用MySQLi回显JOIN SQL表的内容 [英] Echo contents of JOIN SQL tables with MySQLi

问题描述

我在一个系统上工作，并且该模块应该回显数据库的内容.

I'm working on a system, and this module is supposed to echo the contents of the database.

在我添加一些JOIN语句之前，它一直运行良好.

It worked perfectly until I added some JOIN statements to it.

我已经检查并测试了SQL代码，它运行良好.不起作用的是我在其中回显JOINed表的内容的那部分.

I've checked and tested the SQL code, and it works perfectly. What's not working is that part where I echo the content of the JOINed table.

我的代码如下:

$query = "SELECT reg_students.*, courses.*
          FROM reg_students
          JOIN courses ON reg_students.course_id = courses.course_id
          WHERE reg_students.user_id = '".$user_id."'";

$result = mysqli_query($conn, $query);
if (mysqli_fetch_array($result) > 0) {
    while ($row = mysqli_fetch_array($result)) {
       echo $row["course_name"]; 
       echo $row["course_id"];

course_name和course_id既不回显也不给出任何错误消息.

The course_name and course_id neither echo nor give any error messages.

更新:我实际上需要通过联接更多表并更改选定的列来增加查询的复杂性.我需要加入这些表:

UPDATE: I actually need to increase the query complexity by JOINing more tables and changing the selected columns. I need to JOIN these tables:

tutors ，其中包含列: tutor_id ， t_fname ， t_othernames ，电子邮件，电话号码
faculty 包含以下列: faculty_id ， faculty_name ， faculty_code
具有以下列的 courses : course_id ， course_code ， course_name ， tutor_id ， faculty_id

tutors which has columns: tutor_id, t_fname, t_othernames, email, phone number
faculty which has columns: faculty_id, faculty_name, faculty_code
courses which has columns: course_id, course_code, course_name, tutor_id, faculty_id

我想将这些表联接到原始查询中的 reg_students 表中，以便可以按 $ user_id 进行过滤，并希望显示: course_name ， t_fname ， t_othernames ， email ， faculty_name

I want to JOIN these tables to the reg_students table in my original query so that I can filter by $user_id and I want to display: course_name, t_fname, t_othernames, email, faculty_name

推荐答案

我无法想象 user_info 表对加入JOIN有任何好处，因此我将其删除是合理的猜测.我还假设您所需的列都来自 courses 表，因此我用SELECT中的列名来提名表名.

I can't imagine that the user_info table is of any benefit to JOIN in, so I'm removing it as a reasonable guess. I am also assuming that your desired columns are all coming from the courses table, so I am nominating the table name with the column names in the SELECT.

为了便于阅读，我喜欢使用 INNER JOIN 而不是 JOIN .(它们是相同的野兽)

For reader clarity, I like to use INNER JOIN instead of JOIN. (they are the same beast)

将 $ user_id 铸造为整数只是我抛出的最佳做法，以防万一变量是由用户提供/不受信任的输入馈送的.

Casting $user_id as an integer is just a best practices that I am throwing in, just in case that variable is being fed by user-supplied/untrusted input.

您可以使用 mysqli_num_rows()计算结果集中的行数.

You count the number of rows in the result set with mysqli_num_rows().

如果只想使用关联键访问结果集数据，请使用 mysqli_fetch_assoc()生成结果集.

If you only want to access the result set data using the associative keys, generate a result set with mysqli_fetch_assoc().

使用JOIN编写查询时，为每个表声明别名通常会很有帮助.这在很大程度上减少了代码膨胀和读者负担.

When writing a query with JOINs it is often helpful to declare aliases for each table. This largely reduces code bloat and reader-strain.

未经测试的代码:

$query = "SELECT c.course_name, t.t_fname, t.t_othernames, t.email, f.faculty_name
          FROM reg_students r
          INNER JOIN courses c ON r.course_id = c.course_id
          INNER JOIN faculty f ON c.faculty_id = f.faculty_id
          INNER JOIN tutors t ON c.tutor_id = t.tutor_id
          WHERE r.user_id = " . (int)$user_id;
if (!$result = mysqli_query($conn, $query)) {
    echo "Syntax Error";
} elseif (!mysqli_num_rows($result)) {
    echo "No Qualifying Rows";
} else {
    while ($row = mysqli_fetch_assoc($result)) {
        echo "{$row["course_name"]}<br>";
        echo "{$row["t_fname"]}<br>";
        echo "{$row["t_othernames"]}<br>";
        echo "{$row["email"]}<br>";
        echo "{$row["faculty_name"]}<br><br>";
    }
}

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