Use grep to get the occurrences of multiple strings

Question

I have my primary file(log1.txt) strings below.

T202210010515.XX.old:2022-10-01 10:57:56<Other Texts>975673849<0ther Texts>975673849
T202210010515.XX.old:2022-10-01 10:59:56<Other Texts>975673449<0ther Texts>975673449
T202210010515.XX.old:2022-10-01 10:57:58<Other Texts>975673849<0ther Texts>975673849
T202210010515.XX.old:2022-10-01 10:59:51<Other Texts>975673849<0ther Texts>975673849
T202210010515.XX.old:2022-10-01 10:57:52<Other Texts>975673849<0ther Texts>975673849
T202210010515.XX.old:2022-10-01 10:59:59<Other Texts>975673449<0ther Texts>975673449

Each and every set is starting with T202210*. I need to grep the number and return the lines count which that numbers occured. I have another secondary file with all the numbers which I need to search(numbers.txt). Here is what I need as the output, The Number and the lines count of each number.

975673849 -4
975673449 -2

Here is my try.

grep -f numbers.txt log1.txt

But this will produce all the string lines with a highlighted number. Can someone help me to produce the output what I need actually.

Answer 1

That should work:

╰─$ cat log1.txt                                                                                                                                                                                             130 ↵
T202210010515.XX.old:2022-10-01 10:57:56<Other Texts>975673849<0ther Texts>975673849
 T202210010515.XX.old:2022-10-01 10:58:56<Other Texts>975673849<0ther Texts>975673849
╰─$ cat numbers.txt 
975673849
975673449
╰─$ while read -r line; do echo "$line" && cat log1.txt | grep -c "$line"  ; done < numbers.txt
975673849
2
975673449
0