如何从php / mysql中的不同表中获取特定的列数据
[英]How to fetch specific column data from the different tables in php/mysql
问题描述
I'm creating a website on teacher and student database in php, i have done almost all things in mysql database but i want to fetch specific data from the table, like i want to choose "prof. vaidya sir" from column "name" and display professor related all data. 
我正在使用php在教师和学生数据库上创建一个网站，我在mysql数据库中几乎完成了所有工作，但是我想从表中获取特定数据，就像我想从“名称”列中选择“教授vaidya先生”与展示教授相关的所有数据。
Here is my code: 这是我的代码：
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>
<?php
include 'dbh.php';
$name = "Prof. Vaidya Sir";
$class ="BE";
$sqll = "SELECT feedback1.erp,feedback1.class,feedback2.name,feedback2.ability,feedback2.knowledge,feedback2.accessbility,feedback2.attitude FROM feedback1, feedback2 WHERE feedback1.erp=feedback2.erp";
$result = $conn->query($sqll);
?>
<table border="2">
<thead>
<tr><th>erp</th><th>class</th><th>name</th><th>ability</th><th>knowledge</th><th>accessbility</th><th>attitude</th></tr>
</thead>
<tbody>
<?php
if($result->num_rows > 0)
{
while ($row = $result->fetch_assoc()){
echo "<tr><td>{$row["erp"]}</td><td>{$row["class"]}</td><td>{$row["name"]}</td><td>{$row["ability"]}</td><td>{$row["knowledge"]}</td><td>{$row["accessbility"]}</td><td>{$row["attitude"]}</td></tr>\n";
}
else{echo "0 results";
}
?>
</tbody>
</table>
</body>
</html>
This is output of above code 这是上面代码的输出
2 个解决方案
解决方案1
0 2017-02-24 18:34:11
You're looking for mysql joins. 您正在寻找mysql连接。
SELECT table1.name, table2.col2
FROM table1
JOIN table2
ON table1.erp = table2.erp
WHERE table1.name = 'Prof. Vaidya Sir';
This says to first join the two tables on their common columns, then select the specific columns you are looking for where the name matches. 也就是说，首先将两个表连接到它们的公共列，然后选择要在其中查找名称匹配的特定列。
解决方案2
0 2017-02-24 18:37:23
Try something like this disclaimer: I don't usually use mysqli so I think I have it right 试试这样的免责声明：我通常不使用mysqli，所以我认为我做对了
you should use JOINS instead of old way it makes the connection clearer and more human readable 您应该使用JOINS而不是旧方法，它会使连接更清晰，更易读
<?php
include 'dbh.php';
$name = "Prof. Vaidya Sir";
$class ="BE";
$sqll = $mysqli->prepare("SELECT feedback1.erp,feedback1.class,feedback2.name,feedback2.ability,feedback2.knowledge,feedback2.accessbility,feedback2.attitude
FROM feedback1 join feedback2 ON feedback1.erp=feedback2.erp
WHERE feedback2.name =?";
$sqll->bind_param('s', $name);
$sqll->execute();
$result = $sqll->get_result();
var_dump($result);
?>
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