Question

要为我的问题提供更多信息，请修改插入的图片。

https://i.stack.imgur.com/1IuID.jpg

SELECT pro.id
     , pro.featured
     , pro.price_per_night
     , pro.meta_title
     , img.image
  FROM properties AS pro
  INNER JOIN images AS img 
    ON pro.id = img.imageable_id
 WHERE pro.featured = 1

使用此查询，将显示符合查询的所有图像。也就是说，显示图像：image4，image5，image6，image7，image8和image9。

例如，如果我订购LIMIT = 1，则只显示一张图片，更准确地说是＆＃34; image4＆＃34;。不再显示图像。

我想更改查询以显示与查询匹配的字段的第一个图像。更确切地说，我必须展示＆＃34; image4＆＃34;和＆＃34; image7＆＃34;。

再次感谢您的帮助

Answer 1

mysql limit

SELECT pro.id, pro.featured, pro.price_per_night, pro.meta_title, img.image
FROM properties AS pro 
INNER JOIN images AS img ON pro.id = img.imageable_id 
WHERE pro.featured = 1
LIMIT 1

Answer 2

您需要拥有图片ID并首先使用图片FROM properties AS pro。

JOIN images AS img

ON pro.id = img.imageable_id

WHERE image_id=1

#!/bin/bash

# Register signal handler for SIGINT (Ctrl+c)
trap abort INT 

function abort() {
    echo "Sending SIGINT to background process ${pid}"
    # Kill background process
    kill "${pid}"
    # Wait for it to finish after killing it
    wait "${pid}"
    # Exit the script
    echo "Aborting"
    exit 1   
}

# Start long running process in background (&)
sleep 1000 &
# Obtain the pid of that process
pid=$!
# Wait for the background process to finish
wait

read

MySql：从组中选择第一行，两个表之间使用INNER JOIN

2 个答案: