Search MySQL database with LIKE statement - php

Question

I need to return search results of users in my database.
At the moment, I have managed to use the LIKE function in the MySQL query to return a results count, but I need it to echo the usernames of all the results, not just the first result:
<?php
// mysql details
$host="localhost"; // Host name
$username="classified"; // Mysql username
$password="classified"; // Mysql password
$db_name="jaycraftcommunity"; // Database name
$tbl_name="members"; // Table name
// Connect to server and select database.
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");
// form variable to local
$search = $_POST['username'];
// make query
//$sql = "SELECT * FROM $tbl_name WHERE (username='$search' OR verified='$search')"; // sql query
$sql = "SELECT username FROM $tbl_name WHERE username LIKE '%$search%'"; // sql query
$result = mysql_query($sql); // do query
// handle query data
$count = mysql_num_rows($result);
$row = mysql_fetch_row($result);
$array = mysql_fetch_array($result);
echo $count;
foreach( $row as $arrays ): ?>
<tr>
<td><?php echo htmlspecialchars( $arrays ); ?></td>
<br />
</tr>
<?php
endforeach;
?>
The HTML form looks like this:
<h2>Search</h2>
<form method="post" action="search_engine.php">
Search for:
<input type="text" name="username">
<input type="submit" value="Search for username">
</form>
At the moment, if you go to http://dev.jaycraft.co/player/dupe/search.php you can see it in action. Search for individual letters, it show how many results there are with echo $count, but only echos the first result.
Thanks

Answer 1

Every call of mysql_fetch_row($result) will return another row and advance the cursor. You simply need to use a while() to loop over all rows in the result:
while($row = mysql_fetch_row($result)) {
echo $row[0];
}}
For ease of use, it's better to use mysql_fetch_assoc($result); which will return an associative array:
while($row = mysql_fetch_assoc($result)) {
echo $row['username'];
}}

Answer 2

You can
<?php
$sql = mysql_query($sql);
while($row = mysql_fetch_assoc($sql)){
echo $row['username'].'<br />';
}
?>
But please use mysqli instead of mysql! http://www.php.net/manual/en/book.mysqli.php

Answer 3

When you are using the prepare line, the above solutions do not work, given the substitution symbol is the same as the like symbol. Try making the search parameter within the prepare, by preformating it as a string. Then just loop through the available rows by the foreach or for loop.
$Table_Name = $wpdb->prefix.'tablename';
$SearchField = '%'. $YourVariable . '%';
$sql_query = $wpdb->prepare("SELECT * FROM $Table_Name WHERE ColumnName LIKE %s", $SearchField) ;
$rows = $wpdb->get_results($sql_query, ARRAY_A);
if(!empty($rows))
{
$TotalRows = $wpdb->num_rows;
foreach($rows as $row)
{
echo $row['field name'];
}
}

Answer 4

you need a while loop through your rows:
while($row = mysql_fetch_array($result){
foreach( $row as $cell): ?>
<tr>
<td><?php echo htmlspecialchars( $cell); ?></td>
<br />
</tr>
<?php
endforeach;
?>
}
mysql_fetch_row and mysql_fetch_array are almost unique functions, they both get the next row from the result set. you could use mysql_fetch_assoc to fetch the values with as key the database column name. (this is recommended)