How to Call a Function inside of a function

조회 수: 8 (최근 30일)
Bridget
Bridget 2023년 4월 30일
답변: Walter Roberson 2023년 4월 30일
Hi,
I am trying to design a app where the user will input a flowrate and length so the app can calculate the experimental and predicted pressure drop in any of 5 pipes.
I am using a 'switch' function to go through the user's selection of pipe# and however for pipe 4, the values are not calculated they are extracted from an array. I tried to use switch to go through the user's flowrate selection but do i have to input a function prior to this? If this is the case how do I input a function in a function? See below:
% Button pushed function: CalculateButton
function CalculateButtonPushed(app, event)
switch app.PipeDropDown.Value
case '1'
app.PredicteddPEditField.Visible = 'on';
app.PredicteddPEditFieldLabel.Visible = 'on';
app.ExperimentaldPEditField.Visible = 'on';
app.ExperimentaldPEditFieldLabel.Visible = 'on';
'more code here'
case '4'
Exp_dP = {44.58, 91.83, 871.01, 1298.94, 1339.94, 1251.39, 1198.19; 57.21, 124.22, 1119.59, 1616.46, 1534.74, 1551.09, 1493.88; 74.53, 183.87, 1716.98, 2545.96, 2337.25, 2329.90, 2371.43; 81.96, 224.15, 2273.51, 3358.78, 2846.03, 2869.85, 2986.83; 118.77, 312.59, 2901.88, 4273.92, 3682.84, 3744.36, 3913.74; 132.39, 354.67, 3755.95, 5594.69, 4137.18, 4254.86, 4445.09; 194.91, 476.17, 4679.95, 6956.18, 5187.72, 5224.49, 5586.11; 236.99, 580.23, 5769.58, 8551.40, 6222.32, 6397.20, 6818.89};
switch app.FlowrateDropDown.value
case '0.4'
Q = app.FlowrateDropDown.Value*(1/3600);
d = 0.026;
p = 1000;
u = 0.000001;
k = 0.03/1000; 'm';
v = (Q/((3.14)*((d/2)^2)))*(1/3600); 'm/s';
Re = (d*v*p)/u;
F = (3.6*log(6.9/Re)+((k/(3.7*d))^1.11))^-2;
L = app.LegnthDropDown.Value;
if L == 0.7
dP1 = 2*F*(L/d)*(v^2)*p;
app.PredicteddPEditField.Value = dP1;
dP2 = Exp_dP(1,1);
app.ExperimentaldPEditField.Value = dP2;
elseif L == 1.05
dP1 = 2*F*(L/d)*(v^2)*p;
app.PredicteddPEditField.Value = dP1;
dP2 = Exp_dP(1,2);
app.ExperimentaldPEditField.Value = dP2;
I am getting an error "Unrecognized method, property, or field 'value' for class 'matlab.ui.control.DropDown'.". PLEASE HELP

답변 (1개)

Walter Roberson
Walter Roberson 2023년 4월 30일
switch app.FlowrateDropDown.value
Notice you used lower-case V in value . The property names are case-sensitive, so you need to use Value with a capital V

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