MySQL在PHP中检查值等于x并且y不为空

Question

I'm trying to to a check in my database where username=$username and where the name field isn't empty. I'm not sure how to do this, so I am open for some input on this. :)

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PHP CODE:

$isinuse_sql = mysql_query("SELECT * FROM users WHERE username='$username' AND (name IS NOT NULL OR name<>''");
if(mysql_num_rows($isinuse_sql) != 0) {
    $isinuse_row = mysql_fetch_assoc($isinuse_sql);
    $isinuse = $isinuse_row['name'];
    echo "1";
} else {
    echo "0";
}

Answer 1

尝试这个：

"SELECT * FROM users WHERE username='$username' AND (name IS NOT NULL AND name!='')"

Answer 2

Oh I see now, silly me. If it is an empty string it is not null so you get true with an OR when you want false.

"SELECT * FROM users WHERE username='$username' AND name IS NOT NULL AND name<>''"

Answer 3

尝试这个

"SELECT * FROM users WHERE username='$username' AND IFNULL(name, '') != '' "

Answer 4

as NULL value is interpreted as a false you can check value of name like

SELECT * FROM users WHERE username='$username' AND users.name<>''

use mysqli instead of sql query as it is deprecated like David Soussan said

$con=mysqli_connect("localhost","root","","test");

if (mysqli_connect_errno())
{
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$username = 'a';
$sql="SELECT * FROM users WHERE username='$username' AND users.name<>''";
$result=mysqli_query($con,$sql);


$row=mysqli_fetch_array($result,MYSQLI_ASSOC);

print_r($row);
mysqli_close($con);