Question

我需要针对来自users表的user_id显示来自表票证的支持票证。我正在使用MySQLI。这是我写的查询，但对我来说不起作用：

SELECT user_id, subject, message FROM tickets 
INNER JOIN users.user_id WHERE user_id='".$user_id."'"

任何人都可以更正此查询吗？

这里的完整代码

<?php
$servername = "localhost";
$username = "d";
$password = "ddds";
$dbname = "sddd";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

$sql = "SELECT user_id, subject, message FROM tickets INNER JOIN users.user_id USING (user_id) WHERE user_id='".$user_id."'";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
    echo "<table class='table table-bordered'>
         <thead>
            <tr>                                                
              <th>Subject</th>
              <th>Date</th>
              <th>Status</th>
              <th>View</th>
            </tr>
           </thead>";
    // output data of each row
    while($row = $result->fetch_assoc()) {
        echo "<tr><td>".$row["subject"]."</td><td>".$row["date"]." ".$row["status"]."</td><td>".$row["view"]."</td></tr>";
    }
    echo "</table>";
} else {
    echo "0 results";
}
$conn->close();
?>

Answer 1

更正语法：

SELECT T.user_id, T.subject, message FROM tickets T 
INNER JOIN users U ON U.user_id = T.user_id WHERE `T.user_id='".$user_id."'"`

您已忘记ON。

Answer 2

如果您想使用INNER JOIN，则必须指定您正在进行加入的位置（使用on或using），例如以下查询：

SELECT user_id, subject, message 
FROM tickets 
INNER JOIN users.user_id ON tickets.user_id = users.user_id 
WHERE user_id = '".$user_id."'"

OR

SELECT user_id, subject, message 
FROM tickets 
INNER JOIN users.user_id USING (user_id) 
WHERE user_id = '".$user_id."'"

Answer 3

你应该只加入表而不是字段！

SELECT user_id, subject, message FROM tickets 
INNER JOIN users.user_id WHERE user_id='".$user_id."'"

将tickets表加入users.user_id表，该表是users表的字段。应该是：

SELECT user_id, subject, message FROM tickets 
INNER JOIN users USING (user_id) WHERE user_id='".$user_id."'"

您也可以使用ON进行连接（当列名称匹配时使用是一个快捷方式）：

SELECT user_id, subject, message FROM tickets 
INNER JOIN users ON tickets.user_id=users.user_id WHERE user_id='".$user_id."'"

PHP帮助Inner Join问题mysqli

3 个答案: