問題描述

我在互聯網上到處尋找有關此問題的答案，然后出現了準備好的語句和綁定參數(我不知道那是什么東西)

基本上，我有一個逗號分隔的列表

$list = '食物，飲料，烹飪';

好的，現在我想在數據庫的一列中搜索這些項目中的每一項......聽起來很簡單，對吧?

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

然后，

var_dump($variable);

<塊引用>

未定義變量

為什么?我看不出代碼有什么問題.如果我輸入一個特定的值，它會起作用，并且我已經使用 WHERE stuff=$item 對其進行了測試 - 效果很好.

所以不是變量/數據庫，而是IN語句中的錯誤.我不明白為什么它不起作用.

每個元素都需要用引號括起來

$list = "'食物', '飲料', '烹飪'";$query = "SELECT * FROM table WHERE stuff IN ($list)";

或者如果你有一個數組

$array = array("食物","飲料","烹飪");$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.

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).")";

這篇關于MYSQLI - 數組中的位置的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！