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7.8.11. Attribute algorithms 7.8.11.2. Buttons for the creation of equation
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7.8.11.2. Buttons for the creation of equation

"Attribute algorithm" dialog box

$VAR=
This button transfers the variable displayed under Name into the input field.
IF
An algorithm constructed through an IF condition can look as follows:
```
IF ( ) THEN 
   D3 = 
ELSE 
   D3 = 
ENDIF
```
Enter your condition (e.g. "L1.EQ.10") between the brackets ().
Behind THEN D3 =, enter the value which D3 has to accept if the condition is met.
Behind ELSE D3 =, enter the value which D3 has to accept if the condition is not met. End the condition with ENDIF.
Example with values:
```
IF (L1.EQ.10) THEN
D3 = 20
ELSE
D3 = 30
ENDIF
```
ELSEIF
If distinction of cases shall be made, use ELSEIF statements.
Example with values
```
IF (L1.EQ.10) THEN 
D3 = 10 
ELSEIF (L1.EQ.20) THEN 
D3 = 20 
ELSEIF (L1.EQ.30) THEN 
D3 = 30 
ELSE
D3 = 40
ENDIF 
```
Note
ELSEIF cases have to be ended by an ELSE statement.

Several IF conditions set one after another within one Attribute algorithm variable.

If you wanted to solve the same with one single IF condition with ELSEIF cases, this would require much more ELSEIFs.

Structure of conditions with the help of an example:

IF ( KG.EQ.'-')THEN
 NB1 = '$TNR. $TYP.-$KDM.-$HUB.-ZR-$WZ.'
ELSE
 NB1 = '$TNR. $TYP.-$KDM.-$HUB.-ZR-$WZ.-$KG.'
ENDIF
IF ( ZAK.EQ.'-')THEN
 NB1 = '$NB1.'
ELSE
 NB1 = '$NB1.-$ZAK.'
ENDIF
IF( YSR.EQ.'1'.AND.SIE.EQ.'-')THEN
 NB1 = '$NB1. ZUB-C'
ELSEIF( YSR.EQ.'1'.AND.SIE.EQ.'1')THEN
 NB1 = '$NB1. ZUB-CL'
ELSEIF( YSR.EQ.'2'.AND.SIE.EQ.'-')THEN
 NB1 = '$NB1. ZUB-$YSR.C'
ELSEIF( YSR.EQ.'2'.AND.SIE.EQ.'1')THEN
 NB1 = '$NB1. ZUB-$YSR.CL'
ELSEIF( YSR.EQ.'-'.AND.SIE.EQ.'1')THEN
 NB1 = '$NB1. ZUB-L'
ELSE
 NB1 = '$NB1.'
ENDIF

Im  ersten IF wird NB1 initialisiert.



                                    
Im zweiten IF wird an NB1 etwas drangehängt.



                                     
Im dritten IF wird nochmal etwas drangehängt.

Important notes:

The first IF condition has to contain an ELSE alternative, in order for the variable to be initialized. All other IF conditions may contain ELSE alternatives optionally.

Correct:

IF (OPZ1.EQ.'-' )THEN
 LINAALG = '$IDNR./$CX1./$VL./$MOTX./$OM1./$COEL./$DA./$AP1./$AA1.'
ELSE
 LINAALG = '$IDNR./$CX1./$VL./$MOTX./$OM1./$COEL./$DA./$AP1./$AA1./$OPZ1.'
ENDIF
IF (OPZ2.EQ.'-' )THEN
 LINAALG = '$LINAALG./'
ELSE
 LINAALG = '$LINAALG./$OPZ2.'
ENDIF...

Incorrect:

IF (OPZ1.NE.'-' )THEN
 LINAALG = '$IDNR./$CX1./$VL./$MOTX./$OM1./$COEL./$DA./$AP1./$AA1./$OPZ1.'
ENDIF
IF (OPZ2.NE.'-' )THEN
 LINAALG = '$LINAALG./$OPZ2.'
ENDIF...

Also correct: If there is a simple statement at the first position, an "ELSE" can be omitted at all, since the statement will always initialize the variable.

CNSTYPECODE = '$MODEL.$W.-$ST.-$THETA.-$TYPE.-$SPRING.'
IF(K.EQ.1)THEN
 CNSTYPECODE = '$CNSTYPECODE.-K'
ENDIF
IF(FK.EQ.1)THEN
 CNSTYPECODE = '$CNSTYPECODE.-FK'
ENDIF
IF(N.EQ.1)THEN
 CNSTYPECODE = '$CNSTYPECODE.-N'
ENDIF

The first condition, which is always calculated, has to prevent an endless loop. Following "if...endif" always call the first condition and attach an additional value.

So in the first condition do not use the variable of assignment for the ELSE case.

Correct:

IF (OPZ1.EQ.'-' )THEN
 LINAALG = '$IDNR./$CX1./$VL./$MOTX./$OM1./$COEL./$DA./$AP1./$AA1.'
ELSE
 LINAALG = '$IDNR./$CX1./$VL./$MOTX./$OM1./$COEL./$DA./$AP1./$AA1./$OPZ1.'
ENDIF
IF (OPZ2.EQ.'-' )THEN
 LINAALG = '$LINAALG./'
ELSE
 LINAALG = '$LINAALG./$OPZ2.'
ENDIF...

Incorrect: (difference in red)

IF (OPZ1.EQ.'-' )THEN
 LINAALG = '$IDNR./$CX1./$VL./$MOTX./$OM1./$COEL./$DA./$AP1./$AA1.'
ELSE
 LINAALG = '$LINAALG./$OPZ1.'
ENDIF
IF (OPZ2.EQ.'-' )THEN
 LINAALG = '$LINAALG./'
ELSE
 LINAALG = '$LINAALG./$OPZ2.'
ENDIF...

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