从数据库填充下拉列表并设置默认值 [英] Populate dropdown from database and set default value

问题描述

现在，我有一个可行的解决方案，可通过 PHP/MYSQLI 从我的database: listoption的内容填充HTML <select>/<option>-下拉列表.

Right now I have a working solution for populating an HTML <select>/<option>-dropdown with the content through PHP/MYSQLI from my database: listoption.

数据库:

DATABASE NAME: 
# listoption

TABLES:
# ID INT(11) *Primary AI
# listoption_item VARCHAR(255)

这是其他代码(不是mysqli connect，而是之后的所有内容.)

Here's the other code (not the mysqli connect but everything afterwards..)

<?php

    $result = $mysqli->query("select * from listoption");

    echo "<select id='list' name='list'>";

      while ($row = $result->fetch_assoc()) {

        $listoption_item = $row['listoption_item'];

        echo '<option value="'.$listoption_item.'">'.$listoption_item.'</option>';

      }

    echo "</select>";

?>

但是现在的问题是，我希望通过该查询填充的这些选项之一是selected.并且要选择的选项应由URL中的参数确定，例如:index.php?id=1. 因此，现在我需要以某种方式将IF/ELSE和$_GET['id'];添加到代码中，以使其标识数据库中的ID是否与填充的项目相同，然后将其设置为selected.

But the problem is now that I want to have one of these options that are populated through that query to be selected. And the option to be selected should be determed by a parameter in the URL, for example: index.php?id=1. So now I need to somehow add a IF/ELSE and a $_GET['id']; into the code to make it identify if the ID from the database is the same as the populated item and then set it to selected.

有任何想法吗?谢谢！

推荐答案

您可以按照以下步骤进行操作:

You can do that like given below:

<?php

    $result = $mysqli->query("select * from listoption");
    $id = ($_GET['id'])? $_GET['id'] : '';

    echo "<select id='list' name='list'>";

      while ($row = $result->fetch_assoc()) {

        $listoption_item = $row['listoption_item'];
        $sel = ($id == $row['id'])? 'selected="selected"':'';
		
        echo '<option value="'.$listoption_item.'" '.$sel.'>'.$listoption_item.'</option>';  // $sel will deside when to set `selected`
        
      }
    
    echo "</select>";
    
?>

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