MYSQLI-数组中的位置 [英] MYSQLI - WHERE IN array

问题描述

我已经在互联网上寻找了关于该问题的答案，并且准备了语句和绑定参数(我不知道那是什么东西)

基本上，我有一个用逗号分隔的列表

$list = 'food, drink, cooking';

好吧，现在我想在数据库的一列中搜索每个项目...听起来很简单，对吧?

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

然后，

var_dump($variable);

不确定的变量

为什么?我看不到代码有什么问题.如果我输入了一个特定的值，它就可以工作，并且我已经用WHERE stuff=$item测试了它-很好.

这不是变量/数据库，而是IN语句中的错误.我不明白为什么它不起作用.

解决方案

每个元素都需要用引号引起来

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

或者如果您有数组

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

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.

解决方案

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).")";

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