php - 如果 mysql 查询没有返回结果，则返回值

Question

好的，所以我正在尝试这样做，以便如果输入用户名存在以获取该用户名的详细信息，并且如果不存在用户名来告诉我......

到目前为止，它会告诉我用户名是否存在，但如果它不存在则不会...

所以基本上如果查询返回零结果我怎么让它说“嘿，没有匹配”？

到目前为止我的代码...

$user = $_POST['txtUsername'];
$sql = "SELECT * FROM `weaponstreat` WHERE username='$user'";
$rows = $db->query($sql); while ($record = $db->fetch_array($rows)) {

    if ($record['tid'] === NULL) { echo "empty"; } else { echo "full"; }
}

Answer 1

$user = $_POST['txtUsername'];
$sql = "SELECT COUNT(1) rcount FROM `weaponstreat` WHERE username='$user'";
$rows = $db->query($sql); while ($record = $db->fetch_array($rows)) {

    if ($record['rcount'] == 0) { echo "empty"; } else { echo "full"; }
}

Answer 2

只需检查是否返回任何行：

if ($db->num_rows() == 0)
{
    // No results
 }
else
{
    while ($record = $db->fetch_array($rows)) {
        if ($record['tid'] === NULL) { echo "empty"; } else { echo "full"; }
    }
}

Answer 3

$user = $_POST['txtUsername'];
$sql = "SELECT * FROM `weaponstreat` WHERE username='$user'";
$rows = $db->query($sql);
// add a counter variable 
$counterRecords = 0;
while ($record = $db->fetch_array($rows)) {

    $counterRecords++;
}

if($counterRecords==0)
 echo "empty";
else echo "full";

Answer 4

$user = $_POST['txtUsername'];

$query = mysql_query ("SELECT * FROM weaponstreat WHERE username='$user' ORDER BY tid DESC");

if ( mysql_num_rows( $query ) > 0 )
{
    // Process and display username details
}
else
{
    echo "That username does not exist";
}