Simulink.data.dictionary.open

R2026b

Open data dictionary for editing

Syntax

dictionaryObj = Simulink.data.dictionary.open(dictionaryFile)

Description

dictionaryObj = Simulink.data.dictionary.open(dictionaryFile) opens the specified data dictionary and returns a Simulink.data.Dictionary object representing an existing data dictionary identified by its file name and, optionally, file path with dictionaryFile.

Make sure any dictionaries referenced by the target dictionary are on the MATLAB^® path. When the function is unable to open a referenced dictionary, the function displays a warning, but still opens the target dictionary.

Examples

Open Existing Data Dictionary

Open the data dictionary myDictionary_ex_API.sldd and represent it with a Simulink.data.Dictionary object named myDictionaryObj.

myDictionaryObj = Simulink.data.dictionary.open('myDictionary_ex_API.sldd')

myDictionaryObj = 

  Dictionary with properties:

          DataSources: {'myRefDictionary_ex_API.sldd'}
    HasUnsavedChanges: 0
           NumberOfEntries: 4

Input Arguments

dictionaryFile — Target data dictionary

character vector

Target data dictionary, specified as a character vector containing the file name and, optionally, path of the dictionary. If you do not specify a path, Simulink.data.dictionary.open searches the MATLAB path for the specified file. Simulink.data.dictionary.open also supports paths specified relative to the MATLAB working folder.

Example: 'myDictionary_ex_API.sldd'

Example: 'C:\Users\jsmith\myDictionary_ex_API.sldd'

Example: '..\myOtherDictionary.sldd'

Data Types: char

Version History

Introduced in R2015a

R2026b: Warning reported by default when Simulink.data.dictionary.open is unable to open a referenced dictionary

Starting in R2026b, when the Simulink.data.dictionary.open function is unable to open a dictionary referenced by the target dictionary, the function reports a warning, but still opens the target dictionary. The name-value argument SubdictionaryErrorAction has been removed. Previously, the function reported an error unless you called the function with the SubdictionaryErrorAction argument set to "warn".