为什么我的PHP没有显示任何东西? (代码包括在内)

时间:2014-02-12 05:22:36

标签: php mysql ubuntu

这段代码应该给出表猫的表格。但是,它显示一个空白页面。似乎查询部分不起作用。它无法转到“var_dump($results)”。但是,我可以使用已注释的代码创建表。那么问题是什么?

<?php
require_once "login.php";
$db_server = mysql_connect($db_hostname, $db_username, $db_password);
if(!$db_server)
    die("Unable to connect: ".mysql_error());
mysql_select_db($db_database, $db_server) or die("Unable to select a db: ".mysql_error());
/*
$query = "CREATE TABLE cats
(
    id SMALLINT NOT NULL AUTO_INCREMENT,
    family VARCHAR(32) NOT NULL,
    name VARCHAR(32) NOT NULL,
    age TINYINT NOT NULL,
    PRIMARY KEY (id)
)";

mysql_query($query) or die("Create cats DB failed: ".mysql_error());
*/

$query = "DESCRIBE cats";
$results = mysql_query($query)

var_dump($results);

if(!$results)
    die("Error-line22: ".mysql_error());
else 
    echo $results."<br />";

$rows = mysql_num_rows($results);

echo "<table>";
echo "
<tr> <th>Column</th> <th>Type</th> <th>NULL</th> <th>Key</th> </tr>";

for ($j = 0; $j < $rows; ++$j)
{
    $row = mysql_fetch_row($results);
    echo "<tr>";
    for ($k = 0; $k < 4; ++$k)
        echo "<td>$row[$k]</td>";
    echo "</tr>";
}

echo "</table>";

mysql_close($db_server);
?>

1 个答案:

答案 0 :(得分:2)

mysql_query()行缺少';'最后:

$query = "DESCRIBE cats";
$results = mysql_query($query);

var_dump($results);