BASH Mysql输出语句的一部分

Question

I'm trying to run the below script to get a list of Orgs and the amount of users each Org has:

#!/bin/bash

array=(293 182 177 12 85 51 325 225 40 169 357 329 243 349 291 295 22 279 16 69 219 299 301 331 91 281 285 59 283 341 45 289 95 61 77 13 14 201 43 343 223 28 171 26 233 47 303 367 369 339 257 305 353 245 213 87 345 2 71 199 24 179 259 37 35 237)

for i in "${array[@]}"; do   

query=$(export MYSQL_PWD=MYPASS; mysql -e "SELECT COUNT(*) FROM Org.Users WHERE HomeOrgID=$i;")

echo "Org_$i:$query"  

done

I expect the result to be a simple "Org:Number" list:

Org_1:150
Org_25:250
Org_17:64
Org_64:12

But the output is not displayed correctly. It does show part of the MySQL statement:

Org_293:COUNT(*)
1
Org_182:COUNT(*)
0
Org_177:COUNT(*)
8
Org_12:COUNT(*)
0
Org_85:COUNT(*)
1

How can I get the output to NOT display that "COUNT(*)" and display just a simple list?

Thanks in advance!

Answer 1

One quick hack : use bash arrays to store your output.

#!/bin/bash

array=(293 182 177 12 85 51 325 225 40 169 357 329 243 349 291 295 22 279 16 69 219 299 301 331 91 281 285 59 283 341 45 289 95 61 77 13 14 201 43 343 223 28 171 26 233 47 303 367 369 339 257 305 353 245 213 87 345 2 71 199 24 179 259 37 35 237)

for i in "${array[@]}"; do  
    # we store the result of the request as an array
    query=($(export MYSQL_PWD=MYPASS; mysql -e "SELECT COUNT(*) FROM Org.Users WHERE HomeOrgID=$i;"))

    # we echo the 2nd member of the array. ${query[0]} should contain 'count(*)'
    echo "Org_$i:${query[1]}"

done