Question

我有一个简单的“学生” csv文件，其中包含5列和几行，第一列用于名称。我只想提示用户输入一个名称，看看是否在csv文件中有该名称，但是它不起作用，我也不知道为什么。另外，如果不想找到该程序，我也不想退出，只想打印一条消息

read -p "What is your name?: " arg1
if grep -q arg1 "$names.csv"; then
   $ printf "%s\n" "We got you!"
else
   $ printf "%s\n" "Sorry, you are not on the list!"

带有输入的csv文件：

Name,R1, R2 , R3 , R4
James,"35,587","55,800,000,000","1,567,988","491,000,000,000"
Tom,"16,000","11,300,000,000","706,250","120,000,000,000" 
Sarah,"3,069","1,180,000,000","384,490","7,200,000,000"

Answer 1

请您尝试以下操作（由于OP尚未提及示例，因此无法对其进行测试）。

cat script.ksh
read -p "What is your name?: " arg1
if grep -q "$arg1" "names.csv"
then
   printf "%s\n" "We got you!"
else
   printf "%s\n" "Sorry, you are not on the list!"
fi

问题已在OP的代码中解决：

已将$添加到变量arg1中，否则将arg1视为字符串，同时将其grep到names.csv文件中。
已删除$前面不必要的printf。
添加了fi以完成if块，请参见手册页，每个if都应以fi结尾以完成其块。

EDIT1： 已通过自己创建的示例进行了测试：

让我们说以下是我们的Input_file。

cat Input_file
"singh","abc_test","1","2","3"
"abc","singh","1","2","4"
"1","2","3","4","5"

脚本如下：

cat script.ksh
read -p "What is your name?: " arg1
if grep -q "$arg1" "Input_file"
then
   printf "%s\n" "We got you!"
else
   printf "%s\n" "Sorry, you are not on the list!"
fi

现在，当我运行以下代码时，我得到了输出。

./script.ksh
What is your name?: singh
We got you!

EDIT2： ：如果您想在任意字段中匹配 "singh" 之类的精确字符串的csv，然后尝试以下操作。

cat script.ksh
read -p "What is your name?: " arg1
if grep -q "\"$arg1\"" "file"
then
   printf "%s\n" "We got you!"
else
   printf "%s\n" "Sorry, you are not on the list!"
fi

上述脚本的测试：

./script.ksh
What is your name?: singh1
Sorry, you are not on the list!

./script.ksh
What is your name?: singh
We got you!

使用bash在csv文件中搜索特定值

1 个答案: