grep两个字符串/单词的列表，仅返回第一个结果

Question

I have a file (biglist.txt) that I need to search for lines that contain two strings and I need the output to only return the first instance of that search. String one (of two) is variable, meaning I have a file that is another long line-separated (hard returned) list of different strings called (queries.txt). The second string will be a constant word (let's call the word "description"). I need to search for lines with variable string + constant string, and only return the first instance of such a search.

I know something like this will work for a list of single strings.

cat queries.txt | xargs -I{} grep -m 1 {} biglist.txt > output

but I need to add the argument that each of the strings in the queries.txt must also be on a line with the constant word "description". The file "biglist.txt" has multiple lines with each of the strings and the word "description" and I only need to output one of those. Not all of the lines have both the string and the word "description" and often the first line does not have the word "description". This newb appreciates any help.

Answer 1

Without a better description of the input and expected output this might do the job:

grep description biglist.txt | grep -f queries.txt | head -n1

Output the first line in biglist.txt that contains description and one of the queries in queries.txt.