Mysql从表中选择数据。。和echo值，其中id=$id

在Mysql php中，我需要使用

 select * from table_name

及

我有一个包含5列和6行的表，我需要根据列id回显值

对于exm。桌子

id  -- name -- age -- dob -- remarks
1  -- rajju  -- 20 -- 24/12/2015  -- good
2  -- prasad  -- 15  -- 12/2/2010 -- bad
3  -- kranti  -- 22  -- 05/05/2000 -- average

现在我需要选择所有db和

echo value like <`echo $name where id=1` > your health is <`echo $remarks where id=1`> better then <`echo $name where id=2`>

只用

 select * from table_name where id = 1 

---

是否只需要与1个用户进行比较？ 如果是这样，那么你可以这样做

$mainid = 1; //(you can get this value dynamically)
$compareid=2; //(you can get this value dynamically)
$result1=(SELECT * FROM table WHERE id=$mainid);
$result2=(SELECT * FROM table WHERE id=$compareid);

现在您可以使用value

<?php echo Hello $result1->name your health seems better than $result2->name.

---

但是我已经选择了db到数组中，现在我需要回显像name这样的值，其中id=1
$mainid = 1; //(you can get this value dynamically)
$compareid=2; //(you can get this value dynamically)
$result1=(SELECT * FROM table WHERE id=$mainid);
$result2=(SELECT * FROM table WHERE id=$compareid);

<?php echo Hello $result1->name your health seems better than $result2->name.