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Receive post with description containing '%' and two letter together

I have problems with treatment of Post. I received a string with the symbol '%' and two letters together, like 'Geci%de', but on PHP the var_dump I receive a different string, like 'Geci�', if try to use utf8_encode and utf8_decode, however the error continued, the strings resulted were 'GeciÞ' and'Geci?'. How is the better way to convert in the orignal string? I need to use in postgreSQL, it will be in select.
It is uses to treatment:
$data = strip_tags($data);
$data = trim($data);
$data = get_magic_quotes_gpc() == 0 ? addslashes($data) : $data;
$data = preg_replace("#(--|\|)#s", "", $data);
$data = urldecode($data); // especific to Ajax
return utf8_decode($data);

How to add items from a foreach loop into a string with json_encode() in PHP?

I'm trying to insert into a string all emails retrieved from the database, so I can use javascript to check if the email typed by the user into a form field is already registered. I'm trying to use json_encode().
$conectar = mysqli_connect(HOST, USER, PASS, DATABASE);
$listarCorreos = " SELECT userEmail
FROM usuarios
";
$resultado = mysqli_query($conectar,$listarCorreos);
$arrayEmails = mysqli_fetch_array($resultado);
foreach($arrayEmails as $row){
$emails[]=array($row['userEmail']);
}
echo json_encode($emails);
Now, I'm getting this error:
Warning: Illegal string offset 'userEmail' in /home/verificarEmail.php
on line 21
Line 21 is $emails[]=array($row['userEmail']);
What Am I doing wrong?
UPDATE:
I'm also trying:
$resultado = mysqli_query($conectar,$listarCorreos);
$emails = array();
while($row = mysql_fetch_assoc($resultado)) {
$emails[] = $row;
}
echo json_encode($emails);
And I get this error:
Warning: mysql_fetch_assoc() expects parameter 1 to be resource,
object given in /home/verificarEmail.php on line 18
Line 18: while($row = mysql_fetch_assoc($resultado)) {

Use MYSQLI_ASSOC like below
$conectar = mysqli_connect(HOST, USER, PASS, DATABASE);
$listarCorreos = " SELECT userEmail
FROM usuarios
";
$resultado = mysqli_query($conectar,$listarCorreos);
while ( $row = mysqli_fetch_array($resultado, MYSQLI_ASSOC) ) {
$emails[]=array($row['userEmail']);
}
echo json_encode($emails);

There are a few problems there. To get your emails in an array do:
$resultado = mysqli_query($conectar,$listarCorreos);
if(!resultado) die($mysqli_error($conectar));//check for errors
$emails = mysqli_fetch_all($resultado);//fetch all results
echo json_encode($emails);
That should get your code working or show you the error. However, please don't do this.
so I can use javascript to check if the email typed by the user into a
form field is already registered
This is a terrible idea security wise because it will expose all your users' emails to people who are not even registered to your site. You should instead look up the desired username in the DB directly. If it's not there, then it hasn't been registered before. The Javascript side should only get a available or not available response.
The process (pseudo-code):
SELECT userEmail from usarios WHERE userEmail = ? ? is email to look for
execute query and capture $resultado
If $resultado is false, die(mysqli_error($conectar)) to show error
if mysqli_num_rows($resultado) === 0 email is available; else not available
Echo available or not available to Javascript

use the code like below
$conectar = mysqli_connect(HOST, USER, PASS, DATABASE);
$listarCorreos = " SELECT userEmail
FROM usuarios
";
$resultado = mysqli_query($conectar,$listarCorreos);
while ( $row = mysqli_fetch_array($resultado) ) {
$emails[]=$row['userEmail'];
}
echo json_encode($emails);

use the code like below
$conectar = mysqli_connect(HOST, USER, PASS, DATABASE);
$listarCorreos = " SELECT userEmail
FROM usuarios
";
$resultado = mysqli_query($conectar,$listarCorreos);
while ( $row = mysqli_fetch_array($resultado) ) {
$emails[]=$row['userEmail '];
}
echo json_encode($emails);

