在下拉列表中设置所选项目 php sql

Question

我正在创建一个页面，我想在其中显示从数据库捕获的信息。我想显示的项目之一是从下拉列表菜单中选择的值，但默认为选择的 coverid

我使用以下函数来显示下拉列表：

//function to display the cover in the dropdown
//function to display the cover in the dropdown
    public function coverdropdown($companyid, $coverid)
    {

        $mydata = mysql_query("SELECT a.`coverid`, a.`cover_name`, a.`sum_insured`, a.`info`
                    FROM covertype a WHERE a.`companyid`=$companyid");
            while($record = mysql_fetch_array($mydata))
                {
                    echo '<option  value="'.$record['coverid'].'" if ($record['coverid']==$coverid) {selected=selected}">'.$record['cover_name'].' - R'.$record['sum_insured'].'</option>'; 
                }
    }

我希望下拉列表显示所选的 coverid。我知道它是这样的：

<option value="" selected="selected">XYZ cover</option>

但现在我不确定如何将它添加到这一行，因为我添加的 if 语句不起作用，它的语法不正确：

while($record = mysql_fetch_array($mydata))
            {
               echo '<option  value="'.$record['coverid'].'" if ($record['coverid']==$coverid) {selected=selected}">'.$record['cover_name'].' - R'.$record['sum_insured'].'</option>'; 
            }

Answer 1

您应该可以在 <option> 之前分配一个变量。类似于：

while($record = mysql_fetch_array($mydata)) {
        // One thing to note: In XHTML the selected attribute should probably be defined as <option selected="selected">.
        // In regular HTML <option selected> is ok
        $selected   =   ($record['coverid'] == $coverid)? ' selected="selected"':'';
        echo '<option  value="'.$record['coverid'].'"'.$selected.'>'.$record['cover_name'].' - R'.$record['sum_insured'].'</option>'; 
    }

Answer 2

这必须做的工作：

while($record = mysql_fetch_array($mydata)){
    if($record['coverid']==$coverid)
        echo '<option  value="'.$record['coverid'].' selected>'.$record['cover_name'].'</option>';
    else
        echo '<option  value="'.$record['coverid'].'>'.$record['cover_name'].'</option>';
}

在下拉列表中设置所选项目 php sql

setting the selected item on a dropdown php msql

php

mysql

drop-down-menu