我尝试制作一个应该发布的选择并包含来自用户的所有投票
<select name="votacao" size="1">
<option>
<?php
$sql = "SELECT nome_votacao FROM votacoes WHERE user_id = $_SESSION['id']";
$rs = mysql_query($sql);
while($row = mysql_fetch_row($rs))
echo $row['nome_votacao'];
?>
</option>
</select>
答案 0 :(得分:1)
您的<option>
需要在您的循环中:
<select name="votacao" size="1">
<?php
$sql = "SELECT nome_votacao FROM votacoes WHERE user_id = $_SESSION['id']";
$rs = mysql_query($sql);
while($row = mysql_fetch_row($rs)) {
echo sprintf("<option>%s</option>\n", $row['nome_votacao']);
}
?>
</select>
您可能还需要为每个value
添加一个唯一ID作为<option>
属性,以确保真正有用。
答案 1 :(得分:0)
<select name="votacao" size="1">
<?php
$sql = "SELECT nome_votacao FROM votacoes WHERE user_id = $_SESSION['id']";
$rs = mysql_query($sql);
while($row = mysql_fetch_row($rs))
echo "<option>".$row['nome_votacao']."</option>";
?>
</select>
答案 2 :(得分:0)
试试这个
<select name="votacao" size="1">
<option>
<?php
$sql = "SELECT nome_votacao FROM votacoes WHERE user_id = $_SESSION['id']";
$rs = mysql_query($sql);
foreach ($rs as $row)
{ ?>
<option> <?php echo $row['nome_votacao']; ?> </option>
<? } ?>
</option>