Question

我一直试图解决这个问题。我创建了一个带有下拉框的表单，该表单从数据库中获取结果。从此我然后$_POST从那到另一页。从第二页开始，我希望获得ID号，然后获取记录并在屏幕上显示。然后我会把它们放在一张表中，以便更好地组织结果。任何人都可以帮助我实现这一目标。

以下是表单的代码（可以工作并发送$ PlantID）

$sql = "SELECT DISTINCT * FROM PLANTS";
$result = mysqli_query($mysqli,$sql)or die(mysqli_error());

//********************* Botannical name drop down box

echo "<form name='selection' id='selection' action='profile.php' method='post'>";

echo "<select name='flower'>";

while($row = mysqli_fetch_array($result)) {
    $plantid = $row['FlowerID'];
    $plantname = $row['Botannical_Name'];
    $plantcommon = $row['Common_Name'];
   /* $plantheight = $row['Height'];
    $plantav = $row['AV'];
    $plantcolours = $row['Colours'];
    $plantflowering = $row['Flower_Time'];
    $plantspecial = $row['Special_Conditions'];
    $plantfrost = $row['Frost_Hardy'];
    $plantaspect = $row['Aspect'];
    $plantspeed = $row['Growth_Speed'];*/

    echo "<option value=".$plantid.">".$plantname." -> AKA -> ".$plantcommon."</option>";
} 
echo "</select>";
echo "<br />";


//********************* End of form
echo "<input type='submit' name='submit' value='Submit'/>";
echo "</form>";

我创建了此页面以获取ID并在屏幕上显示该ID。你可以告诉我，我可能已经尝试了两种方法来尝试解决这个问题。

$sql = "SELECT * FROM PLANTS";

$result = mysqli_query($mysqli,$sql)or die(mysqli_error());

if(isset($_POST['submit'])){
$selected_val = $_POST['Botannical_Name'];  // Storing Selected Value In Variable
echo "You have selected :" .$selected_val;  // Displaying Selected Value
}
echo "<br />";
echo "well:".$_POST["Botannical_Name"]."<br/>";
echo "now:".$plantquery."<br />";
echo $_POST;



echo "<table>";
 foreach ($_POST as $key => $value) {
        echo "<tr>";
        echo "<td>";
        echo $key;
        echo "</td>";
        echo "<td>";
        echo $value;
        echo "</td>";
        echo "</tr>";
    }
echo "</table>";

非常感谢任何帮助。

Answer 1

if(isset($_POST['submit'])){
     $selected_val = $_POST['flower'];  // Storing Selected Value In Variable
     echo "You have selected :" .$selected_val;  // Displaying Selected Value
     $sql = "SELECT * FROM PLANTS WHERE FlowerID='.$selected_val.'";

     $result = mysqli_query($mysqli,$sql)or die(mysqli_error());
     while ($row=mysqli_fetch_assoc($result))
     {
         echo $row['Botannical_Name'];
     }
}
echo "<br />";
print_r($_POST);


if(!empty(_POST)) {
echo "<table>";
 foreach ($_POST as $key => $value) {
        echo "<tr>";
        echo "<td>";
        echo $key;
        echo "</td>";
        echo "<td>";
        echo $value;
        echo "</td>";
        echo "</tr>";
    }
echo "</table>";
}

Answer 2

您应该使用以下内容来获取所选值

$selected_val = $_POST['flower'];

显示在新页面中$ _POST之后获得的结果

2 个答案: