I am trying to find a way to give current file as argument for module. I couldn't find it anywhere so I'm trying stackoverflow
{
"name": "Python: Synapse",
"type": "python",
"request": "launch",
"module": "projFolder.targetModule",
"console": "integratedTerminal"
}
Manually specifying the module argument such as above example, works. But thinking of doing that for every each file wanted me to automate it. I've tried ${file}, but it gives the file path not the module. So it didn't work. How can I launch current current file as module?
ps: answer from https://stackoverflow.com/a/57416114/6088796 , but I modify a litter
add something like this to your launch.json, but if you want to run multi-module(in a different place, the only way you can do is using multi configuration, or use a plugin(see below)
"configurations": [
{
"name": "Python: Module",
"type": "python",
"request": "launch",
"module": "yourmodule.${fileBasenameNoExtension}",
"stopOnEntry": true
},
You can use the extension Command Variable to get this relative directory with dot separator.
{
"version": "0.2.0",
"configurations": [
{
"name": "Python: Module CmdVar",
"type": "python",
"request": "launch",
"console": "integratedTerminal",
"module": "${command:extension.commandvariable.file.relativeDirDots}.${fileBasenameNoExtension}",
}
]
}
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