PHP SQL 从哪里选择
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PHP SQL Select From Where
提问by Christopher
I am having some difficulty running some SQL code.
我在运行一些 SQL 代码时遇到了一些困难。
What I am trying to do is, find a row that contains the correct username, and then get a value from that correct row.
我想要做的是，找到包含正确用户名的行，然后从该正确行中获取值。
This is my SQL in the php:
这是我在 php 中的 SQL：
mysql_query("SELECT * FROM users WHERE joined='$username' GET name")
As you can see, it looks for a username in users and then once found, it must GET a value from the correct row.
如您所见，它在 users 中查找用户名，然后一旦找到，它必须从正确的行中获取一个值。
How do I do that?
我怎么做？
回答by mimicocotopus
You need some additional PHP code (a call to mysql_fetch_array) to process the result resource returned by MySQL.
您需要一些额外的 PHP 代码（调用 mysql_fetch_array）来处理 MySQL 返回的结果资源。
$result = mysql_query("SELECT name FROM users WHERE joined='$username'");
$row = mysql_fetch_array($result);
echo $row['name'];
回答by Fluffeh
mysql_query("SELECT `name` FROM users WHERE joined='$username' ")
Just select the right column in your 'select clause' like above.
只需像上面一样在“选择子句”中选择正确的列。
Edit: If you are just starting out though, you might want to follow a tutorial like this onewhich should take you through a nice step by step (and more importantly up to date functions) that will get you started.
编辑：如果你是刚刚起步，不过，你可能要遵循这样的教程一个应该带你通过一个漂亮的一步一步（更重要的是最新的功能），这将让你开始。
回答by IEnumerable
mysql_query("SELECT name FROM users WHERE joined='$username'")
回答by asprin
$q = mysql_query("SELECT * FROM users WHERE joined='$username'");
$r = mysql_fetch_array($q);
$name = $r['user_name']; // replace user_name with the column name of your table
回答by elo
mysql_query("SELECT name FROM users WHERE joined='$username' ")
Read documentation : http://dev.mysql.com/doc/refman/5.0/en/select.html
