如何将sql结果存储到变量中？

Why isn't this working for me:

为什么这不适合我:

$item_id =
"SELECT item_id FROM items WHERE item_id=(SELECT max(item_id) FROM items)";

I'm trying to insert the sql query result into the variable item_id.

我正在尝试将sql查询结果插入到变量item_id中。

If i write echo"$item_id"; i get the sentence in between the quotes as an output.

如果我写回“$ item_id”;我把引号之间的句子作为输出。

1 个解决方案

#1

you must have to do the connection out from the var and you need to do a fetch array try this:

你必须要从var进行连接,你需要做一个获取数组试试这个:

// your connection data //

//您的连接数据//

$link = mysql_connect("localhost", "youruser"); mysql_select_db("yourdb", $link);

$ link = mysql_connect(“localhost”,“youruser”); mysql_select_db(“yourdb”,$ link);

// your select //

//你的选择//

$result = mysql_query("SELECT yourvalue FROM yourtable", $link);

$ result = mysql_query(“SELECT yourvalue FROM yourtable”,$ link);

$row = mysql_fetch_array($result)

$ row = mysql_fetch_array($ result)

your item var

你的项目变种

$item_id = $row ["yourvalue"]

$ item_id = $ row [“yourvalue”]

regards

#1