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Error: if condition

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# 1  
Old 10-25-2013
Error: if condition
please follow below code

Code:
if [ eval echo '$lck_'$TABLE != "" ]; then

its giving me an error as "test1.sh: line 3: [: too many arguments"

if i store eval echo '$lck_'$TABLE != "" result into sme variable n then use in if cond its working fine...say

Code:
r=$(eval echo '$lck_'$TABLE != "")
if [ r != "" ]

can anyone tell me how to do inside if condition...with out use any variable..

TIA
Last edited by vbe; 10-25-2013 at 06:28 AM.. Reason: code tags please not icode...
# 2  
Old 10-25-2013
Quote:
Originally Posted by gnnsprapa
please follow below code

if [ eval echo '$lck_'$TABLE != "" ]; then

its giving me an error as "test1.sh: line 3: [: too many arguments"

if i store eval echo '$lck_'$TABLE != "" result into sme variable n then use in if cond its working fine...say

r=$(eval echo '$lck_'$TABLE != "")
if [ r != "" ]

can anyone tell me how to do inside if condition...with out use any variable..

TIA


Try this

Code:
# lck is empty nothing to print
$ echo $lck  

# table is empty nothing to print
$ echo $table

# lck is as not NULL print Yes
$ lck=as
$ if [ $lck'_'$table != "_" ]; then echo "Yes";else echo "No";fi
Yes

# Make lck empty print No
$ unset lck
$ if [ $lck'_'$table != "_" ]; then echo "Yes";else echo "No";fi
No

Last edited by Akshay Hegde; 10-25-2013 at 05:32 AM.. Reason: Comment
# 3  
Old 10-25-2013
You can put the $(...) construct directly in the test, although i'm not sure what you're trying to achieve.
# 4  
Old 10-25-2013
Replace the if statement with
Code:
if [ "$(echo eval '$lck_'$TABLE)" != "" ]; then

# 5  
Old 10-25-2013
Quote:
Originally Posted by krishmaths
Replace the if statement with
Code:
if [ "$(echo eval '$lck_'$TABLE)" != "" ]; then

Your solution fails

Code:
$ lck_=as
$ if [ "$(echo eval '$lck_'$TABLE)" != "" ]; then echo "Yes";else echo "No";fi
Yes
$ unset lck_
$ echo $lck_

$ echo $TABLE

$ if [ "$(echo eval '$lck_'$TABLE)" != "" ]; then echo "Yes";else echo "No";fi
Yes

# 6  
Old 10-25-2013
@ Akshay, you are right. Good catch. But I'm not sure if it is a valid scenario to have a variable like lck_

I presume the variable TABLE would vary and will be attached to lck_ as in

Code:
lck_T1 
lck_T2
lck_T3 and so on

# 7  
Old 10-25-2013
Quote:
Originally Posted by krishmaths
@ Akshay, you are right. Good catch. But I'm not sure if it is a valid scenario to have a variable like lck_

I presume the variable TABLE would vary and will be attached to lck_ as in

Code:
lck_T1 
lck_T2
lck_T3 and so on

variable lck_ is valid, But I am suspecting user might have variable lck and TABLE

if not he can use this

Code:
$ unset lck_
$ unset TABLE
$ if [ -z $lck_$TABLE ];then echo "Empty";else echo "Something is there";fi
Empty
$ lck_=test
$ if [ -z $lck_$TABLE ];then echo "Empty";else echo "Something is there";fi
Something is there

In #4
if [ "$(echo eval '$lck_'$TABLE)" != "" ]; then

Resulting
Code:
$ TABLE=test2
$ echo $(echo eval '$lck_'$TABLE)
eval $lck_test2

eval '$lck_' is treated as string
Last edited by Akshay Hegde; 10-25-2013 at 07:42 AM.. Reason: Color
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