dongzhan5286 2016-03-26 21:38
浏览 220
已采纳

MYSQLI - WHERE IN数组[重复]

This question already has an answer here:

I've looked all over the internet for answers on this one, and prepared statements and bind params come up (I have no idea what that stuff is)

Basically, I have a comma separated list

$list = 'food, drink, cooking';

Ok, now I want to search for each of those items in a column of the database... Sounds simple, right?

$query = "SELECT * FROM table WHERE stuff IN ('$list')";
$runquery = mysqli_query($connection, $query);
while($row = mysqli_fetch_array($runquery,MYSQLI_ASSOC)){
    $variable = $row;
}

Then later on,

var_dump($variable);

UNDEFINED VARIABLE

Why? I can't see anything wrong with the code. It works if I put a particular value, and I have tested it with WHERE stuff=$item - that works fine.

So it's not the variables / database, it's an error in the IN statement. I don't understand why it won't work.

</div>
  • 写回答

3条回答 默认 最新

  • dtcwehta624485 2016-03-26 21:43
    关注

    Each element needs quotes around it

    $list = "'food', 'drink', 'cooking'";
    $query = "SELECT * FROM table WHERE stuff IN ($list)";
    

    Or if you had an array

    $array = array("food","drink","cooking");
    $query = "SELECT * FROM table WHERE stuff IN (".implode(',', $array).")";
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论
查看更多回答(2条)

报告相同问题?

悬赏问题

  • ¥15 如何让企业微信机器人实现消息汇总整合
  • ¥50 关于#ui#的问题:做yolov8的ui界面出现的问题
  • ¥15 如何用Python爬取各高校教师公开的教育和工作经历
  • ¥15 TLE9879QXA40 电机驱动
  • ¥20 对于工程问题的非线性数学模型进行线性化
  • ¥15 Mirare PLUS 进行密钥认证?(详解)
  • ¥15 物体双站RCS和其组成阵列后的双站RCS关系验证
  • ¥20 想用ollama做一个自己的AI数据库
  • ¥15 关于qualoth编辑及缝合服装领子的问题解决方案探寻
  • ¥15 请问怎么才能复现这样的图呀