help with search in date_difference column

help with search in date_difference column

empirerockempirerock Posts: 1Questions: 0Answers: 0
edited August 2016 in Free community support

Greetings.
I am not able to search the column that shows the date_difference.
Can anyone help me what's wrong.
Just is not working the search in this column.
I thank anyone who can help.

```
<?php

include_once("connection.php");

    $params = $columns = $totalRecords = $data = array();

$params = $_REQUEST;

$columns = array( 
    0 =>'`u`.`id`',
    1 =>'`u`.`data_abertura`',
    2 =>'`u`.`data_liberacao`',
    3 =>'`u`.`data_montagem`',
    4 =>'`u`.`codiks`',
    5 => '`u`.`planejador`',

            6 => '`u`.`date_difference`',

);



$where = $sqlTot = $sqlRec = "";

if( !empty($params['search']['value']) ) {   
    $where .=" WHERE ";
    $where .=" ( u.data_abertura LIKE '".$params['search']['value']."%' )";    
    $where .=" OR u.data_liberacao LIKE '".$params['search']['value']."%' ";
    $where .=" OR u.data_montagem LIKE '".$params['search']['value']."%' ";
    $where .=" OR u.codiks LIKE '".$params['search']['value']."%' ";
    $where .=" OR u.planejador LIKE '".$params['search']['value']."%' ";

            $where .=" OR u.date_difference LIKE '".$params['search']['value']."%' ";
}


$sql = "SELECT *, DATEDIFF(data_liberacao, data_abertura) AS date_difference FROM `dia` AS `u` INNER JOIN `estf004` AS `ud` ON (`ud`.`codiks` = `u`.`codiks`)";

$sqlTot .= $sql;
$sqlRec .= $sql;

if(isset($where) && $where != '') {

    $sqlTot .= $where;
    $sqlRec .= $where;
}


$sqlRec .=  " ORDER BY ". $columns[$params['order'][0]['column']]."   ".$params['order'][0]['dir']."  LIMIT ".$params['start']." ,".$params['length']." ";

$queryTot = mysqli_query($conn, $sqlTot) or die("database error:". mysqli_error($conn));


$totalRecords = mysqli_num_rows($queryTot);

$queryRecords = mysqli_query($conn, $sqlRec) or die("error to fetch employees data");


while( $row = mysqli_fetch_array($queryRecords) ) { 


    $result1 = $row[0];
    $result2 = $row[1];
    $result3 = $row[2];
    $result4 = $row[3];
    $result5 = $row[4];
    $result6 = $row[5]; 
    $result7 = $row['date_difference'];

$data[] = array($result1, $result2,  $result3, $result4,  $result5, $result6, $result7);

}   

$json_data = array(
        "draw"            => intval( $params['draw'] ),   
        "recordsTotal"    => intval( $totalRecords ),  
        "recordsFiltered" => intval($totalRecords),
        "data"            => $data   // total data array
        );

echo json_encode($json_data);  // send data as json format]
<?php > ``` ?>
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