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I am getting the notice in PHP code like this. Notice: Undefined index: query on line 2.

Here is my code. 

// gets value sent over search form

$min_length = 1;
// you can set minimum length of the query if you want

if(strlen($query) >= $min_length){ // if query length is more or equal minimum length then

    $query = htmlspecialchars($query); 
    // changes characters used in html to their equivalents, for example: < to &gt;

    $query = mysqli_real_escape_string($con,$query);
    // makes sure nobody uses SQL injection

    $raw_results = mysqli_query($con,"SELECT * FROM dashboard
        WHERE (`company_name` LIKE '%".$query."%')OR (`total_device` LIKE '%".$query."%')OR
        (`working_device` LIKE '%".$query."%')") or die(mysql_error());

    // * means that it selects all fields, you can also write: `id`, `company_name`, `total_device`,`working_device`
    // articles is the name of our table

    // '%$query%' is what we're looking for, % means anything, for example if $query is Hello
    // it will match "hello", "Hello man", "gogohello", if you want exact match use `title`='$query'
    // or if you want to match just full word so "gogohello" is out use '% $query %' ...OR ... '$query %' ... OR ... '% $query'

    if(mysqli_num_rows($raw_results) > 0){ // if one or more rows are returned do following

 while($results = mysqli_fetch_array($raw_results)){ ?>
<div class="wrap">

Working Devices
Total Devices


Any help please..

priya
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    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) – RJParikh Jul 20 '17 at 05:53
  • you doesn't define $query variable – GYaN Jul 20 '17 at 05:55
  • @priya you need to define Your $query Variable before you check in `if(strlen($query) >= $min_length)` condition. – Jaykumar Gondaliya Jul 20 '17 at 05:57
  • Please format your code properly (parts of it are not visible). You are mixing APIs (`mysql_error()`). Show us the code of your form so we can help you. – MrDarkLynx Jul 20 '17 at 05:58
  • You did't define $query variable and trying to access it directly. – akshay saxena Jul 20 '17 at 05:58

1 Answers1

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Thanks to everyone. The problem resolved with this.

$query = isset($_POST['query'])? $_POST['query'] : "";

priya
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