datatables + serverside processing + search filtering

I am following this tutorial:
http://coderexample.com/datatable-demo-server-side-in-phpmysql-and-ajax/
and here is the demo:
http://coderexample.com/demo/datatable-demo-server-side-in-phpmysql-and-ajax/
If i search for ou in the search input i get No matching records found but what I would like returned is this row Airi Satou at least.
this is the code that i have to change, I think, as I have to do the search server side.
<?php
/* Database connection start */
$servername = "localhost";
$username = "root";
$password = "Password1";
$dbname = "test";
$conn = mysqli_connect($servername, $username, $password, $dbname) or die("Connection failed: " . mysqli_connect_error());
/* Database connection end */
// storing request (ie, get/post) global array to a variable
$requestData= $_REQUEST;
$columns = array(
// datatable column index => database column name
0 =>'employee_name',
1 => 'employee_salary',
2=> 'employee_age'
);
// getting total number records without any search
$sql = "SELECT employee_name, employee_salary, employee_age ";
$sql.=" FROM employee";
$query=mysqli_query($conn, $sql) or die("employee-grid-data.php: get employees");
$totalData = mysqli_num_rows($query);
$totalFiltered = $totalData; // when there is no search parameter then total number rows = total number filtered rows.
if( !empty($requestData['search']['value']) ) {
// if there is a search parameter
$sql = "SELECT employee_name, employee_salary, employee_age ";
$sql.=" FROM employee";
$sql.=" WHERE employee_name LIKE '".$requestData['search']['value']."%' "; // $requestData['search']['value'] contains search parameter
$sql.=" OR employee_salary LIKE '".$requestData['search']['value']."%' ";
$sql.=" OR employee_age LIKE '".$requestData['search']['value']."%' ";
$query=mysqli_query($conn, $sql) or die("employee-grid-data.php: get employees");
$totalFiltered = mysqli_num_rows($query); // when there is a search parameter then we have to modify total number filtered rows as per search result without limit in the query
$sql.=" ORDER BY ". $columns[$requestData['order'][0]['column']]." ".$requestData['order'][0]['dir']." LIMIT ".$requestData['start']." ,".$requestData['length']." "; // $requestData['order'][0]['column'] contains colmun index, $requestData['order'][0]['dir'] contains order such as asc/desc , $requestData['start'] contains start row number ,$requestData['length'] contains limit length.
$query=mysqli_query($conn, $sql) or die("employee-grid-data.php: get employees"); // again run query with limit
} else {
$sql = "SELECT employee_name, employee_salary, employee_age ";
$sql.=" FROM employee";
$sql.=" ORDER BY ". $columns[$requestData['order'][0]['column']]." ".$requestData['order'][0]['dir']." LIMIT ".$requestData['start']." ,".$requestData['length']." ";
$query=mysqli_query($conn, $sql) or die("employee-grid-data.php: get employees");
}
$data = array();
while( $row=mysqli_fetch_array($query) ) { // preparing an array
$nestedData=array();
$nestedData[] = $row["employee_name"];
$nestedData[] = $row["employee_salary"];
$nestedData[] = $row["employee_age"];
$data[] = $nestedData;
}
$json_data = array(
"draw" => intval( $requestData['draw'] ), // for every request/draw by clientside , they send a number as a parameter, when they recieve a response/data they first check the draw number, so we are sending same number in draw.
"recordsTotal" => intval( $totalData ), // total number of records
"recordsFiltered" => intval( $totalFiltered ), // total number of records after searching, if there is no searching then totalFiltered = totalData
"data" => $data // total data array
);
echo json_encode($json_data); // send data as json format
?>
Am I right to say that this is the code I have to change?
If so, can anyone advise what I have to do?
I understand this is alot to ask but would appreciate the guidance!

$sql.=" WHERE employee_name LIKE '".$requestData['search']['value']."%' ";
will match search term then anything (due to wild card %)
as you want to match the search term in the middle of the name you need to add the wild card at the begging as well:
$sql.=" WHERE employee_name LIKE '%".$requestData['search']['value']."%' ";
Note this will disable the use of indexes on employee_name which may or may not be a problem for you.
This is not the best search approach over all, you should not check all three fields, but ask the searcher which to use. After all age and salary could have some matching numbers.
a search for 27, could match age 27 or 27000 salary etc. And no one will have an age of bob so its pointless to do that search

Your query need to be updated.
The following SQL statement selects all with a employee_name starting with "search field value":
$sql.=" WHERE employee_name LIKE '".$requestData['search']['value']."%' ";
so, your query should be something like below, to get desired output, as it check whether a particular pattern is present in the employee_name.
$sql.=" WHERE employee_name LIKE '%".$requestData['search']['value']."%' ";

The way you are constructing your Where clause, the fields you are searching must start with the search term. Change your where clauses to
$sql.=" WHERE employee_name LIKE '%".$requestData['search']['value']."%' ";
You are also open to SQL injection by doing what you are doing. You need to use parameterized queries to remove this vulnerability. See How can I prevent SQL injection in PHP? for information

Count query in php

I would like to seek some help with my query problem in php...I want to do is count all (atic) with the same number and if the specific (atic) is equal to 7 my Insert query will execute to that (atic).
The problem is my count query wont work as i wanted....and execute my insert query to all aic even the count is not = 7.
current code:
<?php
mysql_connect("localhost","root","") or die("cant connect!");
mysql_select_db("klayton") or die("cant find database!");
$total = NULL;
$count = "SELECT count(t.atic) as '$total', t2.name FROM app_interview as t, tb_applicants as t2 WHERE t.atic = t2.aic GROUP BY t.atic";
$query = mysql_query($count) or die (mysql_error());
while($rows =mysql_fetch_array($query)){
if($query = 7){
mysql_query("INSERT INTO afnup_worksheet (faic,fnl_name,interview,fregion,ftown,funiq_id,fposition,fsalary_grade,fsalary,dateinputed) SELECT DISTINCT atic, atname,(SELECT sum(inttotal) FROM app_interview t2 WHERE t2.atic = t.atic)/7, region, town, uniq_id, position, salary_grade, salary, CURRENT_TIMESTAMP FROM app_interview t GROUP BY atname HAVING COUNT(DISTINCT atic)");
}
}
?>

$query = 7 means you are assigning value 7 to variable $query
In order to compare you have to use double equal sign ==
or triple equal signs === for same data type.

if($query = 7)
means you assign $query is 7 and it will always true and always executed, you can try this
if( $query == 7 )
That means if $query value is equal to 7 or not

Invalid JSON format from PHP json encode

I have a PHP file that encodes Json data and when i view the JSON output when its a single data block i get a valid json code syntax this is an example :
single data block
But when the JSON results in a multiple data block it generates an invalid JSON format like this: multiple data blocks
This is my PHP code:
<?php
header('Content-Type: application/json; charset=utf-8', true,200);
DEFINE('DATABASE_USER', 'xxxxx');
DEFINE('DATABASE_PASSWORD', 'xxxxxx');
DEFINE('DATABASE_HOST', 'xxxxxxxxxxx');
DEFINE('DATABASE_NAME', 'xxxxxxxx');
// Make the connection:
$dbc = #mysqli_connect(DATABASE_HOST, DATABASE_USER, DATABASE_PASSWORD,
DATABASE_NAME);
$dbc->set_charset("utf8");
if (!$dbc) {
trigger_error('Could not connect to MySQL: ' . mysqli_connect_error());
}
if(isset($_GET['keyword'])){//IF the url contains the parameter "keyword"
$keyword = trim($_GET['keyword']) ;//Remove any extra space
$keyword = mysqli_real_escape_string($dbc, $keyword);//Some validation
$query = "select name,franco,alpha,id,url,songkey,chord from song where name like '%$keyword%' or franco like '%$keyword%'";
//The SQL Query that will search for the word typed by the user .
$result = mysqli_query($dbc,$query);//Run the Query
if($result){//If query successfull
if(mysqli_affected_rows($dbc)!=0){//and if at least one record is found
while($row = mysqli_fetch_array($result,MYSQLI_ASSOC)){ //Display the record
$data = array();
$data = $row;
echo $_GET[$callback]. ''.json_encode($data).'';
}
}else {
echo 'No Results for :"'.$_GET['keyword'].'"';//No Match found in the Database
}
}
}else {
echo 'Parameter Missing in the URL';//If URL is invalid
}
?>

It is because you are JSON-encoding a single line of the result set at at time. This is not a valid JSON structure if the calling client is expecting such.
Likely, you will want to put each row as an entry in an array, and then JSON-encode and echo the resulting array.
Like this:
if($result){//If query successfull
if(mysqli_affected_rows($dbc)!=0){//and if at least one record is found
$array = array();
while($row = mysqli_fetch_array($result,MYSQLI_ASSOC)){ //Display the record
$array[] = $row;
}
echo json_encode($array);
}
}