javascript - Gulp resume task on error? Is this possible

Question

This is my task

export default function glsl() {
  return src(conf["glsl"])
    .pipe(plumber({ errorHandler: false}))
    .pipe(bytediff.start())
    // Minify the file
    .pipe(glslmin())
    // Output    
    .pipe(plumber.stop())
    .pipe(bytediff.stop(byteDiffCB))
    .pipe(dest("./dist"))
}

this is what happens when i run gulp glsl

[16:55:15] Requiring external module ts-node/register
[16:55:17] Using gulpfile ~/Documents/gulp-gameoptimizer/gulpfile.ts
[16:55:17] Starting 'glsl'...
[16:55:17] 'glsl' errored after 97 ms
[16:55:17] Error in plugin "gulp-glsl"
Message:
    glsl/2xbr-hybrid-v5-gamma.glsl: SyntaxError: Expected "(", identifier, or whitespace but "v" found.
Details:
    domainEmitter: [object Object]
    domainThrown: false

What I would ideally want to happen (though i can't figure out how to do it), is when a plugin gives an error, copy the source file to the destination, then resume the task. In other words act as if the plugin weren't there.

question from:https://stackoverflow.com/questions/65872093/gulp-resume-task-on-error-is-this-possible

Answer 1

Waitting for answers