Find files of specified size using bash in Unix

Question

I am looking for a Unix command to print the files with its size. I used this but it didn't work. ``` find . -size +10000k -print. ``` I want to print the size of the file along with the filename/directory.

Accepted Answer

```
find . -size +10000k -exec ls -sd {} +
```

If your version of `find` won't accept the `+` notation (which acts rather like `xargs` does), then you might use (GNU `find` and `xargs`, so `find` probably supports `+` anyway):

```
find . -size +10000k -print0 | xargs -0 ls -sd
```

or you might replace the `+` with `\;` (and live with the relative inefficiency of this), or you might live with problems caused by spaces in names and use the portable:

```
find . -size +10000k -print | xargs ls -sd
```

The `-d` on the `ls` commands ensures that if a directory is ever found (unlikely, but...), then the directory information will be printed, not the files in the directory. And, if you're looking for files more than 1 MB (as a now-deleted comment suggested), you need to adjust the `+10000k` to `1000k` or maybe `+1024k`, or `+2048` (for 512-byte blocks, the default unit for `-size`). This will list the size and then the file name. You could avoid the need for `-d` by adding `-type f` to the `find` command, of course.

Find files of specified size using bash in Unix

Solution

Problem