CODEWITHSUNDEEP
phpmysql
Ben
Ben

Reputation: 2564

mysql passing array into LIKE without loop

$serach=array('abc','def','ghi');

$num=count($search);
for($n=0, $n<$num, $n++){
  $sql .= LIKE % $search[$n] % OR ;
}
$sql=substr_replace($sql,"",-3);//REMOVE LAST OR


SELECT * FROM store_main WHERE name :sql...

i need pass an array into query LIKE, i have use loop to create the statement and put into the query, my question is, is any way do it without loop, so i can make my query more simple.

Upvotes: 0

Views: 67

Answers (2)

Class
Class

Reputation: 3160

Use implode like so:

$serach = array('abc', 'def', 'ghi');
$sql .= "`name` LIKE '%" . implode("%' OR `name` LIKE '%", $serach) . "%' ";

produces:

`name` LIKE '%abc%' OR `name` LIKE '%def%' OR `name` LIKE '%ghi%'

OR you can use REGEXP. This one could be a bit slower depending on how many array elements you have and/or your table/database's efficiency.

$serach = array('abc', 'def', 'ghi');
$sql .= "`name` REGEXP '" . implode("|" $serach) . "' ";

produces:

`name` REGEXP 'abc|def|ghi'

Upvotes: 1

Damodaran
Damodaran

Reputation: 11065

Try

$array = array('lastname', 'email', 'phone');
echo "%".str_replace(",","% OR %",implode(",", $array))."%";

Output

%lastname% OR %email% OR %phone%

Refer implode and str-replace

Upvotes: 1

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