mysql_query()期望参数2为资源，布尔值 [英] mysql_query() expects parameter 2 to be resource, boolean

问题描述

我正在尝试从数据库中获取数据，但是遇到此错误.查询中没有错误，因为它可以在mysql中完美运行.

I am trying to get data from the database but experiencing this error. There is no error in the query as it runs perfectly in mysql.

那我为什么会收到这个错误?

Why i am getting this error then ?

[client 127.0.0.1] PHP Warning:  mysql_query() expects parameter 2 to be resource, boolean given in /var/www/imdb.php on line 35

[client 127.0.0.1] PHP Warning:  mysql_error() expects parameter 1 to be resource, boolean given in /var/www/imdb.php on line 35  

php代码:

php code:

    function getdirector($director)
    {
    global $db;
    $query = 'select name from peopledb where peopleid = '.$director;
    $result = mysql_query($query,$db) or die(mysql_error($db));
    $row = mysql_fetch_assoc($result);
    extract($row);
    return $name;
    }

    function getactor($actor)
    {
    $query = 'select name from peopledb where peopleid = '.$actor;
    $result = mysql_query($query,$db) or die(mysql_error($db));
    $row = mysql_fetch_assoc($result);
    extract($row);
    return $name;
    }


    function getmovietype($type)
    {
    $query = 'select movietype from movietype where movieid = '.$type;
    $result = mysql_query($query,$db) or die(mysql_error($db));
    $row = mysql_fetch_assoc($result);
    extract($row);
    return $movietype;
    }

    $db = mysql_connect('localhost','root','saw123') or die("Connection error");
    $db = mysql_select_db("moviesite",$db) or die(myql_error($db));
    $query = 'select moviename,releaseyear,director,actor,type from movie';
    $result = mysql_query($query,$db) or die(mysql_error($db));

    $table = <<<ENDHTML
    <div style="text-align: center;" >
    <h2>The Ultimate movie database</h2>
    <table border='1' cellpadding='1' cellspacing='2'
           style="width: 70%; margin-left: auto; margin-right: auto;">

    <tr>
    <th>Movie title</th>
    <th>year of release </th>
    <th>Movie director</th>
    <th>Movie Actor</th>
    <th>Movie type</th>
    </tr>
    ENDHTML;

    while ($row = mysql_fetch_assoc($result))
    {
    extract($row);
    $director1 = getdirector($director);
    $actor1 = getactor($actor);
    $movietype1 = getmovietype($type);

    $table .= <<<ENDHTML
    <tr>
    <td>$moviename</td>
    <td>$releaseyear</td>
    <td>$director1</td>
    <td>$actor1</td>
    <td>$movietype1</td>
    </tr>
    ENDHTML;
    }
    $table .=<<<ENDHTML
    </table>
    </div>
    ENDHTML;

    echo $table;


    ?>

* 更新:::无论我在哪里使用mysql_error()函数，它都显示错误.现在，删除所有"$ db"参数后，它就可以工作了.为什么?*

推荐答案

只需这样命名:

$result = mysql_query($query);

如果您查看mysql_connect函数，它会在失败时返回false(布尔值):

If you look at the mysql_connect function it returns false ( a boolean) on failure:

http://php.net/manual/en/function.mysql- connect.php

因此您的连接可能甚至没有建立.

So your connection is likely not even established.

查看此函数: http://php.net/manual/en/function.mysql-select-db.php

您正在呼叫:

$db = mysql_connect('localhost','root','saw123') or die("Connection error");
    $db = mysql_select_db("moviesite",$db) or die(myql_error($db));

第二行将$ db设置为true或false-因为mysql_select_db仅返回布尔值.

The second line sets $db to either true or false - since mysql_select_db only ever returns a boolean.

您可能的意思是:

$db = mysql_connect('localhost','root','saw123') or die("Connection error");
mysql_select_db("moviesite",$db) or die(myql_error($db));

在这种情况下，

$ db不能正确命名；它会更准确地称为$ connection或$ db_connection或类似的东西.

$db in this case would not be appropriately named; it'd be more accurately called $connection or $db_connection or something like that.

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