的Python：如何切串起来，用它的一部分，并丢弃他人Python3

Question

，我想下面的字符串转换：

bpy.types.object.location.* std:label -1 editors/3dview/object/properties/transforms.html#bpy-types-object-location Transform Properties

对此以下字符串：

("bpy.types.Object.location.*", "editors/3dview/object/properties/transforms.html"),

目前，我只有：

with open("objects.tmp") as f: 
    for line in f: 
     if "bpy.types." in line: 
      fw (' ("' + line.rstrip() +'")\n') 

如何在特定点处切断字符串*或/和X空格后？

Answer 1

originalString = 'bpy.types.object.location.* std:label -1 editors/3dview/object/properties/transforms.html#bpy-types-object-location Transform Properties' 

# this separates your string by spaces 
split = originalString.split(' ') 

# first element is ready to go! 
first_element = split[0] # 'bpy.types.object.location.*' 

# second element needs to be cut at "#" 
second_element = split[3] 
second_element = second_element[:second_element.find('#')] # editors/3dview/object/properties/transforms.html 

# now you have your desired output 
output = tuple(first_element, second_element)