编辑:这就是我想要实现的目标:http://i.imgur.com/KE9xx.png
我试图在两列中显示数据库中的结果。我对PHP有点新意,所以我对如何做到这一点没有丝毫的线索。任何人都可以帮我吗?提前谢谢。
这是我目前的代码:
include('connect.db.php');
// get the records from the database
if ($result = $mysqli->query("SELECT * FROM todo ORDER BY id"))
{
// display records if there are records to display
if ($result->num_rows > 0)
{
// display records in a table
echo "<table width='415' cellpadding='0' cellspacing='0'>";
// set table headers
echo "<tr><td><img src='media/title_projectname.png' alt='Project Name' /></td>
<td><img src='media/title_status.png' alt='Status'/></td>
</tr>";
echo "<tr>
<td><div class='tpush'></div></td>
<td> </td>
</tr>"
while ($row = $result->fetch_object())
{
echo "<tr>";
echo "<td><a href='records.php?id=" . $row->id . "'>" . $row->item . "</a></td>";
echo "<td>" . $row->priority . "</td>";
echo "</tr>";
}
echo "</table>";
}
// if there are no records in the database, display an alert message
else
{
echo "No results to display!";
}
}
// show an error if there is an issue with the database query
else
{
echo "Error: " . $mysqli->error;
}
// close database connection
$mysqli->close();
答案 0 :(得分:1)
一个好主意是将数据存储到一个简单的数组中,然后将它们显示在一个2-columned表中,如下所示:
$con = mysql_connect('$myhost', '$myusername', '$mypassword') or die('Error: ' . mysql_error());
mysql_select_db("mydatabase", $con);
mysql_query("SET NAMES 'utf8'", $con);
$q = "Your MySQL query goes here...";
$query = mysql_query($q) or die("Error: " . mysql_error());
$rows = array();
$i=0;
// Put results in an array
while($r = mysql_fetch_assoc($query)) {
$rows[] = $r;
$i++;
}
//display results in a table of 2 columns
echo "<table>";
for ($j=0; $j<$i; $j=$j+2)
{
echo "<tr>";
echo "<td>".$row[$j]."</td><td>".$row[$j+1]."</td>";
echo "</tr>";
}
echo "</table>";
mysql_close($con);
答案 1 :(得分:0)
<table>
<tr>
<td>ProjectName</td>
<td>Status</td>
<td>ProjectName</td>
<td>Status</td>
</tr>
<?php
while($row = $result->fetch_object()) {
echo "<tr>";
echo "<td>".$row->ProjectName."</td>";
echo "<td>".$row->Status."</td>";
echo "<td>".$row->ProjectName."</td>";
echo "<td>".$row->Status."</td>";
echo "</tr>";
}
?>
</table>
这是图片中的事情。使用CSS,您可以操作tds。
答案 2 :(得分:0)
您的功能应该与此类似:
$query = "SELECT *
FROM todo
ORDER BY id";
$result = $mysqli->query($query);
while($row = $result -> fetch_array()) {
$feedback .= "<tr>\n<td>" . $row['item'] . "</td><td>" . $row['priority'] . "</td>\n</tr>";
}
return $feedback;
然后,在您的HTML中已经设置了<table>
,并且您通常会在其中插入<td>
和<tr>
put <?php echo $feedback?>
(其中$ feedback是假定变量on从页面检索$反馈的HTML页面。这不是一个完整的修复程序,您的代码很难阅读,但从这里开始,您应该能够继续填写表格所需的所有额外信息,包括CSS。