如果没有结果显示消息,mysql fetch array
我正在尝试进行数据库调用以显示一条语句,如果没有返回结果,则表示没有找到结果.
I am trying to get a database call to show a statement saying no results found if there are no results returned.
我将如何对我的代码执行此操作:-
How would I go about doing this to my code:-
$getFixtures = mysql_query("SELECT ht.name AS hometeam_name, homescore, awayscore, at.name AS awayteam_name, time, date, week, comp.competition AS comp_name, se.name AS season_name
FROM fixture
JOIN team ht
ON ht.id = fixture.hometeam
JOIN team at
ON at.id = fixture.awayteam
JOIN competition comp
ON comp.id = fixture.competition
JOIN season se
ON se.id = fixture.season
WHERE se.name = '$season' AND comp.competition = '$competitiontitle' AND date >= '$today' AND at.name = '$teamName' OR ht.name = '$teamName' AND se.name = '$season' AND comp.competition = '$competitiontitle' AND date >= '$today'
ORDER BY date ASC
");
while ($fixtureData = mysql_fetch_array($getFixtures))
{
$hfixteamlink = strtolower(str_replace(" ","-",$fixtureData['hometeam_name']));
$afixteamlink = strtolower(str_replace(" ","-",$fixtureData['awayteam_name']));
$date = date("d/m/Y", strtotime($fixtureData['date']));
?>
提前致谢
理查德
推荐答案
这需要一个 IF 语句.
This needs an IF statement.
$rows = mysql_fetch_array($getFixtures);
if(count($rows))
{
while ($fixtureData = $rows)
...
}
else
{
echo 'No results found';
}
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