blocks|key|2996106|text|您需要在MYSQL语句中将变量加引号，如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2996107|$query+=+"SELECT+*+FROM+person+WHERE+email='$email'";|code-block|syntax|javascript|2996108|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|$]]

you need to enquote your variable in your MYSQL statement like this:

<pre><code>$query = "SELECT * FROM person WHERE email='$email'";
</code></pre>

blocks|key|2996278|text|假设您已经正确地定义了$dbc+...|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2996279|首先，您的查询字符串是错误的，您需要用引号将$email括起来|2996280|$query+=+"SELECT+*+FROM+person+WHERE+email='".+$email+.+"'";|code-block|syntax|javascript|2996281|其次，您的SQL是可注入的，您需要转义post字符串：|2996282|$email+=+mysql_real_escape_string($_POST['email']);+|2996283|第三，请不要使用mysql库，因为它们已被弃用，请搜索mysqli并阅读有关它的信息。|2996284|entityMap^0|B|4|0|M|6|0|0|0|0|8|5|R|6|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|X|8|@$9|Y|A|Z|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|10|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|11|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|12|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|13|8|@$9|14|A|15|B|C]|$9|16|A|17|B|C]]|D|@]|E|$]]|$1|S|3|-4|5|6|7|18|8|@]|D|@]|E|$]]]|T|$]]

Assuming that you've defined <code>$dbc</code> correctly...

First of all your query string is wrong, you need quotes around <code>$email</code>:

<pre><code>$query = "SELECT * FROM person WHERE email='". $email . "'";
</code></pre>

Second, your SQL is injectable, you need to escape the post string:

<pre><code>$email = mysql_real_escape_string($_POST['email']); 
</code></pre>

Third, please don't use <code>mysql</code> libs as they're deprecated, search <code>mysqli</code> and read about it.

blocks|key|872007|text|我重写了你的函数，|type|unstyled|depth|inlineStyleRanges|entityRanges|data|872008|public+static+function+find_by_email($email)+{
++++++++global+$dbc;
++++++++$query+=+"SELECT+*+FROM+person+WHERE+email='{$email}'";
++++++++$result+=+mysql_query($query,+$dbc)
++++++++$num_rows+=+mysql_num_rows($result);
++++++++if+($num_rows)+{
++++++++++++if($row+=+mysql_fetch_array($r))+{
++++++++++++++++$person+=+new+Person($row);
++++++++++++++++return+$person;
++++++++++++}
++++++++}else{
++++++++$error+=+"Could+not+find+user";
++++++++++++++++echo+$error;
++++++++return+nil;
++++++++}
}|code-block|syntax|javascript|872009|希望这能有所帮助|872010|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

I rewrote your function , 

<pre><code>public static function find_by_email($email) {
 global $dbc;
 $query = "SELECT * FROM person WHERE email='{$email}'";
 $result = mysql_query($query, $dbc)
 $num_rows = mysql_num_rows($result);
 if ($num_rows) {
 if($row = mysql_fetch_array($r)) {
 $person = new Person($row);
 return $person;
 }
 }else{
 $error = "Could not find user";
 echo $error;
 return nil;
 }
}
</code></pre>

hope this helps

blocks|key|117560|text|。。如果您不习惯检查mysql_error()来查看错误消息，那么可以输出查询并在可视化工具中运行它(比如HeidiSQL--+run+on+Wine+on+Linux，顺便说一下)。只要做..。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|117561|die("SELECT+*+FROM+person+WHERE+email=".+$email);|code-block|syntax|javascript|117562|。。复制输出并运行它。它将向您显示错误消息。|117563|。。现在，在查询中插入该值时似乎没有引用该值，如下所示：|117564|SELECT+*+FROM+person+WHERE+email=foo@bar.com|117565|。。这是无效的语法。你需要用引号把它括起来。虽然你可以简单地通过转义和引用$email变量的值来修复这个错误，但你应该认真考虑在最新的PDO版本中使用它的现代功能，比如准备好的语句，这样你就不需要担心SQL+Injection或损坏的转义函数。|offset|length|style|CODE|117566|。。Amplexos。|117567|entityMap|0|LINK|mutability|MUTABLE|url|http://php.net/manual/en/book.pdo.php|1|http://php.net/manual/en/security.database.sql-injection.php^0|0|0|0|0|0|11|6|1V|3|0|2S|D|1|0|0^^$0|@$1|2|3|4|5|6|7|14|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|15|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|16|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|17|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|18|8|@]|9|@]|A|$E|F]]|$1|M|3|N|5|6|7|19|8|@$O|1A|P|1B|Q|R]]|9|@$O|1C|P|1D|1|1E]|$O|1F|P|1G|1|1H]]|A|$]]|$1|S|3|T|5|6|7|1I|8|@]|9|@]|A|$]]|$1|U|3|-4|5|6|7|1J|8|@]|9|@]|A|$]]]|V|$W|$5|X|Y|Z|A|$10|11]]|12|$5|X|Y|Z|A|$10|13]]]]

