How to do a simple mysqli query for a similar value? - php

Question

I am looking for just a basic query using mysqli to search table for a particular value. For example; If 'a' exist in a table then echo 'already exists'.
Please help as I am still learning to move away from deprecated code.
$mysqli = new mysqli('localhost', 'root', 'pass', 'agents');
$result = $mysqli->query('select * from project');
if ($result)
{
$check = array();
while ($row = $result->fetch_assoc())
{
$check[] = $row['projectname'];
}
}
$a = 'a';
if ($a = $check)
{
echo "<script type='text/jscript'>
alert('already exists.')
</script>";
}

Answer 1

Databases are designed to do searching for you with WHERE clauses. There's no need to read the entire table when you can let the database do it for you.
$result = $mysqli->query("SELECT COUNT(*) AS found FROM project WHERE projectname = '$a'");
$row = $result->fetch_assoc();
if ($row['found'] > 0) {
echo "<script type='text/jscript'>alert('already exists.')</script>";
}

Answer 2

$mysqli = new mysqli('localhost', 'root', 'pass', 'agents');
$result = $mysqli->query("select * from project where projectName = 'a'");
if($result->num_rows > 0)
echo "already exists";

Answer 3

You have a logic error in your code. You try to compare array ($check) with string ($a). Try something like this:
$mysqli = new mysqli('localhost', 'root', 'pass', 'agents');
$result = $mysqli->query('select * from project');
if ($result) {
$check = array();
while ($row = $result->fetch_assoc()) {
$projectname = $row['projectname'];
if($projectname == 'a')
{
echo "<script type='text/jscript'>alert('already exists.')</script>";
break;
}
$check[] = $row['projectname'];
}
print_r($check);
}
else
{
echo "<script type='text/jscript'>alert('mysqli error')</script>";
}
This code will check every projectname in your table and will show alert if it exists projectname = 'a'.
P.s. If you want to contact with your database in really good way read about PDO driver (http://www.phpro.org/tutorials/Introduction-to-PHP-PDO.html)