Parsing an array variable to an array does not work

[dumbenis]

I have array m filled with many rows in 2 columns. I want to compare a string with the content of each row in the first column. That works without any problem. I build a DO-loop to increment the row number and inside the loop I compare the string with the content of the first column in each row:
(%r = row number)
(%m = array m)

DO r = 0 TO %@DEC[%@arrayinfo[m,1]] BY 1 IFF %@WILD["%string%",*%m[%r%,0]*] == 1 THEN ECHO Found ELSE Echo Not found ENDIFF ENDDO

This DO-loop works perfect. Now, I have 2 arrays: array m and t and I want to compare a string with the first column in each row of each array. I build a second DO-loop around the first DO-loop:
(%a = array)
(%r = row number)

DO a IN /L t m DO r = 0 TO %@DEC[%@arrayinfo[%a,1]] BY 1 IFF %@WILD["%string%",*%%a[%r%,0]*] == 1 THEN ECHO Found ELSE Echo Not found ENDIFF ENDDO ENDDO

Variable a become “t” but the parsing does not work. I experimented with more % signs, but I can’t get it to work.

 

[vefatica]

In a nutshell ...

v:\> setarray a[2] & set a[0]=TRUE & set a[1]=FALSE

v:\> set arrayname=a

v:\> echo %[arrayname[0]]
TRUE

v:\> echo %[arrayname[1]]
FALSE

 

[vefatica]

Also (with the set-up above),

v:\> do index=0 to 1 ( echo %[arrayname[%index]] )
TRUE
FALSE

 

[dumbenis]

Thanks Vefatica for your input.
I couldn't get the code working:
IFF %@WILD["%string%",*%[a[%r%,0]]*] == 1 THEN
I'll build a test programm to understand the difficulty.

 

[vefatica]

I don't know. This seems to work.

v:\> setarray a[4,2]

v:\> set a[0,0]=? & set a[1,0]=?? & set a[2,0]=??? & set a[3,0]=????

v:\> set string=xxxx

v:\> do name in /L a ( do row=0 to 3 ( if %@wild[%string,%[name[%row,0]]] == 1 echo YES (row %row) ))
YES (row 3)

 

[dumbenis]

Thanks Vetatica,

I have build a test.btm with:
IFF %@WILD["%string%",*%[a[%r%,0]]*] == 1 THEN
It ran without any problem.
I applied the same fix to my batchfile again and it worked this time without any problem.
So, the bug disappeared. Must be programmerdfailure.
Thanks.