linux学习之——数据操作:添加与查询

时间:2021-11-04 06:55:16

说明:

在linux系统中,利用搭建的服务器,编写两个页面,一个添加信息,一个展现信息;

主要涉及到:php+mysql的操作;

数据添加页面:

<html>
<head>
<meta http-equiv='content-type' content='text/html'; charset='utf-8'>
<title> Test Engineer info table add</title>
</head> <body>
<div align='center'>
<form action='adduser.php' method='post'> <h3> Test Engineer_info Table</h3>
<table width='500' bgcolor='#ccffcc' style='border-color' border='1'>
<tr align=center><td>name</td><td><input type="text" name="tname"/></td></tr>
<tr align=center><td>age</td><td><input type="text" name="tage"/></td></tr>
<tr align=center><td>hobby</td><td><input type="text" name="thobby"/></td></tr>
<tr align=center> <td>job</td><td><input type="text" name="tjob"/></td></tr>
<tr align=center><td colspan='2'><input type="submit" value="add info"></td></tr>
</table>
</form>
</div>
</body>
</html>

数据操作:

<?php

 $mysql_server_name= 'localhost';
$mysql_username='mysqlname';
$mysql_password='mysqlpassword';
$mysql_database='mysql-database'; $tname=$_POST[tname];
$tage=$_POST[tage];
$thobby=$_POST[thobby];
$tjob=$_POST[tjob]; $link = mysql_connect($mysql_server_name,$mysql_username,$mysql_password);
mysql_query("set names'utf8'"); if (!$link)
die('could not connect:' .mysql_error());
else {
/*echo "LAMP is success!MySql is right!Mysql_database is right";*/ mysql_select_db( $mysql_database,$link);
if(!empty($tname)){
$sql_insert="insert into clt_testerinfo values('$tname','$tage','$thobby','$tjob')";
mysql_query($sql_insert);
} $sql_select="select * from clt_testerinfo";
$result=mysql_query($sql_select,$link);
echo "Test Enginner info Table";
echo '</br>';
echo "<table width='800' bgcolor='#ccffcc' style='border-color' border='1'>";
echo "<tr><td>name</td>";
echo "<td>age</td>";
echo "<td>hobby</td>";
echo "<td>job</td></tr>"; while($row = mysql_fetch_array($result))
{
echo "<tr><td>";
echo $row[tname];
echo "</td><td>";
echo $row[tage];
echo "</td><td>";
echo $row[thobby];
echo "</td><td>";
echo $row[tjob];
echo "</td></tr>"; }
echo "</table>"; }
mysql_close($link);
?>