Php 提交后保持所选内容可见
如何使过滤后的选择在请求后仍然显示 如果我有选择1,2和3。当我选择2且数据显示时,我仍然希望2显示,表示数据已通过选项2过滤Php 提交后保持所选内容可见,php,html,mysql,drop-down-menu,Php,Html,Mysql,Drop Down Menu,如何使过滤后的选择在请求后仍然显示 如果我有选择1,2和3。当我选择2且数据显示时,我仍然希望2显示,表示数据已通过选项2过滤 echo "<form name='country_list' method='POST' action='http://opben.com/colombia/familias-de-carteras' >"; echo "<select name='Country' tabindex='1' >"; while($row = mysql_fe
echo "<form name='country_list' method='POST' action='http://opben.com/colombia/familias-de-carteras' >";
echo "<select name='Country' tabindex='1' >";
while($row = mysql_fetch_array($result))
{
echo " <option value='". $row['Fund_Manager_Company_Code'] ."'>". $row['Fund_Manager_Company_Name'] ."</option>";
}
echo "</select>";
echo "<input type='submit' value='Filter' />";
echo "</form>";
echo”“;
回声“;
while($row=mysql\u fetch\u数组($result))
{
echo“$row[“基金经理\公司\名称]”;
}
回声“;
回声“;
回声“;
您需要的是将所选属性添加到选项:
$Country = $_POST['Country'];
$sected = 'selected = "selected" ';
while($row = mysql_fetch_array($result))
{
echo " <option ".($row['Fund_Manager_Company_Code'] == $Country? $selected : '')."value='". $row['Fund_Manager_Company_Code'] ."'>". $row['Fund_Manager_Company_Name'] ."</option>";
}
$Country=$\u POST['Country'];
$sected='selected=“selected”;
while($row=mysql\u fetch\u数组($result))
{
echo“$row[“基金经理\公司\名称]”;
}
选择并发布的值将被选择…您可以执行以下操作:
$country = isset($_POST['Country']) ? $_POST['Country'] : '';
while($row = mysql_fetch_array($result))
{
echo " <option value='". $row['Fund_Manager_Company_Code'] ."' ".(($row['Fund_Manager_Company_Code'] == $country) ? 'selected="selected"' : '').">". $row['Fund_Manager_Company_Name'] ."</option>";
}
$country=isset($\u POST['country'])$_职位[“国家]:”;
while($row=mysql\u fetch\u数组($result))
{
echo“$row[“基金经理\公司\名称]”;
}
类似
echo" <option value='" . $row['Fund_Manager_Company_Code'] . "' " . ((isset($_POST['Country']) && $_POST['Country'] == $row['Fund_Manager_Company_Code'])
? 'selected="selected"' : '') . ">" . $row['Fund_Manager_Company_Name'] . "</option>";
echo”“$第[“基金经理\公司\名称”]行。"";
提交后,您需要在PHP代码中捕获所选内容:
$selection = $_POST['Country'];
echo "<form name='country_list' method='POST' action='http://opben.com/colombia/familias-de-carteras' >";
echo "<select name='Country' tabindex='1' >";
while($row = mysql_fetch_array($result))
{
$selected = "";
if ($row['Fund_Manager_Company_Code'] == $selection) {
$selected = "selected";
}
echo " <option value='". $row['Fund_Manager_Company_Code'] ."' ".$selected.">". $row['Fund_Manager_Company_Name'] ."</option>";
}
echo "</select>";
echo "<input type='submit' value='Filter' />";
echo "</form>";
$selection=$\u POST['Country'];
回声“;
回声“;
while($row=mysql\u fetch\u数组($result))
{
$selected=“”;
如果($row['Fund\u Manager\u Company\u Code']==$selection){
$selected=“selected”;
}
echo“$row[“基金经理\公司\名称]”;
}
回声“;
回声“;
回声“;
您需要检查post数组中的值,并相应地修改html。您需要检查$\u post['Country']是否存在,因为会注意到