Getting error in search script [php, mysqli]

Question

I am trying to create a very simple and basic search script.

Here is what I have done so far:

include('config.php');

$search_token =$_POST['search_token'];

$search_query = mysqli_query($conn, "SELECT Forname FROM idea WHERE post_des LIKE '%$search_token%'");

while($row = mysqli_fetch_array($search_query))
  {

  $idea_body = $row['post_des'];

  echo $idea_body;
}

But when I execute this, I am having the following Warning:

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in F:\Server\xampp\htdocs\p\c3\search.php on line 34

Any help for this situation?

Be alarm about $search_token = $_POST['search_token']. You could SQL-inject yourself :) — Pedram Behroozi
– Pedram Behroozi, Commented Dec 18, 2013 at 7:11

Ignacio Ocampo · Accepted Answer · 2013-12-18 07:00:57Z

1

You probably have a problem in your query, try:

$search_query = mysqli_query($conn, "SELECT Forname FROM idea WHERE post_des LIKE '%$search_token%'") or die(mysqli_error($conn));

To debug your query.

answered Dec 18, 2013 at 7:00

Ignacio Ocampo

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2 Comments

R R Over a year ago

i think this is probably more suitable for comment.

MKM Over a year ago

oh my!! This fixed that. Thanks a lot!

Naveenkumar · Accepted Answer · 2013-12-18 07:05:03Z

0

Your query is failing and returning a false value.

put this after your mysqli_query() to see whats going on.

if (!$search_query) {
    printf("Error: %s\n", mysqli_error($con));
    exit();
}

for more information.

http://www.php.net/manual/en/mysqli.error.php

answered Dec 18, 2013 at 7:05

Naveenkumar

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