通过单击按钮更新信息表 [英] Update a table of information on a button click

问题描述

我有一个表格，它将从数据库中选择信息并显示它，我想知道是否有人可以为我提供一个代码来更新点击一个按钮的列？

示例表：（Access = Boolean）

  ID  - 名称 - 访问
 --- ------------------------ 
 1  -  John  -  1 
 ------------- -------------- 
 2  -  Ben  -  1 
 ----------------------- ---- 
 3  - 特里 -  0 
 --------------------------- 
  

我现有的按钮是基于bootstrap的，

 < button type = \button \id = \passed\class = \btn btn-success btn-flat \>< i class = \ fa fa-check \>< / i>< / button> 
   

我希望得到像这样的东西，ONCLICK button1运行$允许SQL
ONCLICK button2运行$ NotAllowed SQL

  $ allowed = mysqli_query（$ conn，UPDATE users SET Access =1WHERE id ='27' ）; 
 $ notallowed = mysqli_query（$ conn，UPDATE users SET Access =0WHERE id ='453'）; 
  

解决方案

您可以使用post方法和提交具有命名属性的按钮类型并在条件语句中使用这些按钮。

即：

旁注：I删除了转义引号，因为我不确定这些引用是否已经设置在echo中。

HTML表单：

 < form method =postaction =handler.php> 
 
 < button type =submitname =insufficientid =insufficientclass =btn btn-danger btn-flat>< i class =fa fa-times>< / I>< /按钮> 
 
< / form> 
  

PHP：

 < $ c $<？php 
 
 // db连接
 
 if（isset（$ _ POST ['passed']））{
 
 $ allowed = mysqli_query（$ conn，UPDATE users SET Access =1WHERE id ='27'）; 
 
 
 $ b if（isset（$ _ POST ['insufficient']））{
 
 $ notallowed = mysqli_query（$ conn，UPDATE users SET Access =0WHERE id ='453'）; 
 
 
  

• 小心如果这在任何给定的点上都有用户交互的话。

使用 准备好的声明 或 PDO包含准备好的语句 ，它们更安全。 >



---





脚注：
$ b 运行UPDATE查询时，最好使用 mysqli_affected_rows（）来表示绝对真实性。





• < a href =http://php.net/manual/en/mysqli.affected-rows.php =nofollow> http://php.net/manual/zh/mysqli.affected-rows.php






否则，您可能会收到误报。




I have a table which will select information from a database and display it, I'm wondering if anyone can provide a code for me to update a column on the click of a button?

Example Table: (Access=Boolean)

ID   -   Name   -   Access
---------------------------
1   -   John    -    1
---------------------------
2   -   Ben     -    1
---------------------------
3   -   Terry   -    0
---------------------------

My exisiting button is based on bootstrap,

    <button type=\"button\" id=\"passed\" class=\"btn btn-success btn-flat\"><i class=\"fa fa-check\"></i></button>
    <button type=\"button\" id=\"insufficient\" class=\"btn btn-danger btn-flat\"><i class=\"fa fa-times\"></i></button>

I was hoping for something like this, ONCLICK button1 Run $Allowed SQL ONCLICK button2 Run $NotAllowed SQL

$allowed = mysqli_query($conn," UPDATE users SET Access = "1" WHERE id = '27' ");     
$notallowed = mysqli_query($conn," UPDATE users SET Access = "0" WHERE id = '453' ");

解决方案

You can use a form with a post method and a submit type of buttons with named attributes and use those in a conditional statement.

I.e.:

Sidenote: I removed the escaped quotes, as I was not sure if those were already set inside an echo.

HTML form:

<form method="post" action="handler.php">

    <button type="submit" name="passed" id="passed" class="btn btn-success btn-flat"><i class="fa fa-check"></i></button>
    <button type="submit" name="insufficient" id="insufficient" class="btn btn-danger btn-flat"><i class="fa fa-times"></i></button>

</form>

PHP:

<?php 

// db connection

if(isset($_POST['passed'])){

    $allowed = mysqli_query($conn," UPDATE users SET Access = "1" WHERE id = '27' ");

}

if(isset($_POST['insufficient'])){

    $notallowed = mysqli_query($conn," UPDATE users SET Access = "0" WHERE id = '453' ");

}

• Be careful if this has any user interaction at any given point.

• Use a prepared statement, or PDO with prepared statements, they're much safer.

---

Footnotes:

When running an UPDATE query, it's best to use mysqli_affected_rows() for absolute truthness.

• http://php.net/manual/en/mysqli.affected-rows.php

Otherwise, you may get a false positive.

这篇关于通过单击按钮更新信息表的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！