的foreach警告

Question

管理表

+--------------+---------------+ 
| Username  | Password  | 
+--------------+---------------+ 
| JOHN   | 123   | 
| EDWARD  | 123   | 
+--------------+---------------+ 

我的代碼

$result = mysqli_query($connection, $query); 
    echo" <table >"; 
    $row = mysqli_fetch_assoc($result);  
echo "<tr>"; 
    foreach($row as $key => $val){ 
     echo"<th>$key</th> "; 
    echo "</tr>"; 
    /////////////////////////////// 
    $result = mysqli_query($connection, $query); 

     echo"<tr>"; 
    while($row = mysqli_fetch_assoc($result)){ 
    foreach($row as $key => $val){ 
     echo "<td>$val</td>"; 

    } 
    echo "</tr>"; 

    } 


    echo "</table>"; 
} 

那麼，問題是如果在其輸出完全相同的結果超過1行什麼，我想對於例如，如果我說：

SELECT Username from admin; 


     +--------------+ 
     | Username  | 
     +--------------+ 
     | JOHN   | 
     | EDWARD  | 
     +--------------+ 

但如果只有一排它並不顯示列名，並給出此警告

Select Username from admin where Username = 'JOHN'; 

    Warning: Invalid argument supplied for foreach() on line 65 
     +--------------+ 
     | JOHN   | 
     +--------------+ 

Answer 1

偷偷摸摸的錯字/邏輯編碼錯誤：你有一個花括號（大括號），把所有東西搞砸了。看看用來生成標題在foreach結束：

foreach($row as $key => $val){ 
    echo"<th>$key</th> "; 

這應該是：

foreach($row as $key => $val) 
    echo"<th>$key</th> "; 

然後在最後刪除虛假梅開二度。

這工作（只要至少有1列，以獲得heaedrs）：

$result = mysqli_query($connection, $query); 

echo " <table >"; 

$row = mysqli_fetch_assoc($result);  

echo "<tr>"; 
foreach($row as $key => $val) 
    echo"<th>$key</th> "; 

echo "</tr>"; 

$result = mysqli_query($connection, $query); 

echo"<tr>"; 
while($row = mysqli_fetch_assoc($result)) 
{ 
    foreach($row as $key => $val) 
    { 
     echo "<td>$val</td>"; 
    } 
    echo "</tr>"; 

} 

echo "</table>";