shell腳本，如果水珠圖案的任何文件

Question

符合我寫了一個shell腳本來檢查，如果一個二進制存在：

#!/usr/local/bin/bash 

if [ -e "/path/to/a/binary/file/app.bin" ]; then 
    echo "INFO: application has been built successfully" 
else 
    echo "ERROR: application couldn't be built". 
fi 

它工作正常，並給出了預期的效果。但是如果有很多類似的應用程序名稱（比如app1.bin, app2.bin等），我想測試"app*.bin"如果條件失敗：

#!/usr/local/bin/bash 

    if [ -e "/path/to/a/binary/file/app*.bin" ]; then 
     echo "INFO: application has been built successfully" 
    else 
     echo "ERROR: application couldn't be built". 
    fi 

我怎樣才能糾正，如果條件與app爲尋找任何二進制的存在名稱即app*.bin

Answer 1

glob配對不適用於引號內。即使刪除了它，如果從glob匹配中返回多個文件，則條件將失敗，因爲-e僅適用於單個文件。

在bash，你可以做到這一點，

# the nullglob will let the unmatched glob to handled properly 
shopt -s nullglob 
fileList=(/path/to/a/binary/file/app*.bin) 

if [ "${#fileList[@]}" -ge 1 ]; then 
    echo "INFO: application has been built successfully" 
else 
    echo "ERROR: application couldn't be built". 
fi 

# This line is optional. This is just to ensure the nullglob option is 
# reset after our computation 
shopt -u nullglob 

Answer 2

另一種方法是使用compgen -G：它會生成匹配的文件列表，或者如果沒有相匹配，將一個錯誤狀態退出。使用compgen您需要引號以防止擴展。

if compgen -G "/path/to/a/binary/file/app*.bin" > /dev/null; then 
    echo "INFO: application has been built successfully" 
else 
    echo "ERROR: application couldn't be built". 
fi