如何检查表中是否已有名称？ 查询Mysql每天的总和，然后通过PHP填充HTML数据

Question

So, this is my first question here, and I'd really appreciate if someone could help me. I currently have a Mysql database and a web app made with PHP where the users can input how much time they have used on a specific task that day.

Right now I would like to populate an HTML table with the SUM of each days activity in hours. My problem is that I can't find a way to check if the name field is unique, as the data contains many rows with the same name field, one for each day.

I have this Mysql query:

SELECT 
    t_email.name, 
    DATE_FORMAT(input_date,'%d') AS days, 
    SUM(time_worked) AS totalTime
FROM 
    t_transactions 
    LEFT JOIN t_email ON t_transactions.email_id >= t_email.email_id 
WHERE 
    DATE_FORMAT(input_date,'%m') = '02'
    AND DATE_FORMAT(input_date,'%Y') = '2019' 
GROUP BY DATE_FORMAT(input_date,'%d') 
ORDER BY t_email.name

Which returns me a table like this:

name  days  totalTime
james 01    5.00 
james 06    2.00 
jimmy 02    4.0

And so on. I'm using the below PHP:

<?php 
foreach ($monthly_data_per_day as $row): ?>
<tr>
    <td><?= $row['name'] ?></td>
    <?php
    $num_days = date('t');
    for ($i = 1; $i <= $num_days; $i++) {
        if ($row['days'] == $i) {
            echo "<td>" . $row['totalTime'] . "</td>";
        } else {
            echo "<td>" . 0 . "</td>";
        }
    }?></tr>
<?php endforeach;?>

I'd like to have an HTML table where after each name in the database, you could see how many hours that person has done that day. So columns from 1 to how many days there are in the month.

Right now I get a row for each day, with the correct sum of time worked on just one column, as there's one row per column.

Here's a picture of what I'd like the table to look like:

A picture of the table with columns for each day of the month

Answer 1

There is a solution for this using conditional aggregation... that will requires you to write 31 CASE statements, like :

SELECT 
    e.name, 
    SUM(CASE WHEN DAYOFMONTH(t.input_date) = 1 THEN t.time_worked ELSE 0 END) AS day1,
    SUM(CASE WHEN DAYOFMONTH(t.input_date) = 2 THEN t.time_worked ELSE 0 END) AS day2,
    SUM(CASE WHEN DAYOFMONTH(t.input_date) = 3 THEN t.time_worked ELSE 0 END) AS day3,
    ...
FROM 
    t_transactions t
    LEFT JOIN t_email e ON t.email_id = e.email_id 
WHERE YEAR(input_date) = 2019 AND MONTH(t.input_date) = 2
GROUP BY e.name
ORDER BY e.name

Other remarks :

• I guess that you meant t.email_id = e.email_id instead of t.email_id >= e.email_id

• it will probably more efficient to use MySQL built-in date functions such as YEAR , DAYOFMONTH and MONTH , which return integers, rather than manipulating formated string representations of dates

• as not all of the fields in your query were aliase, I made the assumption that columns input_date and time_worked belong to table t_transactions .

• since we SELECT e.name , this column needs to appear in the GROUP BY clause ; on non-ancient versions of MySQL, this is a syntax error