Question

此脚本应该从文件中的变量中删除前导空格和尾随空格，并将它们插入到SQL语句中。这适用于除第一个变量之外的每个变量。它不会从用户标识中删除尾随空格。我附上了用于收集变量数据的示例文件和代码。

示例文件：

"User ID"|First|Last|Password|"E-Mail Address"|CreateDate|RequestDate|Requestor|Title|"Site Location"|"Client Name"|"Requestor Email"|Roles|Tasks|"Correspondance Packages"
"A145078 "|"James "|Joyce|" DO NOT USE"|jjames@job.com|" 2015/06/22 16:30:00"|"2015/06/20 17:00:00"|"John Edwards"|CSR|Aetna|Aetna|bavolt@aetna.com|"CTM Data Analyst, Medical Appeals Status/Status Reasons, RCA Status and Status Reasons, Grievance Status/Status Reasons, CTM Status/Status Reasons"||

正在生成的输出（请参阅user_id之后的空格：

INSERT INTO users(user_id, first, last, password, email, createdate, requestdate, requestby)VALUES('A145078 ','James','Joyce','DO NOT USE','jjames@job.com','2015/06/22 16:30:00','2015/06/20 17:00:00','John Edwards');

使用代码：

    #!/bin/bash
> user_adds.sql

while IFS="|" read -r uid first last password email createdate requestdate requestor title site client requestemail roles tasks packages; do
   #printf '%s %s\n' "$uid" "$first" "$roles"
   uid=$(echo "$uid" | sed 's/\"//g' | sed 's/^[ \t]*//' | sed 's/[ \t]*$//')
   first=$(echo "$first" | sed 's/\"//g' | sed "s/'/ /g" | sed 's/^[ \t]*//;s/[ \t]*$//')
   last=$(echo "$last" | sed 's/\"//g' | sed "s/'/ /g" | sed 's/^[ \t]*//;s/[ \t]*$//')
   echo $uid - "$uid"
   echo $roles

   echo "INSERT INTO users(user_id, first, last, password, email, createdate, requestdate, requestby)VALUES('$uid','$first','$last','DO NOT USE','$email','$createdate','$requestdate','$requestor');" >> user_adds.sql

   #echo "INSERT INTO user_demographics(user_id, title, site_location, comment, client_name, requester_email,last_password_date_changed)select id,'$title','$site_location','$title','$client','$request_email',NULL from users where user_id in('$user_id');" >> demo_adds.sql


done < 'new_users.csv'

Answer 1

$ cat tst.awk        
BEGIN{ FS="[[:space:]\"]*[|][[:space:]\"]*" }
NR>1 {
    gsub(/^[[:space:]"]+|[[:space:]"]+$/,"")
    printf "INSERT INTO users(user_id, first, last, password, email, createdate, requestdate, requestby)VALUES('%s','%s','%s','%s','%s','%s','%s','%s');\n", $1,$2,$3,$4,$5,$6,$7,$8
}

$ awk -f tst.awk file
INSERT INTO users(user_id, first, last, password, email, createdate, requestdate, requestby)VALUES('A145078','James','Joyce','DO NOT USE','jjames@job.com','2015/06/22 16:30:00','2015/06/20 17:00:00','John Edwards');

Answer 2

以下是您可以使用的基本模板：

#!/bin/bash

typeset -a  parts

cat <<'EOF'  |
"A145078 "|"James "|Joyce|" DO NOT USE"|jjames@job.com|" 2015/06/22 16:30:00"|"2015/06/20 17:00:00"|"John Edwards"|CSR|Aetna|Aetna|bavolt@aetna.com|"CTM Data Analyst, Medical Appeals Status/Status Reasons, RCA Status and Status Reasons, Grievance Status/Status Reasons, CTM Status/Status Reasons"||
EOF
while read line; do
    line=$( echo "${line}" | sed -e 's/" /"/g' -e 's/ "/"/g' -e 's/"//g')
    typeset -a parts=( ${line//|/ } )
    echo ${parts[@]}
    uid="${parts[0]}"
    first="${parts[1]}"
    last="${parts[2]}"
    echo "INSERT INTO users(user_id, first, last) VALUES ('$uid','$first','$last');"
done

没有从第一个变量中删除空格

2 个答案: