RAD Studio 10.3.3 无法创建 Java 虚拟机

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英文:

RAD studio 10.3.3 could not create java virtual machine

问题

我已经检查了这个问题,但我没有找到解决方案。

我在Win 10/64位上使用RAD STUDIO 10.3.3,并且已安装了Java 1.8_171/32位

我尝试创建一个新的(示例)Android应用程序32位,并且构建通过。

当我尝试运行它时,会出现错误 无法创建Java虚拟机,随后是
[执行错误] EXEC(1):发生了致命异常。程序将退出。

[执行错误] 命令“\ "C:\ Program Files(x86)\ Java \ jdk1.8.0_171 \ bin \ java.exe\" -genkey -v -keystore \ "C:\ Users \ DELL \ AppData \ Roaming \ Embarcadero \ BDS \ 20.0 \ debug.keystore\" -storepass android -alias androiddebugkey -keypass android -dname \ "cn = 未命名,ou = 移动组织,o = Embarcadero Technologies,l = 旧金山,s = 加利福尼亚州,c = US\" -keyalg RSA -keysize 2048 -validity 10000 \ "已退出,代码为1。”

据我所见,java.exe不支持参数-keypass及其后的所有参数。这是否意味着什么?
请帮我解决这个问题。

英文:

I already checked the issue but i didn't found a solution.

I use RAD STUDIO 10.3.3 in Win 10/64b and i have installed java 1.8_171/32b

I try to create a new (sample) android application 32b and it builds ok.

When i try to run it raises the error could not create java virtual machine followed by
[Exec Error] EXEC(1): A fatal exception has occurred. Program will exit.

and

[Exec Error] The command ""C:\Program Files (x86)\Java\jdk1.8.0_171\bin\java.exe" -genkey -v -keystore "C:\Users\DELL\AppData\Roaming\Embarcadero\BDS\20.0\debug.keystore" -storepass android -alias androiddebugkey -keypass android -dname "cn=Unnamed, ou=Mobile Organization, o=Embarcadero Technologies, l=San Francisco, s=California, c=US" -keyalg RSA -keysize 2048 -validity 10000" exited with code 1.

As i can see java.exe does not support parameter -keypass and all the others following it. Is this meaning something ?
Please help me to solve this problem

答案1

得分: 2

好的,我找到了我的错误。

在工具/选项/部署/SDK管理器/Java/keytool位置中,我将(错误的值)java.exe 错误地设置为了 keytool.exe。

非常感谢您的帮助。

英文:

Ok, I found my mistake.

In tools/options/deployment/SDKmanager/Java/keytool location I had the (wrong value) java.exe instead of keytool.exe

Many thanks for your help

答案2

得分: 1

> 这有什么意思吗?

这些参数是用于 keytool 命令,而不是 java 命令。

请参阅 Keytool 手册页面

如果你的示例 Android 应用实际上正在尝试运行 keytool,那么它的方式是错误的。然而,除非你与我们分享代码以便我们理解它试图做什么,否则我认为我们无法提供进一步的帮助。

英文:

> "Is this meaning something?"

Those arguments are for the keytool command not the java command.

See the Keytool manual page.

If your sample Android app is actually trying to run keytool, it is doing it the wrong way. However, I don't think we can help beyond that unless you share the code with us so that we can understand what it is trying to do.

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  • 本文由 发表于 2020年10月22日 20:07:16
  • 转载请务必保留本文链接:https://go.coder-hub.com/64481883.html
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