Undefined index error in php although i defined it.why?

Question

I am new to php so please guide and don't make it duplicate because i can't get solution from previous solutions.My php code is given below

 <?php 
if($_SERVER['REQUEST_METHOD']=='GET'){
 $city  = $_GET['city'];
 $town  = $_GET['town'];
//$skills=$_POST['skills'];
require_once('DbConnect.php'); 
 //Creating sql query
 $sql = "SELECT FROM employees where city='".$city."' and town='".$town."'";
 
 //getting result 
 $r = mysqli_query($con,$sql);
 
 //creating a blank array 
 $result = array();
 
 //looping through all the records fetched
 while($row = mysqli_fetch_array($r)){
 
 //Pushing name and id in the blank array created 
 array_push($result,array(
 "name"=>$row['name'],
 "phone"=>$row['phone'],
 "skills"=>$row['skills']
 ));
 }
 
 //Displaying the array in json format 
 echo json_encode(array('result'=>$result));
 
 mysqli_close($con);
 }
 ?>

ERROR

Notice: Undefined index: city in C:\xampp\htdocs\getServices\employeesInfo.php on line 3

Notice: Undefined index: town in C:\xampp\htdocs\getServices\employeesInfo.php on line 4

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in C:\xampp\htdocs\getServices\employeesInfo.php on line 17 {"result":[]}

Change your SQL query to `$sql = "SELECT * FROM employees ...` — Rajdeep Paul, Sep 10 '17 at 17:04
Whats the URL or form? You are open to SQL injections, parameterize. Just checking the request method isnt enough. — chris85, Sep 10 '17 at 17:05
Besides the obvious SQL injection issues, if there are no get vars for city or town defined in your URL, then you will get this error — Mark Baker, Sep 10 '17 at 17:06
Use `die` method with `$_GET` to see the contents and verify if city & town keys are present. — Deepansh Sachdeva, Sep 10 '17 at 17:07
@DeepanshSachdeva OP will need `print_r` as well, `die` alone cant print array contents. — chris85, Sep 10 '17 at 17:10
@RajdeepPaul thanks but i am still getting Undefined index error on"city & town".please explain — Abdul Mateen, Sep 10 '17 at 17:11
Also would you ever have a city and town; aren't those the same thing, just different government structures? — chris85, Sep 10 '17 at 17:13
See @chris85's [comment there](https://stackoverflow.com/questions/46143287/undefined-index-error-in-php-although-i-defined-it-why?noredirect=1#comment79248357_46143287). Also, do `var_dump($_GET);` and see what you have in there. — Rajdeep Paul, Sep 10 '17 at 17:13
You're already using an API that supports **prepared statements** with bounded variable input, you should utilize parameterized queries with placeholders (prepared statements) to protect your database against [SQL-injection](http://stackoverflow.com/q/60174/)! Get started with [`mysqli::prepare()`](http://php.net/mysqli.prepare) and [`mysqli_stmt::bind_param()`](http://php.net/mysqli-stmt.bind-param). — Qirel, Sep 10 '17 at 17:34
Possible duplicate of [PHP: "Notice: Undefined variable", "Notice: Undefined index", and "Notice: Undefined offset"](https://stackoverflow.com/questions/4261133/php-notice-undefined-variable-notice-undefined-index-and-notice-undef) — Qirel, Sep 10 '17 at 17:34

score 2 · Accepted Answer · edited Sep 10 '17 at 17:34

You need to select one or more columns, for example by doing select all SELECT * FROM.., the query would look like this

$sql = "SELECT * FROM employees where city='".$city."' and town='".$town."'";

Update_Code:

<?php

if($_SERVER['REQUEST_METHOD']=='GET' && !empty($_GET['city']) && !empty($_GET['town'])){

 $city  = $_GET['city'];
 $town  = $_GET['town'];

//$skills=$_POST['skills'];
require_once('DbConnect.php'); 

 //Creating sql query 
 $sql = "SELECT * FROM employees where city='".$city."' and town='".$town."'";

 //getting result 
 $r = mysqli_query($con,$sql);

 //creating a blank array 
 $result = array();

 //looping through all the records fetched
 while($row = mysqli_fetch_array($r)){

 //Pushing name and id in the blank array created 
 array_push($result,array(
 "name"=>$row['name'],
 "phone"=>$row['phone'],
 "skills"=>$row['skills']
 ));
 }

 //Displaying the array in json format 
 echo json_encode(array('result'=>$result));

 mysqli_close($con);
 }
 else{

 $message = " Fill all the details First";

 }

if(isset($message) && !empty($message) ){

echo "$message";
}

?>

Undefined index error in php although i defined it.why?

1 Answers1