如何从下拉列表中获得选定的用户名是URL的var - how to get selected username from dropdown list to be the var for URL

how to get selected username from dropdown list to be the var for URL

本文关键字：用户 var URL 下拉列表 | 更新日期: 2023-09-27

如何让所选变量显示在URL中<select>…

我的代码是:

print "<select name='"assignedby'" multiple size='"10'">";
while ($data = dbResult($qh)) {
        print "<option name='"$data[name]'"";
        print ">$data[name]</option>'n";
}
print "</select>";
print "<br><a href='".$_SERVER['PHP_SELF']."?action=add'>Add</a> || <a href='".$_SERVER['PHP_SELF']."?origname=$data[name]'>Edit</a>'n";

当有人点击"编辑"链接时，它显示为:http://www.site.com?origname=我想让它与下拉列表中实际选中的origin一起显示。如:http://www.site.com?origname=美元selecteduser-fromdroplist

请帮忙!

为什么不直接使用form method = get呢?

<form id="select_name" action="" method="get">
    <select name="origname">
    <?php while ($data = dbResult($qh)): ?>
        <option value="<?php echo $data[name]; ?>"><?php echo $data[name]; ?></option>
    <?php endwhile; ?>
    </select>
    <a href="<?php echo $_SERVER['PHP_SELF']; ?>?action=add">Add</a> || <input type="submit" name="submit" value="Edit">
</form>

但至于为什么你的代码不工作，你调用$data[name']项目外的while循环