Substituting a String in Place of Variable in Shell

Replace a string in shell script using a variable

If you want to interpret $replace, you should not use single quotes since they prevent variable substitution.

Try:

echo $LINE | sed -e "s/12345678/${replace}/g"

Transcript:

pax> export replace=987654321
pax> echo X123456789X | sed "s/123456789/${replace}/"
X987654321X
pax> _

Just be careful to ensure that ${replace} doesn't have any characters of significance to sed (like / for instance) since it will cause confusion unless escaped. But if, as you say, you're replacing one number with another, that shouldn't be a problem.

Replace one substring for another string in shell script

To replace the first occurrence of a pattern with a given string, use ${parameter/pattern/string}:

#!/bin/bash
firstString="I love Suzi and Marry"
secondString="Sara"
echo "${firstString/Suzi/"$secondString"}"    
# prints 'I love Sara and Marry'

To replace all occurrences, use ${parameter//pattern/string}:

message='The secret code is 12345'
echo "${message//[0-9]/X}"           
# prints 'The secret code is XXXXX'

(This is documented in the Bash Reference Manual, §3.5.3 "Shell Parameter Expansion".)

Note that this feature is not specified by POSIX — it's a Bash extension — so not all Unix shells implement it. For the relevant POSIX documentation, see The Open Group Technical Standard Base Specifications, Issue 7, the Shell & Utilities volume, §2.6.2 "Parameter Expansion".

substituting a string in place of variable in shell

The syntax for this is:

${!VAR}

Example:

$ function hello() { echo ${!1}; }
$ hello HOME
/home/me

How to replace a string in a variable with another string in a file using shell script

The default delimiter of sed clashes with the same character's present in your WORKPLACE. As mentioned in the comments, changing the delimiter to a character not present in any of your input or escaping the slashes would work for that issue.

However, you also have a single quoting issue. Not only would you get an error, the variables will not expand.

This sed should work

$ sed -i.bak s"|$WORKSPACE|/home/Projects/XXX/$PASE_PRODUCT/common/XXX-$SUB_PRODUCT|" polyspaceFiles_"$PASE_SUB_PRODUCT"_tmp.opts

how to replace a variable in shell script string

You are missing the end of that single-quote pair in your script.

Change from:

echo $SQL | sed -e "s/'$BATCH_END/$BATCH_END/g"

To:

echo $SQL | sed -e "s/\$BATCH_END/$BATCH_END/g"

Updated - as per followup comment:

To save the result of the above replacement back into $SQL, do either of the following:

# Preferred way
SQL=$(echo $SQL | sed -e "s/\$BATCH_END/$BATCH_END/g")

# Old way
SQL=`echo $SQL | sed -e "s/\$BATCH_END/$BATCH_END/g"`

This is called command substitution. Either syntax ($(...) vs. enclosure by backticks) works, but the preferred one allows you to do nesting.

The preferred-preferred way: Herestring

This is probably a bit more advanced than what you care about, but doing it in the following way will save you a subprocess from having to use echo unnecessarily:

SQL=$(sed -e "s/\$BATCH_END/$BATCH_END/g" <<< $SQL)

Replace one character with another in Bash

Use inline shell string replacement. Example:

foo="  "

# replace first blank only
bar=${foo/ /.}

# replace all blanks
bar=${foo// /.}

See http://tldp.org/LDP/abs/html/string-manipulation.html for more details.

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