. . If you don't feel comfortable checking the mysql_error() to see the error message, you can output the query and run it in a visual tool (like HeidiSQL--runs on Wine on Linux as well, btw). Just do...

<pre><code>die("SELECT * FROM person WHERE email=". $email);
</code></pre>

. . Copy the output and run it. It will give you the error message.

. . Right now it looks like you're not quoting the value when inserting it in the query, like:

<pre><code>SELECT * FROM person WHERE email=foo@bar.com
</code></pre>

. . It's invalid syntax. You need to enclose it in quotes. Altough you can fix the error simply by escaping and quoting the value of <code>$email</code> variable, you should seriously consider using <a href="http://php.net/manual/en/book.pdo.php" rel="nofollow">PDO</a> in most recent PHP versions for its modern features, such as prepared statements, so that you don't need to worry about <a href="http://php.net/manual/en/security.database.sql-injection.php" rel="nofollow">SQL Injection</a> or broken escaping functions.

. . Amplexos.

blocks|key|1262940|text|让我们假设$dbc已经打开了到PDO的连接|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1262941|public+static+function+find_by_email($email)+{
++++global+$dbc;
++++$query+=+"SELECT+*+FROM+person+WHERE+email=?";
++++$stm+=+$dbc->prepare($query);
++++$stm->execute(array($email));
++++if+($row+=+$stm->fetch())+{
++++++++return+new+Person($row);
++++}
}|code-block|syntax|javascript|1262942|请注意，您不应该直接从函数回显任何内容。|1262943|检查返回值，然后决定是否回显任何内容。|1262944|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/tags/pdo/info^0|F|3|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|C|$]]|$1|K|3|L|5|6|7|10|8|@]|9|@]|C|$]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

Lets assume that $dbc is already opened connection to <a href="https://stackoverflow.com/tags/pdo/info">PDO</a>

<pre><code>public static function find_by_email($email) {
 global $dbc;
 $query = "SELECT * FROM person WHERE email=?";
 $stm = $dbc-&gt;prepare($query);
 $stm-&gt;execute(array($email));
 if ($row = $stm-&gt;fetch()) {
 return new Person($row);
 }
}
</code></pre>

note that you should never echo anything directly from the function. 
check the returned value and then decide if you going to echo anything.

So im working on this php program which has a <code>Person</code> Class. There is a method inside that class called <code>find_by_email($email)</code>. I try to find a user by email, using an email address that I know for certain is in my db. After querying the db, it should return a person object. 

I've already checked to make sure my table names and columns are correct. After testing it appears that the code does not pass the first if statement: <code>if ($r = mysql_query($query, $dbc))</code>. The only thing I can think of is that my query must be incorrect. Is this not the right syntax?

here's how I call the method:

<pre><code>$email = $_POST['email']; 
$person = Person::find_by_email($_POST['email']);
</code></pre>

here is my <code>Person::find_by_email()</code> method:

<pre><code> public static function find_by_email($email) {
 global $dbc;
 $query = "SELECT * FROM person WHERE email=". $email;
 if ($r = mysql_query($query, $dbc)) {
 if($row = mysql_fetch_array($r)) {
 $person = new Person($row);
 return $person;
 }
 }else{
 $error = "Could not find user";
 echo $error;
 return nil;
 }
}
</code></pre>

Stuck on a MySQL query

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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所以我正在做这个php程序，它有一个Person类。该类内部有一个名为find_by_email($email)的方法。我尝试通过电子邮件查找用户，使用的电子邮件地址肯定在我的数据库中。查询db之后，它应该返回一个person对象。我已经检查过以确保我的表名和列是正确的。经过测试，代码似乎没有通过第一个if语句：if ...

问在MySQL查询上卡住
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