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。我试图让它打印查询。我知道如何打印结果但是在打印连接时遇到了问题。
SELECT user.name, course.name
FROM `user`
INNER JOIN `course` on user.course = course.id;
在我在 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);
?>
感谢您的任何见解...
最佳答案
你不需要使用 $row['user.name']
它应该是 $row['name']
但因为你有两列相同name as name
所以你必须改变将它们分配给其他变量。
[提示]:检查新查询和 while 循环中的 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);
?>
关于php - MySQL Join 和 Echo 结果,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/25850741/