mysql_query always return empty

Question

I am a little new to php, and I am trying to retrieve a raw from the database and the result is always empty. I have used the following code to insert a record in the table and it worked.

INSERT INTO `registeration`.`cars` (`lnumber`, `type`, `model`, `motor`, `owner`
,`expdate`) VALUES ('123654', 'Nissan maxima', '2008', 
'1500', 'name name', '2013-11-30');

The result was 1 raw inserted

The code used in the php file is:

if (isset($_GET["lnumber"])) {
$lnumber = $_GET['lnumber'];
echo $lnumber;
$result = mysql_query("SELECT * FROM cars WHERE lnumber = $lnumber");
if (!empty($result)) {
    // check for empty result
    if (mysql_num_rows($result) > 0) {
        ....code....

    }
} else {
    // no car found
    $response["success"] = 0;
    $response["message"] = "No cars found";

    // echo no users JSON
    echo json_encode($response);
    }
    } else {
    // required field is missing
    $response["success"] = 0;
    $response["message"] = "Required field(s) is missing";

    // echoing JSON response
    echo json_encode($response);
    }

However, the results are always: 123654{"success":0,"message":"No cars found"}

Answer 1

Your code is incorrect. The mysql function return boolean false on FAILURE, eg when an error occurs. mysql_query either return a result handle (query ran successfully) or a boolean FALSE (something blew up).

A result set which has no rows is NOT an error. It's simply an empty set. The code should be more like

$result = mysql_query($sql) or die(mysql_error()); // handle any error conditions
if (mysql_num_rows($result) == 0) {
  echo 'no records found';
} else {
  ... build your json here ...
}

And as others have said above, the mysql_*() functions are obsolete and deprecated. You should be using mysqli or PDO instead.

And again as others have said, you are vulnerable to SQL injection attacks . Read up and learn about how to prevent those before you do any more work with PHP/sql

Answer 2

First, you'll want to look into moving away from mysql_* and into mysqli_* or PDO. Second, you need to fetch the information appropriately:

$result = mysql_query("SELECT * FROM cars WHERE lnumber = '$lnumber'");

if(mysql_num_rows($result) > 0)
{
    while($row = mysql_fetch_array($result))
    {
        echo $row['car'];
    }
}
else
{
    // No cars.
}

Then do whatever it is you're going to do with the database information.

Answer 3

Just remove this test if (!empty($result)) and add die(mysql_error());

$result = mysql_query("SELECT * FROM cars WHERE lnumber = $lnumber") or die(mysql_error());

// check for empty result
if (mysql_num_rows($result) > 0) {
    ....code....

} else {
    // no car found
    $response["success"] = 0;
    $response["message"] = "No cars found";

    // echo no users JSON
    echo json_encode($response);
}

Answer 4

谢谢大家的宝贵意见和建议，我不想以错误的方式开始，我是一名Android开发人员，我需要为正在工作的项目学习php，很高兴知道这些示例已经很老了。