blocks|key|2055445|text|试试这个：：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2055446|Select+
u.fullname,u.id
from+
`users`+u
INNER+JOIN+msgs+m+ON+(u.id=m.receiver)
WHERE+email+=+'$email'|code-block|syntax|javascript|2055447|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Try this::

<pre><code>Select 
u.fullname,u.id
from 
`users` u
INNER JOIN msgs m ON (u.id=m.receiver)
WHERE email = '$email'
</code></pre>

blocks|key|4372046|text|您应该只检索所需的行，而不是所有行(如果数据库中有1.000.000条消息怎么办?)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4372047|这可以通过使用"JOIN“的一个查询来完成：http://dev.mysql.com/doc/refman/5.0/en/join.html|offset|length|4372048|entityMap|0|LINK|mutability|MUTABLE|url|http://dev.mysql.com/doc/refman/5.0/en/join.html^0|0|M|1C|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|O|8|@]|9|@$D|P|E|Q|1|R]]|A|$]]|$1|F|3|-4|5|6|7|S|8|@]|9|@]|A|$]]]|G|$H|$5|I|J|K|A|$L|M]]]]

You should only retrieve the lines you want, not all the lines (what if you have 1.000.000 messages in your database ?).

This can be done with only one query using "JOIN" : <a href="http://dev.mysql.com/doc/refman/5.0/en/join.html" rel="nofollow">http://dev.mysql.com/doc/refman/5.0/en/join.html</a>

blocks|key|2055546|text|使用join子句|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|2055547|SELECT+m.*,+u.email,+u.name+FROM+`users`+AS+u
JOIN+`msgs`+AS+m+ON+m.receiver+=+u.id
WHERE+u.email+=+'$email'|code-block|syntax|javascript|2055548|这为您提供了例如|2055549|%7C+id+%7C+sender+%7C+receiver+%7C+++link+++%7C++++++email+++++++%7C++++++name+++++++%7C++
%2B----%2B--------%2B----------%2B----------%2B------------------%2B-----------------%2B
%7C++1+%7C++++1+++%7C+++++2++++%7C+++adfd+++%7C+receiver's+email+%7C+receiver's+name+%7C|2055550|您可以将结果提取到关联数组并使用命名索引，如下所示：|2055551|$row['link']
$row['name']|2055552|或者普通的索引数组(正确的索引可以通过打印整个数组来找到)|2055553|print_r($row)

$row[3]+//+should+be+link
$row[5]+//+should+be+name|2055554|entityMap|0|LINK|mutability|MUTABLE|url|http://dev.mysql.com/doc/refman/5.0/en/join.html^0|2|4|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@$A|13|B|14|1|15]]|C|$]]|$1|D|3|E|5|F|7|16|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|17|8|@]|9|@]|C|$]]|$1|K|3|L|5|F|7|18|8|@]|9|@]|C|$G|H]]|$1|M|3|N|5|6|7|19|8|@]|9|@]|C|$]]|$1|O|3|P|5|F|7|1A|8|@]|9|@]|C|$G|H]]|$1|Q|3|R|5|6|7|1B|8|@]|9|@]|C|$]]|$1|S|3|T|5|F|7|1C|8|@]|9|@]|C|$G|H]]|$1|U|3|-4|5|6|7|1D|8|@]|9|@]|C|$]]]|V|$W|$5|X|Y|Z|C|$10|11]]]]

use <a href="http://dev.mysql.com/doc/refman/5.0/en/join.html" rel="nofollow">join</a> clausule

<pre><code>SELECT m.*, u.email, u.name FROM `users` AS u
JOIN `msgs` AS m ON m.receiver = u.id
WHERE u.email = '$email'
</code></pre>

this gives you e.g.

<pre><code>| id | sender | receiver | link | email | name | 
+----+--------+----------+----------+------------------+-----------------+
| 1 | 1 | 2 | adfd | receiver's email | receiver's name |
</code></pre>

you can fetch result to associative array and use named indexes like:

<pre><code>$row['link']
$row['name']
</code></pre>

or to normal indexed array (correct indexes you can find out by printing whole array)

<pre><code>print_r($row)

$row[3] // should be link
$row[5] // should be name
</code></pre>

blocks|key|2055584|text|在你的例子中，循环的代价很高，而且没有什么用处。使用LEFT+JOIN+(如果任何列为NULL，则JOIN或INNER+JOIN将不返回任何内容)；|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2055585|SELECT+
++u.*,
++m.sender,+m.receiver,+m.link
FROM+users+u
LEFT+JOIN+msgs+m+ON+(m.receiver+=+u.id)
WHERE+email+=+'$email'|code-block|syntax|javascript|2055586|entityMap^0|Q|9|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Loop in your case is much costly and not useful. Use <code>LEFT JOIN</code> (JOIN or INNER JOIN will return nothing if any column is NULL);

<pre><code>SELECT 
 u.*,
 m.sender, m.receiver, m.link
FROM users u
LEFT JOIN msgs m ON (m.receiver = u.id)
WHERE email = '$email'
</code></pre>

I have 2 tables users and msgs : 
<code>users</code> contain <code>id</code>,<code>name</code>,<code>email</code>,<code>password</code> 
and <code>msgs</code> contains <code>id</code>, <code>sender</code>, <code>receiver</code>, <code>link</code> 
i want to select all the rows where receiver is equal to the same id of the user how can i loop it to get them all ?

i try doing something but nevermind

<pre><code>$name_query = mysql_query("SELECT fullname,id FROM users WHERE email = '$email'");
$name = mysql_fetch_object($name_query);

$user_id = $name-&gt;id;

$result = mysql_query("SELECT receiver FROM mytable");

while ($row = mysql_fetch_array($result, MYSQL_NUM)) {
 printf("ID: %s Name: %s", $row[0], $row[1]); 
}

mysql_free_result($result);

?&gt;
</code></pre>

looping through databse in my sql to get all results and displaying them

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我有两个表users和msgs：users包含id、name、email、passwordmsgs包含id、sender、receiver、link我想选择receiver等于用户的相同id的所有行，如何循环才能获得所有行？我试着做一些事，但不要紧$name_query = mysql_query("SELECT fu...

问在我的sql中遍历数据库以获取所有结果并显示它们
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问在我的sql中遍历数据库以获取所有结果并显示它们EN