這就是我的表格。如何在php中輸出不同類型的數據?
用於創建和表中的返回代碼: 我想測試運動ID值的值。如果是1,則返回足球。如果是2回網球,如果是3迴游泳。如果離開列是TRUE,我希望它輸出離開,否則輸出HOME。
$sql = "CREATE TABLE fixtureDetails
(
fixtureID INT(5) NOT NULL AUTO INCREMENT,
opponent VARCHARD(30) NOT NULL,
date DATE
away BOOLEAN,
sportID INT
refereeID INT,
PRIMARY KEY (fixtureID),
FOREIGN KEY (sportID) REFERENCES sport(sportID),
FOREIGN KEY (refereeID) REFERENCES referee(refereeID)
)";
$sql1 = "SELECT * FROM fixtureDetails";
if ($result = mysqli_query($link, $sql)){
if(mysqli_num_rows($result)>0){
echo "<table>";
echo "<tr>";
echo "<th>fixtureID</th>";
echo "<th>opponent</th>";
echo "<th>date</th>";
echo "<th>away</th>";
echo "<th>sportID</th>";
echo "<th>refereeID</th>";
echo"</tr>";
while($row = mysqli_fetch_array($result)){
echo "<tr>";
echo "<td>" . $row['fixtureID'] . "</td>";
echo "<td>" . $row['opponent'] . "</td>";
echo "<td>" . $row['date'] . "</td>";
echo "<td>" . $row['away'] . "</td>";
echo "<td>" . $row['sportID'] . "</td>";
echo "<td>" . $row['refereeID'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_free_result($result);
} else {
echo "No records matching your query found.";
}
} else {
echo "ERROR: could not execute $sql1. " . mysqli_error($link);
}
你應該考慮增加一個表,包含了'1 = football.2 =網球場和3 =游泳 – RiggsFolly
,另一個做同樣的事情'家庭和離開' – RiggsFolly
I有這張桌子。它被稱爲體育與列sportID和名稱 –