我有以下MySQL查询
$jokerQuery = mysql_query("SELECT `Joker sport`,
COUNT(`Joker sport`) AS jokerCount
FROM Profiles
WHERE `CompetitorID` = 5
GROUP BY `Joker sport`
ORDER BY COUNT(`Joker sport`) DESC
LIMIT 1
");
在phpMyAdmin中返回以下结果
Joker sport | jokerCount
8 | 8
我认为以下php会显示结果,但它不起作用。我应该写什么来回应结果?
$jokerResult = mysql_fetch_array($jokerQuery);
echo $jokerResult['Joker sport'];
echo $jokerResult['jokerCount'];
答案 0 :(得分:1)
试试这个,在mysql_fetch_array中添加MYSQL_ASSOC
作为const:
$jokerResult = mysql_fetch_array($jokerQuery, MYSQL_ASSOC);
echo $jokerResult['Joker sport'];
echo $jokerResult['jokerCount'];
答案 1 :(得分:1)
您也可以尝试:
$jokerResult = mysql_fetch_array($jokerQuery, MYSQL_ASSOC);
print_r($jokerResult);
查看列名。