MySQL with PHP - LIKE is not working properly

Question

I'd like to search a MySQL database to match keywords passed by the user. I heard that using LIKE is the fastest option but can't find an example of a full simple query using LIKE in PHP code.

This is what I am tring:

$value = 'Fire';
$result = mysql_query("SELECT * FROM effects WHERE title LIKE 'value%'");

I know there is an row in the database with Fire as the title but the query is returning null.

Can someone please give me an example of how to perform a MySQL LIKE search or alternative to find rows by keyword(s).

Thank you.

'value%' should be '$value%'.. you also need to sanitize user inputs to avoid SQL injections or use PDO prepared statements. — tradyblix
– tradyblix, Commented Jul 1, 2012 at 6:31
ok so this is the new example $value = mysql_real_escape_string('fire'); $result = mysql_query("SELECT * FROM effects WHERE title LIKE '$value%'"); — codelove
– codelove, Commented Jul 1, 2012 at 6:37
you can read about it here.. that way you can avoid using mysql_* functions. — tradyblix
– tradyblix, Commented Jul 1, 2012 at 6:42
@codelove : Say there is title like "My Fire Effect". You don't want this in search result? — Fahim Parkar
– Fahim Parkar, Commented Jul 1, 2012 at 7:03

Yan Berk · Accepted Answer · 2012-07-01 06:31:26Z

3

You need to add the dollar sign of the variable:

$result = mysql_query("SELECT * FROM effects WHERE title LIKE '$value%'");

answered Jul 1, 2012 at 6:31

Yan Berk

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6 Comments

Anonymous Over a year ago

This query is not valid. Use Fahim's answer.

MaxSem Over a year ago

@Bailey: I've tweaked the answer. You can do it yourself, too.

Yan Berk Over a year ago

@Bailey The original question consisted only of the word 'Fire'.

MaxSem Over a year ago

@Bailey: my edit is waiting for review, you're unable to see it yet.

Anonymous Over a year ago

Still waiting for review? '$value%' isn't valid. It's '%$value%'

|

Fahim Parkar · Accepted Answer · 2012-07-01 07:05:43Z

2

You are missing $ before variable value.

Your query should look like

$value = 'Fire';
$result = mysql_query("SELECT * FROM effects WHERE title LIKE '$value%'");

Say there is title like My Fire Effect. You don't want this in search result? If you want to display My Fire Effect in result too then you should use % before $value

$value = 'Fire';
$result = mysql_query("SELECT * FROM effects WHERE title LIKE '%$value%'");

Hope this helps you.

For Differences, see demo with live example

Note, first query in demo returns me 5 rows, however second query returns me 8 rows which is PERFECT.

answered Jul 1, 2012 at 7:05

Fahim Parkar

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