Fix class name : change "string" to "String"
Change case of "s" letter : "string" does not compile but "String" does (for gdScript)
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@ -43,7 +43,7 @@
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You can access a dictionary's values by referencing the appropriate key. In the above example, [code]points_dict["White"][/code] will return [code]50[/code]. You can also write [code]points_dict.White[/code], which is equivalent. However, you'll have to use the bracket syntax if the key you're accessing the dictionary with isn't a fixed string (such as a number or variable).
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You can access a dictionary's values by referencing the appropriate key. In the above example, [code]points_dict["White"][/code] will return [code]50[/code]. You can also write [code]points_dict.White[/code], which is equivalent. However, you'll have to use the bracket syntax if the key you're accessing the dictionary with isn't a fixed string (such as a number or variable).
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[codeblocks]
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[codeblocks]
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[gdscript]
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[gdscript]
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export(string, "White", "Yellow", "Orange") var my_color
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export(String, "White", "Yellow", "Orange") var my_color
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var points_dict = {"White": 50, "Yellow": 75, "Orange": 100}
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var points_dict = {"White": 50, "Yellow": 75, "Orange": 100}
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func _ready():
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func _ready():
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# We can't use dot syntax here as `my_color` is a variable.
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# We can't use dot syntax here as `my_color` is a variable.
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