PHP PDO MySQL 查询 LIKE -> 多个关键字

Question

I have a users table in MySQL and would like a search by name. Right now I have the following code:

<?php
$search = @$_GET['q'];
$search = strtoupper($search);
$search = strip_tags($search);
$search = trim($search);

$query = $con->prepare('SELECT * FROM `users` WHERE name LIKE ?');
$query->execute(array('%'.$search.'%'));

$result = $query->rowCount();
echo $result;
?>

The problem is that I want to have multiple keywords. Let's say someone types "Here should be a nice name of a person" then it would search for "here", "should", "be" etc. and display results for every row where there words are in 'name' column. I searched on the web and read that it is possible to do "OR name LIKE ?" as many times as the keywords, but I could not really get it working and I'm not sure if it is optimized enough with ~20 words (in case they search with that many words). If it should be used, can you help me change my code so it would search for every word independently?

Thank you!

EDIT:

I was able to fix this issue by one guy who posted in this thread. The following solution works for me:

<?php
$search = isset($_POST['q']) ? $_POST['q'] : '';
$search = strtoupper($search);
$search = strip_tags($search);
$search = trim($search);

$split_words = explode(" ", $search);

if(count($split_words) > 0) {

    $sql = "SELECT * FROM `users` WHERE ";

    for($i=0 ; $i < count($split_words); $i++){
        $sql .= " name LIKE ? OR";
    }

    $sql = substr($sql , 0, -3); //Remove last 3 characters OR with space
    array_walk($split_words, "addPercentage");


    $query = $con->prepare($sql);
    $query->execute($split_words);
}

function addPercentage(&$value, $key) {
    $value = '%'.$value.'%';
}
?>

Answer 1

You shouldn't use @ to silence errors it is a bad practice, check if the value is set. The example below should work, but the results might not be all that relevant.

$search = isset($_GET['q']) ? $_GET['q'] : ''; 
$search = strtoupper($search);
$search = strip_tags($search); 
$search = trim($search);
$words = explode(' ', $search);
$words_condition = array();
$arguments = array();
foreach ($words as $word) {
    $words_condition[] = 'name LIKE ?';
    $arguments[] = '%'.$word.'%';
}

$query = $con->prepare('SELECT * FROM `users` WHERE '.implode(' OR ', $words_condition));
$query->execute($arguments);

$result = $query->rowCount();
echo $result;

Answer 2

$words = explode(" ", $search);

$i = 0;
while($words[i] != null)
{
    //Query where name LIKE words[i]
}

Answer 3

Please check the below code.

<?php
$search = isset($_GET['q']) ? $_GET['q'] : '';
$search = strtoupper($search);
$search = strip_tags($search);
$search = trim($search);

$split_words = explode(" ", $search);

$query = "SELECT * FROM `users`";

if(count($split_words) > 0){
 $query .= " WHERE "
 for($i=0 ; $i < $split_words; $i++){
   $query .= " name LIKE ? OR ";
 }
 $query = substr($query , 0, -3); //Remove last 3 characters OR with space


 array_walk($split_words,"addPercentage");

 $query->execute($split_words);
}else{
 $query->execute();
}

$result = $query->rowCount();
echo $result;

function addPercentage(&$value,$key)
{
  $value = "%".$value."%" ;
}
?>