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); 
}

來源

2017-01-09 Faayo Bedrudin

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 –

echo "<td>" . $row['fixtureID'] . "</td>"; 
echo "<td>" . $row['opponent'] . "</td>"; 
echo "<td>" . $row['date'] . "</td>"; 
if($row['away']) 
    echo "<td>Away</td>"; 
else 
    echo "<td>Home</td>"; 

switch($row['sportID']){ 
    case 1: echo "<td>Football</td>"; 
     break; 
    case 2: echo "<td>Tennis</td>"; 
     break; 
    case 3: echo "<td>Swimming</td>"; 
     break; 
    default: echo "Error!"; 
} 
echo "<td>" . $row['refereeID'] . "</td>";

來源

2017-01-09 15:21:19

這段代碼去哪裏> –

@FaayoBedrudin - 我剛更新了它。檢查。 –

謝謝你是一位救生員!! SM Iftakhairul –

請查看：

echo "<td>" . (isset($row['away']) && $row['away']) ? 'AWAY' : 'HOME' . "</td>";

來源

2017-01-09 15:15:06

這只是答案的一半...... – Andy

你不顯示Sport表的結構，所以我假設有名爲Name表中列。您可以更改此部分以反映您的實際表格。

我不會用PHP來檢索這些值，如果他們已經在數據庫中（通過在CREATE聲明下面的暗示）：

FOREIGN KEY (sportID) REFERENCES sport(sportID),

取而代之的是SELECT * FROM fixtureDetails;查詢您所使用，我會用，而不是下面的查詢：

Select F.FixtureId, 
     F.Opponent, 
     F.Date, 
     Case When F.Away = 1 Then 'Away' Else 'Home' End As Away, 
     S.Name As Sport, -- Change S.Name to the actual column in Sport. 
     F.RefereeId 
From FixtureDetails F 
Join Sport   S On S.SportId = F.SportId

來源

2017-01-09 15:23:26 Siyual

這可以節省我把它寫出來 – RiggsFolly

如何在php中輸出不同類型的數據？

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