用于过滤HTML表格中的数据的SQL查询 [英] Sql query to filter data in HTML table

问题描述

我需要创建一个搜索表单，用户可以使用任何关键字进行搜索。但是现在当用户搜索它不起作用时。



我尝试通过简单查询在搜索部分sql查询中测试我的代码，例如

  SELECT  *  FROM `course` < span class =code-keyword> WHERE  CONCAT（`id`，`name`，`department_id`，`course_desc`，`hidden_​​id`） LIKE  ' ％。$ valueToSearch。％' 

，它的工作原理。所以，我知道我的问题是在搜索部分的SQL查询。但我仍然不知道如何解决我的问题。任何人都可以帮助我。



我尝试过：



< pre lang =sql> if （isset（ $ _ POST [' search']））

{
$ valueToSearch = $ _ POST [' valueToSearch' ];

$ sql = 选择
s1.id作为id，
s1.name作为名称，
s2.name作为department_id，
s1.course_desc作为course_desc，
s3.name作为hidden_​​id
从
当然s1
离开加入部门为s2
在s1.department_id = s2.id
左加入隐藏为s3
在s1.hidden_​​id = s3.id
WHERE CONCAT（`id`，`name`，`department_id`，`course_desc`，`hidden_​​id`）
LIKE'％。 $ valueToSearch 。 ％'
LIMIT $ start_from，$ per_page< /跨度>;
$ result = $ mydb-> query（ $ sql ）;
}

解决方案

valueToSearch。％'

，它工作。所以，我知道我的问题是在搜索部分的SQL查询。但我仍然不知道如何解决我的问题。任何人都可以帮助我。



我尝试过：

  if （isset（ 

_POST [' search']））

{

valueToSearch =

I need to create a search form where users can search using any keyword. But right now when user search its does not work.

I try test my code at search part sql query by simple query like

SELECT * FROM `course` WHERE CONCAT(`id`, `name`, `department_id`, `course_desc`,`hidden_id`) LIKE '%".$valueToSearch."%' 

and it work. so, i know my problem is at sql query for search part only. but still i don't know how to solve my problem. Can anyone help me.

What I have tried:

if(isset($_POST['search']))

    {
      $valueToSearch = $_POST['valueToSearch'];

      $sql = "select  	
			s1.id		as id,
			s1.name		as name,
			s2.name 	as department_id,
			s1.course_desc 	as course_desc,
			s3.name as hidden_id
	from 		
			course  as s1
	left join	department as s2
		on	s1.department_id=s2.id
	left join hidden as s3
		on	s1.hidden_id=s3.id
      WHERE CONCAT(`id`, `name`, `department_id`, `course_desc`,`hidden_id`) 
      LIKE '%".$valueToSearch."%' 
      LIMIT $start_from, $per_page";
      $result = $mydb->query($sql);
    }

解决方案

valueToSearch."%'

and it work. so, i know my problem is at sql query for search part only. but still i don't know how to solve my problem. Can anyone help me.

What I have tried:

if(isset(

_POST['search'])) {

valueToSearch =

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