我正在尝试从我的MySQL数据库中显示我的表。它似乎不起作用,我想这很可能是一件简单的事情。如果可能的话,我希望桌子看起来像一张没有多余装饰的普通桌子;只是一个边框,并能够适合正常大小的页面。这就是我到目前为止所做的:
<html>
<head>
<title>Profile Database </title>
</head>
<body>
<?
$connection = "connect.php";
require $connection;
mysql_select_db("bank");
$sql = "SELECT * FROM profiles";
$result = mysql_query($sql);
?>
<table border="2" style= "background-color: #84ed86; color: #761a9b; margin: 0 auto;" >
<thead>
<tr>
<th>ID</th>
<th>First Name</th>
<th>Last Name</th>
<th>E-Mail</th>
<th>DOB</th>
<th>Age</th>
<th>City</th>
<th>State</th>
<th>Zip Code</th>
<th>Offence</th>
<th>Notes</th>
</tr>
</thead>
<tbody>
<?php
while( $row = mysql_fetch_assoc( $result ))
{
echo <tr>
<td>{$row\['id'\]}</td>
<td>{$row\['fname'\]}</td>
<td>{$row\['lname'\]}</td>
<td>{$row\['email'\]}</td>
<td>{$row\['dob'\]}</td>
<td>{$row\['age'\]}</td>
<td>{$row\['city'\]}</td>
<td>{$row\['state'\]}</td>
<td>{$row\['zip'\]}</td>
<td>{$row\['offence'\]}</td>
<td>{$row\['notes'\]}</td>
</tr>\n;
}
?>
</tbody>
</table>
<?php
mysql_close($connection);
?>
</body>
</html>
答案 0 :(得分:0)
放手一搏:
<table>
<tr>
<th>ID</th>
<th>First Name</th>
<th>Last Name</th>
<th>E-Mail</th>
<th>DOB</th>
<th>Age</th>
<th>City</th>
<th>State</th>
<th>Zip Code</th>
<th>Offence</th>
<th>Notes</th>
</tr>
<?php
require "connect.php";
mysql_select_db("bank");
$result = mysql_query("SELECT * FROM profiles");
while($row = mysql_fetch_array($result, MYSQL_NUM)) {
echo '<tr>';
foreach($row as $column) {
echo '<td>', $column, '</td>';
}
echo '</tr>';
}
?>
</table>
答案 1 :(得分:0)
WHILE($row = mysql_fetch_array($result))
应该是
WHILE($row = mysql_fetch_array($result, MYSQL_ASSOC))
OR
WHILE($row = mysql_fetch_assoc($result))
此外,不推荐使用mysql_ *函数,并且非常不鼓励使用它们。
请检查以下讨论: -
Deprecated MySql Functions
How to successfully rewrite old mysql-php code with deprecated mysql_* functions?
http://php.net/manual/en/migration55.deprecated.php
使用MySQLi
或PDO
。
答案 2 :(得分:0)
<center><table border="1">
<tr>
<th>ID</th>
<th>First Name</th>
<th>Last Name</th>
<th>E-Mail</th>
<th>DOB</th>
<th>Age</th>
<th>City</th>
<th>State</th>
<th>Zip Code</th>
<th>Offence</th>
<th>Notes</th>
</tr>
<?
$connection = "connect.php";
require $connection;
mysql_select_db("bank");
$sql = "SELECT * FROM profiles";
$result = mysql_query($sql);
WHILE($row = mysql_fetch_array($result))
{
$id = $row ['id'];
$firstname = $row ['fname'];
$lastname = $row ['lname'];
$email = $row ['email'];
$dob = $row ['dob'];
$age = $row ['age'];
$city = $row ['city'];
$state = $row ['state'];
$zip = $row ['zip'];
$offence = $row ['offence'];
$nots = $row ['notes'];
echo '<tr>'; // change here
echo '<td>'.$id.'</td>';
echo '<td>'.$firstname.'</td>';
echo '<td>'.$lastname.'</td>';
echo '<td>'.$email.'</td>';
echo '<td>'.$dob.'</td>';
echo '<td>'.$age.'</td>';
echo '<td>'.$city.'</td>';
echo '<td>'.$state.'</td>';
echo '<td>'.$zip.'</td>';
echo '<td>'.$offence.'</td>';
echo '<td>'.$notes.'</td>';
echo '</tr>'; // change here
}
mysql_close();
?>
</table></center>
答案 3 :(得分:0)
试试这段代码:
<?php
$username = "your_name";
$password = "your_password";
$hostname = "localhost";
$dbconnect = mysql_connect($hostname, $username, $password) or die("Unable to connect to MySQL");
$select=mysql_select_db("bank",$dbconnect) or die("Could not select bank");
$sql = "SELECT * FROM profiles";
$result = mysql_query($sql);
?>
<center><table border="1">
<tr>
<th>ID</th>
<th>First Name</th>
<th>Last Name</th>
<th>E-Mail</th>
<th>DOB</th>
<th>Age</th>
<th>City</th>
<th>State</th>
<th>Zip Code</th>
<th>Offence</th>
<th>Notes</th></tr>
<?php
while($row = mysql_fetch_array($result))
{
$id = $row ['id'];
$firstname = $row ['fname'];
$lastname = $row ['lname'];
$email = $row ['email'];
$dob = $row ['dob'];
$age = $row ['age'];
$city = $row ['city'];
$state = $row ['state'];
$zip = $row ['zip'];
$offence = $row ['offence'];
$nots = $row ['notes'];
echo '<tr>';
echo '<td>'.$id.'</td>';
echo '<td>'.$firstname.'</td>';
echo '<td>'.$lastname.'</td>';
echo '<td>'.$email.'</td>';
echo '<td>'.$dob.'</td>';
echo '<td>'.$age.'</td>';
echo '<td>'.$city.'</td>';
echo '<td>'.$state.'</td>';
echo '<td>'.$zip.'</td>';
echo '<td>'.$offence.'</td>';
echo '<td>'.$notes.'</td>';
echo '</tr>';
}
?>
</table></center>
<?php
mysql_close();
?>
答案 4 :(得分:0)
更正代码:
<html>
<head>
<title>Profile Database </title>
</head>
<body>
<?php
$connection = "connect.php";
require $connection;
mysql_select_db("bank");
$sql = "SELECT * FROM profiles";
$result = mysql_query($sql);
?>
<table border="2" style= "background-color: #84ed86; color: #761a9b; margin: 0 auto;" >
<thead>
<tr>
<th>ID</th>
<th>First Name</th>
<th>Last Name</th>
<th>E-Mail</th>
<th>DOB</th>
<th>Age</th>
<th>City</th>
<th>State</th>
<th>Zip Code</th>
<th>Offence</th>
<th>Notes</th>
</tr>
</thead>
<tbody>
<?php
while( $row = mysql_fetch_assoc( $result ))
{
?>
<tr>
<td><?php echo $row['id'];?></td>
<td><?php echo $row['fname'];?></td>
<td><?php echo $row['lname'];?></td>
<td><?php echo $row['email'];?></td>
<td><?php echo $row['dob'];?></td>
<td><?php echo $row['age'];?></td>
<td><?php echo $row['city'];?></td>
<td><?php echo $row['state'];?></td>
<td><?php echo $row['zip'];?></td>
<td><?php echo $row['offence'];?></td>
<td><?php echo $row['notes'];?></td>
</tr>
<?php
}
?>
</tbody>
</table>
<?php
mysql_close($connection);
?>
</body>
</html>