Question

我有问题。我需要制作动态SELECT标签，其中OPTIONS是从mySQL DB动态填充的。我有这个代码，但它不起作用。因此，如果你有一些纠正或想法如何以其他方式，我会非常感谢:)非常感谢！

<SELECT>
<?php
 $con = mysql_connect("localhost", "root", "123456");
mysql_select_db("aplikace",$con);
$result = ("SELECT * kod FROM protokoly".
"ORDER BY kod");
$rs = mysql_query($result);
$nr = mysql_num_rows($rs);
for ($i=0; $i<$nr; $i++) {
    $r = mysql_fetch_array($rs);
    echo "<OPTION VALUE=\"".$r["kod"]."\">".$r["kod"]."</OPTION>";
}
?>
</SELECT>

Answer 1

正如@Michael Berkowski在评论中所说，protokoly和ORDER BY之间目前没有空格。这将导致MySQL错误，但您没有看到导致难以进行故障排除的错误。

如果出现错误，您可以修改代码以输出错误，如下所示：

$rs = mysql_query($result) or die('Error during query: ' . mysql_error());

如果mysql_query()失败，die()调用mysql_error()将输出最新MySQL错误的内容，帮助您解决问题。

修改

为了说明die()的使用情况，我编辑了您的代码，如下所示：

<?php // Place at top of file error_reporting(E_ALL); ini_set('display_errors', '1'); ?> <SELECT> <?php $con = mysql_connect("localhost", "root", "123456") or die('Connection failed: ' . mysql_error()); mysql_select_db("aplikace", $con) or die('Cannot select database: ' . mysql_error()); $result = ("SELECT * kod FROM protokoly ORDER BY kod"); $rs = mysql_query($result) or die('Query failed: ' . mysql_error()); $nr = mysql_num_rows($rs) or die('Failed to get num_rows: ' . mysql_error()); for ($i = 0; $i < $nr; $i++) { $r = mysql_fetch_array($rs); echo "<OPTION VALUE=\"{$r['kod']}\">{$r['kod']}</OPTION>"; } ?> </SELECT>

请注意，使用mysql_...()函数的语句包括mysql_error()内的die()，以便输出任何MySQL错误。我还在echo语句的双引号字符串中用括号括起数组引用，以简化它并消除多个连接。

Answer 2

<SELECT>
<?php

//added error checking.
if (!($con = mysql_connect("localhost", "root", "123456")))
{
    die("Could not connect to database server. " . mysql_errno() . " => " . mysql_error());
}

//added error checking.
if (!mysql_select_db("aplikace",$con))
{
    die("Could not select database. " . mysql_errno() . " => " . mysql_error());
}

//removed *, pushed ORDER BY onto same line
$sql = "SELECT kod FROM protokoly ORDER BY kod";

//added error checking. added connection parameter
if (!($rs = mysql_query($sql,$con)))
{
    die("Could not execute query. " . mysql_errno() . " => " . mysql_error());
}

//added condition
if (($nr = mysql_num_rows($rs)) == 0)
{
    die("No options to give!");
}

//changed to while loop that checks result of function return
while ($r = mysql_fetch_array($rs))
{
    //changed " to ' to get rid of \"
    echo '<OPTION VALUE="' . $r["kod"] . '">' . $r["kod"] . '</OPTION>';
}
?>
</SELECT>

从MYSQL DB填充的动态SELECT TAG

2 个答案: