Affected shell variable in awk

Hi
I try to assign a variable in awk previously defined in my script, but I can not find the correct syntax:

Code:

---

/ bin / bash
var = 2

awk-F '' '
    BEGIN (
        tata = $ 1
        $ var = tata
    )
    END '

# # Here, the script continues, and var is inside tata.

---

Could you help me with line number 7?

Hi
I think your calls are not clear:

Code:

---

varBash = 2
awk-F ''-v varAwk = $ varBash '(uneVar = $ 1; varBash = uneVar; print uneVar, varBash, varAwk)' <<< "this is an example"

---

Think you should correct this and I suggest you to keep your code clean so that it can to readable to you in future.

Hi
Note that the $ are used for variables "columns" ($ 0 is the complete line, $ 1, $ 2 ... being the columns) so when you do $ var = tata, it's like if you replaced it by var = $ 0.
You have to do is simply

Code:

---

var = tata to affect the contents of the variable in the variable tata
and do var = "tata" to assign the string variable in tata var

---

Hope this helps you. If nay more queries then do post back.

Hello,
I said my application in simplifying the example:

Code:

---

/ bin / bash
var = 2
awk-F '' '
    BEGIN (
        var = 3
    )
    END '

# # Here, the script continues, and var is 3.

---

Is this clearer now? Any advice now?

Hello
You can also do it like say, block the last statement END of your procedure AWK is add the following line:

Code:

---

END (
....
....
....
printf ( "% s", var);
)

---

And in your script you call your awk

Code:

---

VAR = `awk-f script.awk ....`
echo "VAR = $ VAR"

---

This will help for sure!

Hi
Thank you for your solution. I confess that I use the same exit (which limits the return has a numeric value less than 256), but I was looking if there was another way of doing things. I do not think this is possible, by the fact that the awk process can access the memory space of the script. Thanks again.