如何从 MySQL 数据库中获取特定数据到我的 PHP 表？
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How to fetch specific data from MySQL database to my PHP table?
提问by strellson
I want to fetch data from MySQL database to my table in .php file. In every row of my table I want to display title, text and attachment($name) that user can download. The problem is when I display that, I get all attachments from database shown in list in every table row in my .php file. So I want only to display only one attachment per row, from database, that have same ID as title and text of database table row.
我想从 MySQL 数据库中获取数据到我在 .php 文件中的表。在表格的每一行中，我想显示用户可以下载的标题、文本和附件（$name）。问题是当我显示它时，我从 .php 文件中每个表行的列表中显示的数据库中获取所有附件。所以我只想每行只显示一个附件，来自数据库，与数据库表行的标题和文本具有相同的 ID。
This is my database table:
这是我的数据库表：
CREATE TABLE IF NOT EXISTS `table` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`title` varchar(30) NOT NULL,
`text` varchar(30) NOT NULL,
`name` varchar(30) NOT NULL,
`type` varchar(30) NOT NULL,
`size` int(11) NOT NULL,
`content` longblob NOT NULL,
PRIMARY KEY (`id`) )
My code:
我的代码：
<?php
include('config.php');
$sqlget="SELECT * FROM table ORDER BY timestamp DESC";
$sqldata= mysqli_query($dbcon, $sqlget) or die ('error');
echo"<table>";
while ($row=mysqli_fetch_array($sqldata,
MYSQLI_ASSOC)) {
echo "<tr><td>";
echo"<b><font color='#DF01A5'> Title: ".$row['title']."</font></b>";
echo "<br/>";
echo $row['text'];
echo "<br/>";
echo "<b><font color='#DF01A5'>Attachment: </font>";
?>
<?php
$con = mysql_connect('localhost', 'root', 'pass') or die(mysql_error());
$db = mysql_select_db('database', $con);
$query = "SELECT id, name FROM table";
$result = mysql_query($query) or die('Error, query failed');
if(mysql_num_rows($result) == 0)
{
echo "Database is empty <br>";
}
else
{
while(list($id, $name) = mysql_fetch_array($result))
{
?>
<a href="download.php?id=<?php echo urlencode($id);?>"
><?php echo urlencode($name);?></a> <br>
<?php
}
}
mysql_close();
echo"</table";
?>
回答by moskito-x
Use only one while loop
只使用一个 while 循环
$sqlget="SELECT * FROM table ORDER BY timestamp DESC";
$sqldata= mysqli_query($dbcon, $sqlget) or die ('error');
echo"<table>";
while ($row=mysqli_fetch_array($sqldata,MYSQLI_ASSOC)) {
echo "<tr><td>";
echo"<b><font color='#DF01A5'> Title: ".$row['title']."</font></b>";
echo "<br/>";
echo $row['text'];
echo "<br/>";
echo "<b><font color='#DF01A5'>Attachment: </font>";
echo "<a href=\"download.php?id=".$row['id']."\">".$row['name']."</a><br />\n";
...
}
echo '</table>';
回答by Unexpected Pair of Colons
Assuming your two tables are related on the ID alone:
假设您的两个表仅与 ID 相关：
$sql = $db_conn->query("SELECT t1.title, t1.text, t2.ID AS AID, t2.name FROM table1 AS t1 LEFT JOIN table2 AS t2 ON t1.ID = t2.ID ORDER BY t1.timestamp");
echo '<table><thead><tr><th>Title</th><th>Description</th><th>Attachment</th></tr></thead>';
while($row = $sql->fetch_array())
{
echo '<tr>';
echo '<td>' . $row['title'] . '</td>';
echo '<td>' . $row['text'] . '</td>';
echo '<td><a href="download.php?id=' . urlencode($row['AID']) . '">' . $row['name'] . '</a></td>';
echo '</tr>';
}
echo '</table>';
Not tested, but start with this.
未测试，但从这里开始。
回答by Andrew Therin
We'll need a sample output from this table. I'm still not sure how those attachments are being stored. In the meantime, this PHP function should return only one row: http://www.php.net/manual/en/mysqli-result.fetch-row.php
我们需要这个表的样本输出。我仍然不确定这些附件是如何存储的。同时，这个 PHP 函数应该只返回一行：http: //www.php.net/manual/en/mysqli-result.fetch-row.php
Edit: unnecessary comment.
编辑：不必要的评论。
回答by Subedi Kishor
For your particular condition, this will help help:
对于您的特定情况，这将有助于：
$sqlget = "SELECT `title`, `text`, `name` FROM `table` WHERE `title` = `text`";
Enjoy...
享受...
