Unable to get result from querying a MySQL database

Question

I was working with a database in MySQL, where I have a table say ABC.

I want to query the database for a given search keyword, and display the output in a table. When I am querying with the following code, it is working fine:(Games Played 1 is a column name)

if($_POST['searchby'] == "Games Type"){
          $query_Recordset1 = "SELECT * FROM users_entity WHERE `Games Played 1' LIKE '%".$searchword."%'";

However, when I am trying to check whether the keyword is present in two columns, Games Played 1 or Games Played 2, it is not working. I am using the following code for that.

if($_POST['searchby'] == "Games Type"){
          $query_Recordset1 = "SELECT * FROM elgg_users_entity WHERE `Games Played 1` OR `Games Played 2` LIKE '%".$searchword."%'";

Can anyone suggest where I am doing wrong?? Also, when I run the webpage with the above code, it gives a warning as "The MySQL server has gone away"

You are missing the first condition: Select * from table WHERE condition OR condition. You have to use 'Games Played 1' LIKE 'SOMETHING' OR 'Games Played 2' LIKE 'SOMETHING'. Run the final generated query outside php and see. — FirstOne
– FirstOne, Commented Dec 27, 2015 at 16:50
Also, it looks like you are vulnerable to SQL Injection. Take a look at How can I prevent SQL-injection in PHP?. — FirstOne
– FirstOne, Commented Dec 27, 2015 at 16:56

Vojko · Accepted Answer · 2015-12-27 16:52:33Z

3

if($_POST['searchby'] == "Games Type"){
          $query_Recordset1 = "SELECT * FROM elgg_users_entity WHERE `Games Played 1` LIKE '%".$searchword."%' OR `Games Played 2` LIKE '%".$searchword."%'";

answered Dec 27, 2015 at 16:52

Vojko

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Rajdeep Paul · Accepted Answer · 2015-12-27 16:52:07Z

2

Your query should be like this:

$query_Recordset1 = "SELECT * FROM elgg_users_entity WHERE `Games Played 1` LIKE '%".$searchword."%' OR `Games Played 2` LIKE '%".$searchword."%'";

answered Dec 27, 2015 at 16:52

Rajdeep Paul

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Abdulla Nilam · Accepted Answer · 2015-12-27 16:54:02Z

2

Errors

Wrong WHERE condition.

So final code is

if($_POST['searchby'] == "Games Type"){
      $query_Recordset1 = "
      SELECT * 
      FROM elgg_users_entity 
      WHERE table_coumn1 
        LIKE '%$searchword%'  
      OR table_coumn2 
        LIKE '%$searchword%' ";

For Your Information

Don't use spaces between table column names(Games Played 1). Add it with _separated word. (Games_Played_1).

Hence make it all small caps as best practices

answered Dec 27, 2015 at 16:54

Abdulla Nilam

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1 Comment

Abdulla Nilam Over a year ago

@Arnab if my answer helpful grateful by accepting it or Vote it. Thank you

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