這個簡單代碼中的錯誤在哪裏？

Question

哪裏出錯？ 24線，並沒有看到任何東西..任何人都可以查明錯誤？它會如此感激...它說的錯誤是在線24

錯誤消息

Parse error: syntax error, unexpected '' (T_ENCAPSED_AND_WHITESPACE), expecting identifier (T_STRING) or variable (T_VARIABLE) or number (T_NUM_STRING) in C:\Program Files (x86)\EasyPHP-DevServer-13.1VC9\data\localweb\database.php on line 24 

<?php 

$mysqli = new mysqli('localhost', 'root', '', 'test'); 


if ($mysqli->connect_error) { 
    die('Connect Error (' . $mysqli->connect_errno . ') ' 
      . $mysqli->connect_error); 
} 


$query = "SELECT fname, lname,name FROM persons INNER JOIN foods ON persons.id = food_id"; 
$result = $mysqli->query($query); 

echo "<table>"; 
echo "<tr>"; 
echo "<td> <b> First Name </b> </td>"; 
echo "<td> <b> Last Name </b> </td>"; 
echo "<td> <b> Food </b> </td>"; 
echo "</tr>"; 

while ($row = $result->fetch_assoc()) { 

echo "<td> $row['fname'] </td>"; 
echo "<td> $row['fname'] </td>"; 
echo "<td> $row['name'] </td>"; 

} 

echo "</tr>"; 
echo "</table>"; 

?> 

Answer 1

試試這個

echo "<td>" . $row['fname'] . "</td>"; 

Answer 2

更改while循環此

while ($row = $result->fetch_assoc()) { 

    echo "<td>". $row['fname'] ."</td>"; 
    echo "<td>". $row['fname'] ."</td>"; 
    echo "<td>". $row['name'] ."</td>"; 

} 

Answer 3

Read the manual for the correct syntax用於在字符串中嵌入數組值，"$arr['key']"是而不是正確的兩種方法之一。

這是不是

"$arr[key]" 

或

"{$arr['key']}" 

Answer 4

你應該行使用此代碼24

echo "<td> $row[fname] </td>"; 
    echo "<td> $row[fname] </td>"; 
    echo "<td> $row[name] </td>"; 

Answer 5

您也可以使用此：

while ($row = $result->fetch_assoc()) { 

    echo "<td>{$row['fname']}</td>"; 
    echo "<td>{$row['fname']}</td>"; 
    echo "<td>{$row['name']}</td>"; 

}