XYZ Employees” in the first line. 4. Store the lines pertaining to the directors, d.g.m and g.m into three separate files. 5. Using address store first 4 lines into a file Empupdate. 6. Find the pattern “account” in the database and replaces that with “accounts”. 7. Select those lines which do not have a pattern “g.m”. 8. Insert a blank line after every line in the database.
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Hi, all Scripting professional. Q. I have written script1.sh and calling script2.sh in script1.sh file using bash shell as interpreter and my log in shell also bash shell.My code like Script1 #!/bin/bash echo "My script2 call" . script2.sh Here script2.sh file run successfully but when I have changed my interpreter bash to ksh like #!/bin/ksh Error are comming script2.sh command not found. Guid me how to call other script in our main script.
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