简体   繁体   English

如何从mysql-db中选择正确的选择/选项

[英]How to get the right select/option selected from mysql-db

I have a edit page where I can edit the status of a message. 我有一个编辑页面,可以在其中编辑消息的状态。

<select name='cm_status'>
option value='Open'>Open</option>
<option value='Closed'>Closed</option>
</select>

Then I change the status to closed, but when i go into the page again, the open is selected even when the status is set to closed, and I wonder how to get the closed one selected. 然后,我将状态更改为已关闭,但是当我再次进入页面时,即使状态设置为已关闭,也会选择打开,并且我想知道如何选择已关闭的状态。

Thanks in advance, 提前致谢,

Kristian 克里斯蒂安

You will need to query the database when the page loads and create the form after evaluating the results. 您需要在页面加载时查询数据库,并在评估结果后创建表单。

You will set the attribute selected=selected on the option value depending on the value received from the database. 您将根据从数据库接收的值,在选项值上设置selected=selected属性。

Here is a simple version of what you will need: 这是您需要的简单版本:

$query = mysql_query("SELECT cm_status FROM some_table");
$row = mysql_fetch_array($query)
echo "<select name='cm_status'>";
if($row == "Open")
echo "<option value='Open' selected>Open</option><option value='Closed'>Closed</option>";
else
echo "<option value='Open'>Open</option><option value='Closed' selected>Closed</option>";
echo "</select>"

This form will need a form handling PHP script that will update your MySQL database table using $sql="UPDATE mytable SET mycolumn = 1 WHERE mytable_id = 1"; 该表单将需要一个处理PHP脚本的表单,该脚本将使用$sql="UPDATE mytable SET mycolumn = 1 WHERE mytable_id = 1";来更新MySQL数据库表$sql="UPDATE mytable SET mycolumn = 1 WHERE mytable_id = 1";

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

相关问题 使用php mysql错误从数据库中选择的值进入下拉选择框选项 - selected value get from db into dropdown select box option using php mysql error 我们如何重新使用从任何 MySQL-DB 表中删除的 id? - How can we re-use the deleted id from any MySQL-DB table? PHP使用mySQL-db中的下拉值 - PHP work with dropdown values from mySQL-db 从JavaScript函数将PHP发布到MySql-db - Post PHP to MySql-db from JavaScript function 如何在一个 select 选项中获取选定值,并在此基础上,从 MySQL 表中获取数据,以相同形式显示另一个 select 选项 - How to get selected value in one select option and based on that, fetch data from MySQL table to show another select option in the same form 如何从选择选项中获取价值而不在Laravel中选择选项 - How to get value from select option without selected the option in Laravel 如何从多个选择选项中获取所有选择的值? - How to get all selected values from multiple select option? 使用 php mysql 将选定的值从 db 获取到数据列表选项 - selected value get from db into Data list option using php mysql 如何获得SELECT OPTION的选定值 - How to get selected value of SELECT OPTION 从动态选择中运行Ajax,并选择从DB中检索的选项 - Run Ajax from a dynamic select with option selected retrieved from DB
 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM