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how to send a variable which have not just usual string, from Javascript to PHP and so to be able using in SQL query

I want to send my variable from javascript to PHP using post method that needs to be used in SQL query, but my variable value has a special character like " / " that server doesn't perform it,
it's my sample code:
js:
var my_var = "ok1 / ok2_4"; // its my table name in database
$.post(
"part/order/order_from_js.php",
{
name: my_var,
},
function(data) {
// using data ....
}
php:
<?php
session_start();
$db=mysqli_connect("localhost","user_db","","db_name");
$name = mysqli_real_escape_string($db, $_POST['name']);// this is my main problem somehow
$select_query = "SELECT * FROM $name";
$users_query = $db->query($select_query);
if ($users_query ->num_rows > 0) {
$i=0;
while($row = $users_query->fetch_assoc()) {
$array[$i]= $row;
$i=$i+1;
}
}
mysqli_close($db);
echo json_encode($array);
?>
I have an error and nothing will pass back to javascript. I want to persist using mysqli_real_escape_string for avoid SQL injection attack.

AJAX function for retrieving postgres data not working

I have a simple AJAX function bound to a button that should execute a PostgreSQL query. However, when I click the button that I bound the ajax query to, all I get is the confirmation that the database connection was successful. Nothing seems to happen withe the ajax result (should be printing to console in the handleAjax() function. What am I doing wrong?
This is the javascript code (with jquery):
$(document).ready(function() {
function sendAjax() {
$.ajax({
url: "db/database.php",
success: function (result) {
handleAjax(result);
}
});
}
function handleAjax(result) {
console.log(result);
}
$("#submit-button").on("click", sendAjax);
});
And this it the contents of database.php:
<?php
function dbconn(){
ini_set('display_errors', 1); // Displays errors
//database login info
$host = 'localhost';
$port = 5432;
$dbname = 'sms';
$user = 'postgres';
$password = 'postgres';
// establish connection
$conn = pg_connect("host=$host port=$port dbname=$dbname user=$user password=$password");
if (!$conn) {
echo "Not connected : " . pg_error();
exit;
} else {
echo "Connected.";
}
}
$conn = dbconn();
$sql = "SELECT * FROM numbers;";
$result = pg_query( $sql ) or die('Query Failed: ' .pg_last_error());
$count = 0;
$text = 'error';
while( $row = pg_fetch_array( $result, null, PGSQL_ASSOC ) ) {
$text = $row['message'];
//echo $text;
}
pg_free_result( $result );
?>

The problem is in the database.php file, all you get is "Connected." because you don't print your result at the end. Ajax only receive the output of the php file.
So at the end of your php file you should add :
echo $text;
And you also should remove the echo "Connected.";

AJAX is not a magic wand that in magic way reads PHP code. Let's say AJAX is a user. So what does user do.
Open web page
Wait until PHP execute code and display data
Tells you what he sees
If you don't display anything, ajax can't tell you what he saw.
In thi's place is worth to say that the best way to communicate between PHP and AJAX is using JSON format.
Your code generally is good. All you have to do is to display your data. All your data is in your $text var. So let's convert your array ($text) to JSON.
header('Content-Type: application/json');
echo json_encode($text);
First you set content-type to json, so ajax knows that he reads json. Then you encode (convert) your PHP array to js-friendly format (JSON). Also delete unnecessary echoes like 'Conntected' because as I said, AJAX reads everything what he sees.

You should return $conn from dbconn()
if (!$conn) {
echo "Not connected : " . pg_error();
exit;
} else {
echo "Connected.";
return $conn;
}

mysql query from php isnt working but it works in a different php file

I'm trying to make a query from my MySQL database here is the code
<?php
$link = new MySQLi('localhost','root','Rrtynt','copy');
if(isset($_POST['id'])){
$name = $_POST['id'];
$profile = 'profile';
$thestring = $name.$profile;
//echo $thestring;
$result = $link->query("SELECT Email,Name,idauth FROM user WHERE Email = '$name'");
echo $result;
}
?>
the code for the query
$result = $link->query("SELECT Email,Name,idauth FROM user WHERE Email = '$name'");
works in a different php script the same exact code but it keeps giving me http 500 error, I'm using this to post from a javascript file
$.post("/getfirstfolder.php", { id: value1 }, function (data) {
cop = data;
console.log("Data: " + data);
});
if I take out
$result = $link->query("SELECT Email,Name,idauth FROM user WHERE Email = '$name'");
and just echo $thestring it works fine, I cant figure out the problem so thank you for your time and your help is greatly appreciated

The fact that you mentioned $thestring, makes me think that perhaps you should be using $thestring as the parameter to the query like this:
$result = $link->query("SELECT Email,Name,idauth FROM user WHERE Email = '$thestring'");

php string comparison search only returning one entry not returning correct entries

