我正在处理一项任务,该任务要求我创建一个将信息发送到数据库的表单。我在连接到服务器时遇到问题。
我尝试过使用mysql_connect和mysqli_connect,但两者都不起作用。
mysql_connect错误:警告: mysql_connect():连接超时
mysqli_connect错误:致命错误:调用未定义的函数mysqli_connect()
$connect=mysql_connect("server.com","username","password","database")
or die("Failed to connect to server");
mysql_query("INSERT INTO client (title,fname,lname,organization,email,phone,tshirt) VALUES ($title,$fname,$lname,$org,$email,$phone,$tshirt)");
$tabledata = mysql_query("SELECT * FROM client");
echo "<table>
<tr>
<th>Title</th>
<th>First Name</th>
<th>Last Name</th>
<th>Organization</th>
<th>Email</th>
<th>Phone</th>
<th>TShirt</th>
</tr>";
while($row = mysql_fetch_array($tabledata)){
echo "<tr>";
echo "<td>" . $row['title'] . "</td>";
echo "<td>" . $row['fname'] . "</td>";
echo "<td>" . $row['lname'] . "</td>";
echo "<td>" . $row['org'] . "</td>";
echo "<td>" . $row['email'] . "</td>";
echo "<td>" . $row['phone'] . "</td>";
echo "<td>" . $row['tshirt'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysql_close($connect);信息来自PHP信息
mysql
MySQL支持已启用活动持久链接0活动链接0客户端API版本mysqlnd 5.0.10 - 20111026 - $Id: e707c415db32080b3752b232487a435ee0372157 $
PDO
启用PDO支持的PDO驱动程序sqlite
pdo_sqlite
支持SQLite 3.x的SQLite库3.7.7.1的PDO驱动程序
sqlite3
启用SQLite3支持的SQLite3模块版本0.7 SQLite库3.7.7.1
指令本地值主值sqlite3.extension_dir无值无值
发布于 2013-11-19 23:18:54
首先,确保mysql正在运行,如果已经在运行,那么无论如何我都会尝试为您的php环境安装mysql扩展,只是为了确保:
sudo apt-get install php5-mysql libapache2-mod-auth-mysql发布于 2013-11-19 23:18:46
请使用或,但在您的问题中
尝试:
$connect=mysql_connect("server.com","username","password");
mysql_select_db("database");发布于 2013-11-19 23:20:43
首先,您是在本地还是从外部连接到数据库?如果您在本地连接,则mysql_connect();的参数将如下所示:mysql_connect('localhost', 'username', 'password');
然后:
mysql_select_db('your_db') https://stackoverflow.com/questions/20075193
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