如何使用 shell 脚本在空白字符串之前获取第一个字符串？

Question

I have string lists

Facades core/app/Firstmedia/Facades/FmSettingFacades.php
core/app/Firstmedia/FmSettingFacades.php
core/app/Http/Controllers/Frontpage/ContentController.php
core/app/Http/Controllers/Frontpage/HomeController.php
core/app/Http/Controllers/Frontpage/OTPController.php
core/resources/views/admin/pages/solution/form.blade.php
core/resources/views/frontpage/pages/content.blade.php
core/routes/web.php
backend core/resources/views/backend/pages/content.blade.php

I just want get first string before blank string

Expected output:

Facades
backend

I try this code

#!/bin/bash
filename='copy.md'
n=1
tanggal=$(date +%d-%m-%Y)
tanggalWaktu=$(date +"%d-%m-%Y %H:%M:%S")
rm -rf ./update/$tanggal
mkdir -p ./update/$tanggal
mkdir -p ./logs

while read line; do
    echo "$line" | awk '{sub(/:.*/,x)}1'
done < $filename

Answer 1

If a row contains more than 1 column, output first column:

awk 'NF>1{print $1}'

awk uses spaces and tabs as default field separator.

---

See: 8 Powerful Awk Built-in Variables – FS, OFS, RS, ORS, NR, NF , FILENAME, FNR

Answer 2

Assuming that blank string means one or more spaces:

cut -f 1 -d ' ' "$filename"