Web Development || Node JS, Angular, PHP

Thursday, November 17, 2016

Copy Selected fields from table2 to table1

If we want to set employeeCode in employeeMaster1 from employeeMaster2 for corresponding employeeId

A join:

UPDATE employeeMaster1 AS t1
INNER JOIN employeeMaster2 AS t2 ON t1.employeeId = t2.employeeId

SET t1.employeeCode = t2.employeeCode

By Using left join:

UPDATE employeeMaster1 AS t1
LEFT JOIN employeeMaster2 AS t2 ON t1.employeeId = t2.employeeId
SET t1.employeeCode = t2.employeeCode

A subquery:

UPDATE employeeMaster1
SET employeeCode = (
SELECT employeeCode
FROM employeeMaster2
WHERE employeeId = employeeMaster1.employeeId

)

Saturday, April 23, 2016

Upload Image Using JQuery. Insert Into Database.

How To insert image into Database.
Please follow the below Steps:-

(1) Create a database

CREATE TABLE IF NOT EXISTS `userinfo` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`name` varchar(222) NOT NULL,
`picture` varchar(222) NOT NULL,
`createdDateTime` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP,
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8 AUTO_INCREMENT=1 ;

(2) create a folder structure inside your web application folder.

upload

(3) Create a file with the name index.php

a. Create HTML form
name,userImage
b. load jquery "https://ajax.googleapis.com/ajax/libs/jquery/1.12.0/jquery.min.js"

c. use id "uploadTopic" for form.

d. Create DIV as follows

e. Place following code at the bottom

4. Create a file upload.php and place the following code:-

mysql_connect("localhost","root","root");
mysql_select_db("test");

if(!empty($_POST))
{
$name = $_POST['name'];
if(isset($_FILES['userImage']['name']) && $_FILES['userImage']['name']!='')
{
$desired_path = 'upload/';

$rand=rand(0,999);
$desired_file_name = $rand.'_'.str_replace(' ','_',strtolower($_FILES['userImage']['name']));

move_uploaded_file($_FILES['userImage']['tmp_name'], $desired_path . $desired_file_name);
}

$query_insert="INSERT INTO userInfo SET name = '".$name."',picture='".$desired_file_name."'";
mysql_query($query_insert);
}
?>

Thursday, March 1, 2012

10 Second Highest Salary of Employee

SELECT * FROM employee E1
WHERE 1 = ( SELECT COUNT( DISTINCT salary )
FROM employee E2
WHERE E1.salary < E2.salary
)
LIMIT 0,10

Friday, February 17, 2012

Sorting a Multi-Dimensional Array with PHP

//an array of some songs I like
$employee =  array(
  '1' => array('name'=>'Hanzala Subhani', 'job_title'=>'Software Developer'),
  '2' => array('name'=>'Safdar Subhani', 'job_title'=>'Software Developer'),
  '3' => array('name'=>'Shamsh Jawed', 'job_title' =>'Manager')
 );

function sort_employee($a,$subkey) {
 foreach($a as $k=>$v) {
  $b[$k] = strtolower($v[$subkey]);
 }
 asort($b);
 foreach($b as $key=>$val) {
  $c[] = $a[$key];
 }
 return $c;
}


$employee = sort_employee($employee,'artist'); 
print_r($employee);