PHP If / Else Argument和PDO对象 [英] PHP If/Else Argument and PDO Objects

问题描述

如果/ else使用以下内容，我该怎么做一个参数。我想将变量$ condition定义为表A和表B的交集，并最终在这条件为真的情况下，然后执行一些代码。

How can I make an argument if/else with the following. I would like to define the variable $condition as the intersection of Table A and Table B, and ultimately, where this condition is true, then execute some code.

这是我有PHP（我试图匹配table1.field1和table2.field1，如果是真的，那么我将添加一个CSS类到< tr> ）

Here is the PHP I have (I am trying to match up the table1.field1 and table2.field1 and where true, then I will add a CSS class to the <tr>)

$result = $dbh->query("SELECT field1, field2, field3, field4 FROM table1 ORDER BY field1 DESC");
$result->setFetchMode(PDO::FETCH_ASSOC);

$match = $dbh->query("SELECT table1.field1 FROM table1 INNER JOIN table2 WHERE table2.field1 = table1.field1");

$condition = **???????**

此外，有没有办法简化此代码？我觉得它太久了。马克吐温有一句话......我很抱歉这封信太长了，我没有时间把它缩短。我希望我可以说这是相关的，但我已经在这一段时间了！任何帮助将不胜感激。

Also, is there a way to simplify this code? I feel like its too long. Mark Twain has a saying... "I'm sorry this letter is too long, I didn't have time to make it shorter". I wish I could say that this is relevant here, but I've been at this for a while! Any help would be greatly appreciated.

推荐答案

如何（在您的第一次和当时的查询中）

What about something like (in your first and then-only query)

$result = $dbh->query("
    SELECT
        t1.field1,
        t1.field2,
        t1.field3,
        t1.field4,
        IF(t2.field1 IS NULL, 0, 1) `match`
    FROM
    table1 t1
        LEFT JOIN table2 t2 ON t2.field1 = t1.field1
");

（你应该添加反引号（`）围绕匹配，因为它是一个MySQL关键字）

并且在你的循环中

(you should put backticks (`) around match, since it's a MySQL keyword)
and in your loop

<tr class="<?=$v["match"] ? "match" : ""?>">

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