SQL查询无法从数据库中接收正确的值？

Question

Solution: I asked this question with little knowledge in PHP and SQL, the solution to my answer was to wrap up the variable like this '$PLATE_NUMBER' since it is a string. This is also answered in the first comment below.

As you can see from the code, when I click search, I receive "There was no search results!" There is something wrong with my query, because if I change this to

"SELECT * FROM `tires` WHERE PLATE_NUMBER LIKE '%$PLATE_NUMBER%'"

then I receive the data inside the mySQL database?

Btw: I have also tested both query in mySQL, and it works. So what can be the problem here? I wish to search for instance "CV33610", and receive the row from my database..

$query = mysql_query("SELECT * FROM `tires` WHERE PLATE_NUMBER='%$PLATE_NUMBER%'", $connection) or die("Could not search!"); 
$count = mysql_num_rows($query); // Returns an integer of how many rows in your table that is picked up.

if($count == 0){

    $output = 'There was no search results!';

} else {

    // Collect the data into an array
    while($row = mysql_fetch_array($query)){

        $platenumber = $row['PLATE_NUMBER'];
        $id = $row['id'];

        // View the parameters to the screen. 
        $output .= '<div>'.$id.' '.$platenumber.'</div>';

    } // End while statement
} // End if statement

Answer 1

Since we're dealing with a string value CV33610, you need to wrap your $PLATE_NUMBER variable using single quotes.

$query = mysql_query("SELECT * FROM `tires` 
WHERE PLATE_NUMBER = '$PLATE_NUMBER'", $connection) 
or die(mysql_error());

Having added or die(mysql_error()) to your query would have signaled the syntax error.

Sidenote:

• Your present code is open to SQL injection. Use prepared statements, or PDO with prepared statements, they're much safer.

---

Add error reporting to the top of your file(s) which will help find errors.

<?php 
error_reporting(E_ALL);
ini_set('display_errors', 1);

// rest of your code

Sidenote: Error reporting should only be done in staging, and never production.