我想获得5位数据,其中5位来自tbl_playerstats
这些专栏是
StatsID
score
Kills
Deaths
Rank
我正在努力将StatsID链接到另一个名为tbl_playerdata的表
在这个表中,PlayerID与StatsID相同,它们的值是数字,在这个tbl_playerdata中也是SoldierName,这是我试图代替StatsID,最终目标是只有这些显示:
SoldierName
score
Kills
Deaths
Rank
// Create connection
$conn = new MysqLi($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection Failed: " . $conn->connect_error);
}
$sql = "SELECT StatsID, score, Kills, Deaths, Rounds FROM tbl_playerstats Limit 10";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "<br> Soldier: ". $row["StatsID"]. "<p> score: ". $row["score"]. "<p> Kills: ". $row["Kills"]. " <p>Deaths: " . $row["Deaths"] . $row["Rounds"] . "<br>";
}
} else {
echo "0 results";
}
$conn->close();
我已经google了,看起来使用JOIN是要走的路,我发现的例子对我来说没有用,这可能是因为我不知道这叫做我想做什么.
解决方法:
使用StatsID和PlayerID加入两个表
<?PHP
$sql = "SELECT tps.*, tpd.* FROM tbl_playerstats tps, tbl_playerdata tpd WHERE tps.StatsID = tpd.PlayerID Limit 10";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
echo "<br> Soldier: ". $row["SoldierName"]. "<p> score: ". $row["score"]. "<p> Kills: ". $row["Kills"]. " <p>Deaths: " . $row["Deaths"] . $row["Rounds"] . "<br>";
}
} else {
echo "0 results";
}?>