Question

我正在使用来自数据库的值填充在php的选择输入框中，但是，当我单击“提交”按钮时，先前选择的值消失了，并且仅显示默认值。

我已经尝试了其他开发人员分享的一些帖子，但是我的似乎无法正常工作。

  <select name="pickup">

  echo "<option value='area'>Select where you want the cab to pick you from</option>";

   <?php $sql1 = "SELECT area FROM places ORDER BY area ASC;";
   $resultSet1 = mysqli_query($conn, $sql1);

    while($rows1 = mysqli_fetch_assoc($resultSet1)) {

    $area1 = $rows1['area'];

  echo "<option value='$area1'>$area1</option>";


   }?>

我没有任何错误，只是不保留先前选择的值。

Answer 1

尝试以下代码片段，可能会解决您的查询-

<?php
@session_start(); /*This line would be on top of the page*/
    if(isset($_POST["demo"]))  {
        $_SESSION["selected_val"] = $_POST["demo"];
    }else{
        $_SESSION["selected_val"] = ""; 
    }

    echo "<select name="demo" > <option value='area'>Select where you want the cab to pick you from</option>";

    $sql1 = "SELECT area FROM places ORDER BY area ASC;";
    $resultSet1 = mysqli_query($conn, $sql1);

    while($rows1 = mysqli_fetch_assoc($resultSet1)) {
        $area1 = $rows1['area'];
        $selected = ($area1 == $_SESSION["selected_val"]) ? 'selected' : "" ;
        echo "<option value='$area1' ".$selected.">$area1</option>";
    }
    echo "</select>";    
?>

Answer 2

在会话中设置选定的选项值，然后在重载后匹配选项和会话值并设置选择

在php中提交后如何在下拉菜单中保持选定的值

2 个答案: