PHP-下拉菜单(选择)无法正常运行至while循环 [英] PHP - Dropdown menu(select) not working properly to while loop

问题描述

我希望我的dropdown-textbox-select(我很困惑，它实际上是什么)可以查询要显示的适当表中的所有数据.但是它将显示所有数据，并带有单独的dropdown-textbox-select.我只希望数据在内部.到目前为止，这是我正在做的事情.

I want my dropdown-textbox-select(I'm confused what it really is) to query all the data in the appropriate table I want to display. But it is displaying all the data with separated dropdown-textbox-select. I just want the data to be inside. Here's what I'm doing so far.

<?php   


    $sel_admin = "SELECT * FROM author";
    $rs_admin = mysql_query($sel_admin);
    while($row = mysql_fetch_array($rs_admin))
                {
                echo '<select class="form-control">';
                echo"  <option value='volvo'>" . $row['author_firstname'] . $row['author_lastname'] ."</option>";
                echo'</select>';
                }
                ?>

PS.如果用户单击提交，我希望保存author_id而不是名称.怎么可能呢?

PS. If the user click submit, I want the author_id to be save, instead of the name. How can this be also?

推荐答案

您的select标记应位于while循环之外:您只需要一个选择字段，其中包含所有已提取行的选项.

Your select tag should be outside of the while loop : you want only one select field, containing options with all the fetched rows.

echo '<select class="form-control">';
while($row = mysql_fetch_array($rs_admin))
{
    echo"  <option value='volvo'>" . $row['author_firstname'] . $row['author_lastname'] ."</option>";
}
echo'</select>';

如果要从此表单获取author_id，只需将此值放在选项的value属性中即可.

If you want to get the author_id from this form, just put this value in the option's value attribute.

echo"  <option value='". $row['author_id'] ."'>" . $row['author_firstname'] . $row['author_lastname'] ."</option>";

别忘了给select标记加上name.

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