我目前有如下表中存储的信息：
+----+------+-------+-----------+-----------+
| id | name | state | xcoord | ycoord |
+----+------+-------+-----------+-----------+
| 1 | lake | CA | 36.746585 | 22.234564 |
| 2 | pond | TX | 26.123123 | 12.456789 |
+----+------+-------+-----------+-----------+
我的网页上有一个HTML表格，显示：
<html>
<head>
<title>Locations</title>
</head>
<body>
<table border=1>
<tr>
<th>ID</th>
<th>Name</th>
<th>State</th>
<th>X Coord</th>
<th>Y Coord</th>
</tr>
</table>
</body>
</html>我想在我的网页上传播我的数据库中存储的值的表。目前，当我尝试以编程方式这样做时，结果是：
"; ?>
ID Name State X Coord Y Coord
{$row['id'] {$row["name"] {$row["state"]} {$row["xcoord"]} {$row["ycoord"]}
目前这些是我正在使用的文件：
db.inc.php
<?php
/*
* db.inc.php
* These are the DBMS credentials and the database name
*/
$hostName = "xxxx";
$databaseName = "yyyy";
$username = "zzzz";
$password = "wwww";
// Show an error and stop the script
function showerror()
{
if (mysql_error())
die("Error " . mysql_errno() . " : " . mysql_error());
else
die ("Could not connect to the DBMS");
}
?>
location.html
<html>
<head>
<title>Locations</title>
</head>
<body>
<table border=1>
<tr>
<th>ID</th>
<th>Name</th>
<th>State</th>
<th>X Coord</th>
<th>Y Coord</th>
</tr>
<?php
include 'db.inc.php';
// Connect to MySQL DBMS
if (!($connection = @ mysql_connect($hostName, $username, $password)))
showerror();
// Use the location database
if (!mysql_select_db($databaseName, $connection))
showerror();
// Create SQL statement
$query = "SELECT * FROM locations"; // the table name is "locations"
// Execute SQL statement
if (!($result = @ mysql_query ($query, $connection)))
showerror();
// Display results *** My inclination is that something is awry with the following echo
while ($row = @ mysql_fetch_array($result))
echo "<tr>
<td>{$row["id"]}</td>
<td>{$row["name"]}</td>
<td>{$row["state"]}</td>
<td>{$row["xcoord"]}</td>
<td>{$row["ycoord"]}</td>
</tr>";
?>
</table>
</body>
</html>
如何正确显示HTML表中的信息？
如果你的文件中有php代码，为了执行-文件扩展名必须是'.php'，即：location.php而不是location.html
mysql_select_db接受单个参数。它是数据库名称。
<?php
if (!mysql_select_db($databaseName))
showerror();
?>