PHP query error

Question

I am using LIKE to do my searching, i try it in phpMyAdmin and return the result but when i use it in php it return empty result.

$search = "ip";
$start = 0;
$query = "SELECT * FROM product WHERE product_name LIKE '%$search%' LIMIT $start,30";
$result = mysql_query($query);
if(empty($result))
    $nrows = 0;
else
    $nrows = mysql_num_rows($result);

It will return result when i using phpMyAdmin to run this query but when i use it in php, it return empty.

Update:

Sorry guys,

I just found out the problem is i didn't connect database as well. anyway, thanks for helping.

If you check the value of $query, is it identical to what you used in phpmyadmin? — Adam Bergmark, Apr 27 '11 at 09:19
Did you try to print out `$query` before execute it? Are you sure it's exactly written as you need? — Marco, Apr 27 '11 at 09:19
What does `$search` contain? What does `$start` contain? Also note you will need to sanitize both values to protect against SQL injection http://stackoverflow.com/questions/601300/what-is-sql-injection — Pekka, Apr 27 '11 at 09:20
try this `$query = "SELECT * FROM product WHERE product_name LIKE %".$search."% LIMIT $start,30";` — ianace, Apr 27 '11 at 09:22
You haven't returned the result anywhere here in your code.. $results only stores the object of the resultset and you have either user mysql_fetch_array or mysql_fetch_object .. Make sure you included that.. — Vijay, Apr 27 '11 at 09:26

Dipesh · Answer 1 · 2016-10-04T14:09:10.800

2

Try This

 $query = "SELECT * FROM `product` WHERE `product_name` LIKE '%".$search."%' LIMIT 0, 30";

edited Oct 04 '16 at 14:09

answered Oct 04 '16 at 14:00

Dipesh

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5
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ImBhavin95 · Answer 2 · 2016-10-05T08:21:05.643

0

$search = "ip";
$start = '0';
$query = "SELECT * FROM product WHERE product_name LIKE '%".$search."%' LIMIT $start,30";
$result = mysql_query($query)or die(mysql_error());
if(mysql_num_rows($result) == 0){
    $nrows = 0;
} else{
    $nrows = mysql_num_rows($result);
}

//use mysql_num_rows($result) instead of empty($result) because in this situation $result is every time not empty so use inbuilt PHP function mysql_num_rows($result);

edited Oct 05 '16 at 08:21

answered Oct 04 '16 at 14:05

ImBhavin95

1,403
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25

Although this code may help to solve the problem, it doesn't explain _why_ and/or _how_ it answers the question. Providing this additional context would significantly improve its long-term educational value. Please [edit] your answer to add explanation, including what limitations and assumptions apply. – Toby Speight Oct 04 '16 at 15:03
@TobySpeight if you satisfy with my answer edit reply positively. thx bro. – ImBhavin95 Oct 05 '16 at 08:23

score 0 · Answer 3 · answered Apr 27 '11 at 09:26

And if the sole purpose of your code is to get the number of products with the searched-for name, use SELECT COUNT(*) instead of doing a mysql_num_rows() on all your data. It will decrease your querytime and the amount of data that is (unnecessarily) fetched.

score 0 · Answer 4 · answered Apr 27 '11 at 09:31

I am not sure why this is not working, as the query seems to be correct to me. I would like to suggest you writing query this way

$query = <<<SQL

 SELECT * FROM product WHERE product_name LIKE "%$search%" LIMIT $start,30

SQL;

please note that there should not be any space or any character after SQL;

score 0 · Answer 5 · answered Apr 27 '11 at 09:34

0

$query = "SELECT * FROM product WHERE product_name LIKE '%" . $search . "%' LIMIT " . (int) $start. ",30";

answered Apr 27 '11 at 09:34

Flask

4,891
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score 0 · Answer 6 · answered Apr 27 '11 at 09:38

0

you can use directly mysql_num_rows()

but here is right code

$query = "SELECT * FROM product WHERE product_name LIKE '%".$search."%' LIMIT $start,30";

answered Apr 27 '11 at 09:38

Noor Khan

542
1
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PHP query error

6 Answers6