Script to Read lines and print

UNIX AIX
1

Dears,

I am trying to write a script to read the lines from a file in AIX Server.

Below are the contents of the file. To explain, first part before = is path and second part is number.

i am trying to write the script to print the path and count where count>100. any help must be appreciated

Output for my below data is

/Slave/13578/temp/1 = 100
 ./Slave/13578/temp/1=300 
 ./2390/15248/Test/Test1/20160331220000= 131 
 ./2390/15248/Test/Test1/20160331220000= 131

FILE:

./KING/12546/Test= 3 

 ./Queen/11980= 6 

 ./KING/12844/Test= 0 

 ./Slave/13578/temp/1=300 

 ./Queen/11982= 0 

 ./Queen/11984= 0 

 ./KING/12567/log= 0 

 ./2390/15248/Test/Test1/20160331220000= 131 

 ./Slave/4455/log/1= 0 

 ./Queen/11989= 0
2

Please use code tags as required by forum rules!

Any attempts from your side? Any preferred tool to deploy?

And, is it safe to assume "count" is the field you call "number" in your explanation? Where does the first line in your desired output come from? It is not in your input file, and, if it were, it wouldn't satisfy the condition. And, why the duplicate output of the last line?

3 
awk -F " *= *" '$2>100 {sub(" +","",$1); print "path="$1, "count="$2}' file
4

The output requested could almost be achieved just using:

awk -F '= *' '$2>=100' FILE

To get the output specified in the textual requirements, that would be:

awk -F '= *' '$2>100' FILE

but that would just produce the output:

 ./Slave/13578/temp/1=300 
 ./2390/15248/Test/Test1/20160331220000= 131

since 100 is not greater than 100.

Also note that in either case the last line shown above is only output once (not twice as requested) because that line only appears in the input once.