我试图做一个简单的JOIN,当尝试加载页面时,它是空白的。当我输入查询时。我正试图让它打印查询。我知道如何打印结果,但是在打印联接时遇到了问题。
SELECT user.name, course.name
FROM `user`
INNER JOIN `course` on user.course = course.id;
在PHPmyAdmin上的SQL中,它返回
NAME Course
Alice HTML5
Bob HTML5
Caroline CSS3
<?php
$con=mysqli_connect("URL,"USERNAME","PASS","DB");
// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT user.name, course.name
FROM `user`
INNER JOIN `course` on user.course = course.id;");
while($row = mysqli_fetch_array($result)) {
echo $row['user.name'] . " " . $row['course.name'];
echo "<br>";
}
mysqli_close($con);
?>
感谢您的真知灼见。。。
您不需要使用$row['user.name']
,它应该是$row['name']
,但由于您有两个与name
同名的列,因此您必须更改并将它们分配给其他变量。
[HINT]:检查while循环中的新查询和echo
。请进行以下更改:
<?php
$con=mysqli_connect("HOST","USERNAME","PASS","DB");
// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT user.name as username, course.name as coursename
FROM `user`
INNER JOIN `course` on user.course = course.id;");
while($row = mysqli_fetch_array($result)) {
echo $row['username'] . " " . $row['coursename'];
echo "<br>";
}
mysqli_close($con);
?>