Question

我是一个Python noob，我正在尝试编写一个脚本来获取两个Intel hex文件（一个用于我的应用程序代码，一个用于引导程序），从第一个文件中删除EOF记录，附加第二个文件到第一个的剥离版本，并保存为新文件。我已经完成了所有工作，但后来又决定变得更加高兴：我想确保第一个文件的最后一行真正与英特尔EOF记录格式相匹配。但是我似乎无法正确地获取此条件的语法。

appFile = open("MyAppFile.hex", "r")
lines = appFile.readlines()
appFile.close()

appStrip = open("MyAppFile and BootFile.hex",'w')

if appStrip.readline[:] == ":00000001FF":  #Python complains about "builtin_function_or_method" not subscriptable here
    appStrip.writelines([item for item in lines[:-1]])
    appStrip.close()
else:
    print("No EOF record in last line. File may be corrupted.")

appFile = open("MyAppFile and BootFile", "r")
appObcode = appFile.read()
appFile.close()

bootFile = open("MyBootFile", "r")
bootObcode = bootFile.read()
bootFile.close()

comboData = appObcode + bootObcode

comboFile = open("MyAppFile and BootFile", "w")
comboFile.write(comboData)
comboFile.close()

对于更清洁或更安全版本的任何其他建议也是受欢迎的。

更新：添加了一行来打印最后一行;我得到了预期的输出，但每次比较仍然失败。这是当前的完整计划：

appFile = open("C:/LightLock/Master/Project/Debug/Exe/Light Lock.hex")

appLines = appFile.readlines()

appFile = open("MyAppFile.hex").read()

EOF = appLines[len(appLines)-1]

print(appLines[len(appLines)-1])

if not EOF == (":00000001FF"):
    print("No EOF record in last line of file. File may be corrupted.")
else:
    with open("MyAppFile Plus Boot", "a") as appStrip:
        appStrip.writelines([item for item in appLines[:-1]])

    with open("MyAppFile Plus Boot.hex", "r") as appFile:
        appObcode = appFile.read()

    with open("MyBootFile.hex", "r") as bootFile:
        bootObcode = bootFile.read()

    comboData = appObcode + bootObcode

    with open("MyAppFile Plus Boot.hex", "w") as comboFile:
        comboFile.write(comboData)

UPDATE2：尝试修改支票以包括回车和换行，如下所示：

EOF = appLines[len(appLines)-1]

print(EOF)

if EOF != (":00000001FF","\r","\n"):
    print("No EOF record in last line of file. File may be corrupted.")

仍然没有运气。

Answer 1

这是一个更简单的版本：

app = open("app.hex").read()
if not app.endswith(":00000001FF"):
   print("No EOF")
combo = open("combo.hex","w")
combo.write(app)
boot = open("boot.hex").read()
combo.write(boot)
combo.close() # it's automatic after program ended

Answer 2

最后想通了：我写了一些测试代码来输出Python正在读取的字符串的长度。事实证明它只有12个字符，但只显示了11个字符。所以我知道其中一个“看不见的”字符必须是回车或换行。试过两个;原来是换行（新线）。

这是最终的（有效但“未优化”）代码：

appFile = open("MyAppFile.hex")

appLines = appFile.readlines()

appFile = open("MyAppFile.hex").read()

EOF = appLines[len(appLines)-1]

if EOF != (":00000001FF\n"):
    print("No EOF record in last line of file. File may be corrupted.")
else:
    with open("MyAppFile and Boot.hex", "a") as appStrip:
        appStrip.writelines([item for item in appLines[:-1]])

    with open("MyAppFile and Boot.hex", "r") as appFile:
        appObcode = appFile.read()

    with open("MyBootFile.hex", "r") as bootFile:
        bootObcode = bootFile.read()

    comboData = appObcode + bootObcode

    with open("MyAppFile and Boot.hex", "w") as comboFile:
        comboFile.write(comboData)

将最后一行文件与Python中的字符串进行比较

2 个答案: