OR function in indicator don’t work…

Robin81 · 2022-12-16T20:45:37+00:00

I made a script for finding equal lows on the DAX. I look for differences of 0, 0.5 and 1 points, because the price sometimes deviates slightly. Now I have made the script below, so that a line is placed on the graph. The problem is that the script can’t handle searching for 0, 0.5 and 1 simultaneously. I’ve used the function with OR and it doesn’t seem to work. I can of course put the formula in the script again to create three conditions, but it seems to me that there is a smarter way to do it. If I enter 1 value in Af1 it works perfectly, but with multiple values using the OR function it doesn’t work. Who knows how I can cleverly include this in the script with Af1= 0 OR 0.5 OR 1. What am I overlooking? 🙂 //DAX same lows //26-11-22 Af1 = 0 OR 0.5 OF 1 //Same lows for p = 4 to 100 do if ABS(low[1] – low[p]) = Af1 THEN x1 = barindex[1] y1 = low[1] x2 = barindex[p] y2 = low[p] endif next DRAWSEGMENT (x1,y1,x2,y2)coloured(255,10,10) RETURN

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12/16/2022 at 9:45 PM #205926 quote

Robin81
Participant

Junior

I made a script for finding equal lows on the DAX. I look for differences of 0, 0.5 and 1 points, because the price sometimes deviates slightly.

Now I have made the script below, so that a line is placed on the graph. The problem is that the script can’t handle searching for 0, 0.5 and 1 simultaneously. I’ve used the function with OR and it doesn’t seem to work. I can of course put the formula in the script again to create three conditions, but it seems to me that there is a smarter way to do it.

If I enter 1 value in Af1 it works perfectly, but with multiple values using the OR function it doesn’t work.

Who knows how I can cleverly include this in the script with Af1= 0 OR 0.5 OR 1. What am I overlooking? 🙂

//DAX same lows
//26-11-22

Af1 = 0 OR 0.5 OF 1

//Same lows
for p = 4 to 100 do
if ABS(low[1] – low[p]) = Af1 THEN
x1 = barindex[1]
y1 = low[1]
x2 = barindex[p]
y2 = low[p]
endif
next
DRAWSEGMENT (x1,y1,x2,y2)coloured(255,10,10)
RETURN

12/16/2022 at 9:58 PM #205927 quote

JC_Bywan
Moderator

Master

Hi,

“or” is for booleans, not for choosing among various numbers.

You could say you want anything between 0 and 1, with 1 included, by keeping a simple Af1=1 , and by using <= (it means inferior or equal to) this way:

if ABS(low[1] – low[p]) <= Af1 THEN

12/16/2022 at 10:24 PM #205929 quote
JS
Participant

Master
Hi @Robin81

You can try this one…
```
//DAX same lows
//26-11-22

Af1 = -0.5

//Same lows
While Af1<>1 do
Af1 = Af1 + 0.5
for p = 4 to 100 do
if ABS(low[1] - low[p]) = Af1 THEN
x1 = barindex[1]
y1 = low[1]
x2 = barindex[p]
y2 = low[p]
DRAWSEGMENT (x1,y1,x2,y2)coloured(255,10,10)
endif
next
WEnd
RETURN
```
Schermafbeelding-2022-12-16-om-22.21.48.png
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