<?php
mysql_connect("localhost", "root","") or die(mysql_error());
mysql_select_db("tnews2") or die(mysql_error());
$query = "SELECT name,id FROM categories ORDER BY ID DESC LIMIT 0,6";
$result = mysql_query($query) or die(mysql_error()."[".$query."]");?>
<select name="categories">
<?php while ($row = mysql_fetch_array($result)){
?>
<option value=" <?php $row['path']; ?> ">
<?php echo $row['name']; ?>
</option>
<?php
}
?>
</select>?>
所以这是选择选项菜单,它是我从数据库中读取的值,但是当我尝试获取所选值时,我只得到第一个。
<?php
mysql_connect("localhost", "root","") or die(mysql_error());
mysql_select_db("tnews2") or die(mysql_error());
if(!empty($_POST['title']) && !empty($_POST['date']) && !empty($_POST['txt']) && !empty($_POST['image'])){
$TITLE=$_POST['title'];
$DATE=$_POST['date'];
$TXT=$_POST['txt'];
$IMAGE=$_POST['image'];
$CATEGORIES=$_POST['categories'];
echo $CATEGORIES;
$ANSWER=$_POST['main'];
$MAINPAGE=0;
你能帮我一个想法来获得选定的选项
编辑:(减少你的代码)
<?php
mysql_connect("localhost", "root","") or die(mysql_error());
mysql_select_db("tnews2") or die(mysql_error());
$query = "SELECT name,id,path FROM categories ORDER BY ID DESC LIMIT 0,6";
$result = mysql_query($query) or die(mysql_error()."[".$query."]");
?>
<select name="categories">
<?php
while ($row = mysql_fetch_array($result))
{
echo "<option value='".$row['path']."'>'".$row['name']."'</option>";
}
?>
</select>
我更喜欢使用 mysqli 这样做
<?php
/* Database connection settings */
$host = 'localhost';
$user = 'root';
$pass = '';
$db = 'tnews2';
$mysqli = new mysqli($host,$user,$pass,$db) or die($mysqli->error);
/* Your query */
$result = $mysqli->query("SELECT name,id,path FROM categories ORDER BY ID DESC LIMIT 0,6";) or die($mysqli->error);
?>
然后将元素添加到 html 中,如下所示:
<select name="categories">
<option value="Select School">Select Shool</option>
<?php
while ($row = mysqli_fetch_array($result)) {
echo "<option value='" . $row['path'] . "'>'" . $row['name'] . "'</option>";
}
?>
</select>
<?php
include("conect.php");
if ($db) {
$res=mysql_query("select bolao_concurso,bolao_data from Bolao");
if ($res) {
$lin_inic=1;
$lines=mysql_num_rows($res);
echo "<select name='conc'>";
echo "<p>YOUR CHOICE : ";
while ($lin_inic<=$linesS) {
$row=mysql_fetch_array($res);
$concurso =$row['bolao_concurso'];
$data_conc=$row['bolao_data'];
echo "<option value='$concurso'>$concurso $data_conc</option>";
$lin_inic=$lin_inic+1;
}
echo "</select>";
}
}
?>
我建议确保对SQL查询中的字段进行排序以匹配所需的顺序,然后使用:
$query = "SELECT path, name, id FROM categories ORDER BY ID DESC LIMIT 0,6";
echo "<option value='".$row[0]."'>'".$row[1]."'</option>";
然后,您的代码将作为"代码片段"更具可重用性,您需要做的就是编写一个新的SQL查询