1

我正在自学 PHP 和 MySQL，我正在尝试从我的数据库中检索一些信息并将其放入表中。

到目前为止，只显示了表格列标题，并且没有显示每列的信息。PHP 文件也需要很长时间才能显示。

请您指出我的代码的问题。

<?php
mysql_connect("localhost",$username,$password);
mysql_select_db($dbname) or die("Unable to select Database");
$query = "SELECT * FROM table_1";
$result = mysql_query($query);
$numcount = mysql_num_rows($result);
echo "<h2>$numcount rows in table_1.</h2>";
mysql_close();
?>

<table border="0" cellspacing="4" cellpadding="2">
<tr>
<th><font face="Futura">Type |</font></th>
<th><font face="Futura">Name |</font></th>
<th><font face="Futura">Street |</font></th>
<th><font face="Futura">Address1 |</font></th>
<th><font face="Futura">Address2 |</font></th>
<th><font face="Futura">Town |</font></th>
<th><font face="Futura">County |</font></th>
<th><font face="Futura">Postcode |</font></th>
<th><font face="Futura">Number |</font></th>
<th><font face="Futura">Latitude,Longitude</font></th>
</tr>

<?php
$i=0;
while ($i < 843) {
$type = mysql_result($result,$i,"type");
$name = mysql_result($result,$i,"name");
$street = mysql_result($result,$i,"street");
$addr1 = mysql_result($result,$im,"address1");
$addr2 = mysql_result($result,$im,"address2");
$town = mysql_result($result,$im,"town");
$county = mysql_result($result,$im,"county");
$postcode = mysql_result($result,$im,"postcode");
$number = mysql_result($result,$im,"number");
$latlong = mysql_result($result,$im,"latlong");
}
?>

<tr>
<td><font face="Futura"><?php echo $type;?></font></td>
<td><font face="Arial, Helvetica, sans-serif"><?php echo $name;?></font></td>
<td><font face="Arial, Helvetica, sans-serif"><?php echo $street;?></font></td>
<td><font face="Arial, Helvetica, sans-serif"><?php echo $addr1;?></font></td>
<td><font face="Arial, Helvetica, sans-serif"><?php echo $addr2;?></font></td>
<td><font face="Arial, Helvetica, sans-serif"><?php echo $town;?></font></td>
</tr>
<?php
$i++;
?>
<?
echo "</table>"; 
?>
</body>
</html>
4

2 回答 2

2

  1. mysql_result() 在您之前执行 mysql_close() 时有效吗？

  2. 为什么不在这里使用 mysql_fetch_row() ？

为@XcodeDev 编辑更多信息：

当您需要单个结果时，您可以使用 mysql_result()，例如：

$query = mysql_query("SELECT COUNT(id) FROM users");

然后

$count = mysql_result($query, 0);

当您期望具有多个数据的单行结果时，使用

$result = mysql_fetch_row($query); => $result[0], $result[1], $result[2] etc

或者

$result = mysql_fetch_assoc($query); => $result['type'], $result['name'] etc

当您期望具有多个数据的多行结果时，使用

while ($result = mysql_fetch_row($query)) {
    => $result[0], $result[1], $result[2] etc
}

或者

while ($result = mysql_fetch_assoc($query)) {
    => $result['type'], $result['name'] etc
}
于 2011-03-13T23:04:50.743 回答
0

首先，您有一个无限循环，因为 $i 没有在循环内递增。如果你增加 $i，它应该可以解决问题。$im 是什么？

代码应如下所示：

<?php
$i=0;
while ($i < 843) {
    $type = mysql_result($result,$i,"type");
    $name = mysql_result($result,$i,"name");
    $street = mysql_result($result,$i,"street");
    $addr1 = mysql_result($result,$i,"address1");
    $addr2 = mysql_result($result,$i,"address2");
    $town = mysql_result($result,$i,"town");
    $county = mysql_result($result,$i,"county");
    $postcode = mysql_result($result,$i,"postcode");
    $number = mysql_result($result,$i,"number");
    $latlong = mysql_result($result,$i,"latlong");

?>

<tr>
<td><font face="Futura"><?php echo $type;?></font></td>
<td><font face="Arial, Helvetica, sans-serif"><?php echo $name;?></font></td>
<td><font face="Arial, Helvetica, sans-serif"><?php echo $street;?></font></td>
<td><font face="Arial, Helvetica, sans-serif"><?php echo $addr1;?></font></td>
<td><font face="Arial, Helvetica, sans-serif"><?php echo $addr2;?></font></td>
<td><font face="Arial, Helvetica, sans-serif"><?php echo $town;?></font></td>
</tr>
<?php
    $i++;
} // this is where the loop is ending.
?>
<?
mysql_close(); // close mysql connection after reading data
echo "</table>"; 
?>
于 2011-03-13T22:59:58.510 回答