暂且不论事情像检查，如果一个数组是空，你可以为它创建一个一般方法和用它在你的具体情况，如：

 公共字符串[] concatAll（字符串[] jobsA，字符串[] jobsB，字符串[] jobsC，字符串[] jobsD）
    {
        返回generalConcatAll（jobsA，jobsB，jobsC，jobsD）;
    }    公众的String [] generalConcatAll（字符串[] ...作业）{
        INT LEN = 0;
        对于（最终的String []的工作：作业）{
            LEN + = job.length;
        }        最终的String []结果=新的String [LEN]        INT currentPos = 0;
        对于（最终的String []的工作：作业）{
            System.arraycopy（工作，为0，因此，currentPos，job.length）;
            currentPos + = job.length;
        }        返回结果;
    }
 

I'm wondering how I can concatenate 4 string arrays in Java.

There is a question about this already. How to concatenate two arrays in Java?

But I tried to replicate it but it does not work for me.

This is what my code looks like:

Calling the method:

concatAll(jobs1, jobs2, jobs3, jobs4);

The method itself:

public String[] concatAll(String[] jobsA, String[] jobsB, String[] jobsC, String[] jobsD) {
    int totalLength = jobsA.length;
    for (String[] array : jobsD) {
        totalLength += array.length;
    }

    String[] result = Arrays.copyOf(jobsA, totalLength);

    int offset = jobsA.length;

    for (String[] array : jobsD) {
        System.arraycopy(array, 0, result, offset, array.length);
        offset += array.length;
    }
    return result;
}

Putting aside things like checking if an array is null, you can create a general method for it and use it in your specific case, like this:

    public String[] concatAll(String[] jobsA, String[] jobsB, String[] jobsC, String[] jobsD) 
    {
        return generalConcatAll (jobsA, jobsB, jobsC, jobsD);
    }

    public String[] generalConcatAll(String[]... jobs) {
        int len = 0;
        for (final String[] job : jobs) {
            len += job.length;
        }

        final String[] result = new String[len];

        int currentPos = 0;
        for (final String[] job : jobs) {
            System.arraycopy(job, 0, result, currentPos, job.length);
            currentPos += job.length;
        }

        return result;
    }