I'm using ajax to send a search string to a php script that executes a mysql like function to find all related entries with the username like the string being sent for friend searching. I have two current entries in the database zukeru and zukeru2. when i search z i only get zukeru returned in my console output. When i search 2 i still get zukeru and im really not sure why.
Also how to i remove a specific field from a php nested tupple. I don't want to include the password field for obvious reason. Sorry im new to php learning as i go so far its not as bad as I thought it would be kinda similar to python.
returned object when searching the number 2, but i get zukeru and not zukeru2 doesn't make sense.
Object {0: "2", 1: "you wish you could see", 2: "zukeru", 3: "deleted for security", 4: "grant", id: "2", email: "deleted for security", username: "zukeru", password: "deleted for security", name: "grant"}
this is the search string i used for the above result. You can see i searched 2 and got back zukeru and not zukeru2
profile.php:92 searchstring=2
<?php
$db = new mysqli(security reasons removed.);
extract($_POST);
//I think i can remove this session start ?
session_start();
$serach_string = $_POST['searchstring'];
$fetch=$db->query("SELECT * FROM users WHERE username LIKE '%$serach_string%'");
$friends=mysqli_fetch_array($fetch);
//echo $search_string
echo json_encode($friends);
?>
Here is my jquery incase you wanted to see
function search(){
var url = "search_friends.php";
$.ajax({
type: "POST",
url: url,
data: $("#search_friends").serialize(), // serializes the form's elements.
success: function(data)
{
//console.log(data);
var returned_friends = JSON.parse(data);
var html_built = '<br>';
console.log(returned_friends);
console.log($("#search_friends").serialize());
if (returned_friends){
$.each( returned_friends, function( key, value ) {
if (key =="username"){
html_built += '<li><a href="#"><button class="btn btn-primary" style="width:100%;" id="'+value+'" onClick="add_friend(this.id)"> Send '+value+' A Friend Request</button></li>';
}
});
}
html_built += ""
document.getElementById("list_friends").innerHTML = html_built;
}
});
return false;
}
this is what im currently using and I get undefined method. It cant find fetch_all(); and im using php 5.4
here is the console error returned.
<br />
<b>Fatal error</b>: Call to undefined method mysqli_result::fetch_all() in <b>/home/gzukel/public_html/search_friends.php</b> on line <b>7</b><br />
<?php
$db = new mysqli();
extract($_POST);
session_start();
$serach_string = $_POST['searchstring'];
if($fetch=$db->query("SELECT username FROM users WHERE username LIKE '%$serach_string%'")){
$friends=$fetch->fetch_all();
echo json_encode($friends);
}else{
echo 'no results';
}
?>

so something like this?
<?php
$db = new mysqli();
extract($_POST);
session_start();
$serach_string = $_POST['searchstring'];
$fetch=$db->query("SELECT * FROM users WHERE username LIKE '%$serach_string%'");
$friends=[]
while($row = $fetch->fetch_array())
{
$rows[] = $row;
}
foreach($rows as $row)
{
array_push($friends,$row['username']);
}
//echo $search_string
echo json_encode($friends);
?>

You Could use fetch all:
if($fetch=$db->query("SELECT username FROM users WHERE username LIKE '%$serach_string%'")){
$friends= $fetch->fetch_all();
echo json_encode($friends);
}else{
echo 'no results';
}

Php, js redirecting after successful registering, It inserts the user but won't redirect

<?php
$con = mysqli_connect("localhost", "root", "", "" ) or die("Neuspjelo spajanje");
function InsertUser(){ global $con;
if(isset($_POST['sign_up'])){
$name = mysqli_real_escape_string($con, $_POST['u_name']);
$pass = mysqli_real_escape_string($con,$_POST['u_pass']);
$email = mysqli_real_escape_string($con,$_POST['u_email']);
$country = mysqli_real_escape_string($con,$_POST['u_country']);
$gender = mysqli_real_escape_string($con,$_POST['u_gender']);
$b_day = mysqli_real_escape_string($con,$_POST['u_birthday']);
$date = date("m-d-Y");
$status = "unverified";
$posts = "No";
$get_email = "select * from users where user_email='$email'";
$run_email = mysqli_query($con, $get_email);
$check = mysqli_num_rows($run_email);
$insert = "insert into users (user_name, user_pass, user_email, user_country, user_gender, user_b_day,
user_image, register_date, last_login, status, posts) values
('$name','$pass', '$email', '$country', '$gender', '$b_day', 'default.jpg',
'$date', '$date', '$status', '$posts')";
$run_insert = mysqli_query($con, $insert);
$result = mysql_query($insert);
if($result){
echo "<script>alert ('You're successfully registered!')</script>";
echo "<script>window.open('home.php', '_self')</script>";
}
}
}
?>

You can't echo javascript and run it in a page that's already loaded. This would need to be the result of an ajax call on the client side with your redirects occuring from your ajax callbacks.
If you're ok with ditching the alert, you can just issue a redirect from php:
header('Location: home.php');
To do it ajaxy:
$.ajax({
type: "GET",
url: "your_insert_user.php"
}).success(function(xhr) {
alert ("You're successfully registered!");
window.open('home.php', '_self');
}).fail(function (jqXHR, status, errorThrown) {
//something else here
});
But, why would you want to issue an ajax call just to redirect?
Additionally, you need to issue the appropriate responses from your insert script:
if ($result) { echo ""; } //issues a "200 OK"
else { header("HTTP/1.1 422 Unprocessable Entity"); } //fires the failure callback in ajax
I would pass a conditional GET or POST paramater to home.php with some value flag and display your message there.

Based on what you post above, you are dealing with two separate issues here.
You say "it inserts" so I'm assuming that means that the mysql query to insert the new row into your database completes successfully. Then you send some HTML code, containing a (somewhat mangled) Javascript snippet, to the browser, which is supposed to issue a redirect request to the client's web browser, which doesn't have the desired result, seeing as you write that it "won't redirect".
Keep in mind that redirection is performed by the browser, is dependent on the browser's capabilities and/or settings, and requires proper javascript in the first place.
How do properly request a redirect from the browser has been discussed before on SO.

First of all,remove this line $result = mysql_query($insert); then modify your code and add this, hope it will work:
$run_insert = mysqli_query($con, $insert);
if($run_insert){
echo "<script>alert ('You\'re successfully registered!')</script>";
echo "<script>window.open('home.php', '_self')</script>";
}