PHP使用DOM宽度进行计算
我想为产品创建一个响应部分，但是当我尝试获得DOM OffsetWidth时，我遇到了一个问题。
基本上，我有一个SQL语句，该语句应该显示所选项目，但我想limit结果数量。 限制应通过乘积偶有宽度的容器宽度和划分为172（=产品容器的宽度。这些容器具有固定的172px宽度）。
我唯一的问题是我无法弄清楚如何使用PHP获得DOM宽度。我设法使用JavaScript OffsetWidth获得了宽度，但我无法将此值传递给PHP SQL语句。
知道如何将DOM元素宽度或将JavaScript值传递给PHP？
$offset = (round(1302/176)*2)-1
SELECT * FROM table WHERE condotion LIKE "TRUE" LIMIT $offset;
PS：$ offset计算：1302：主容器的宽度，该宽度除以172（固定值）和*** 2 - ＆gt;它使结果翻了一番，所以我有两行。 -1是为了删除最后一个元素，因为我想用其他somethins替换它。
I want to create a responsive section for products but I ran into a problem when I tried to get DOM offsetWidth.
Basically I have an SQL statement which should show the selected items but I want to LIMIT the number of results. The LIMIT should be calculated by the container width and division of the product cointainers width which is 172 (= the width of the products container. These containers have the fixed 172px width).
My only problem is that I couldn't figure out how to get the DOM width using PHP. I have managed to get the width using javascript offsetWidth but I couldn't pass this value to the PHP SQL statement.
Any idea how to get the DOM elements width or pass javascript values to PHP?
$offset = (round(1302/176)*2)-1
SELECT * FROM table WHERE condotion LIKE "TRUE" LIMIT $offset;
PS: The $offset calculation: 1302 : the width of the main container which is divided by 172 (fixed value) AND *2 -> it doubles the result so I'll have two rows. AND the -1 is for to delete the last elements because I want to replace it with somethins else.
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