Question

哪里可能出错？???在第24行，没有看到任何东西..任何人都可以找出错误？我会非常感激...它说错误在第24行

错误消息

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

将循环更改为此

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

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

}

Answer 2

试试这个

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

Answer 3

Read the manual for the correct syntax用于在字符串中嵌入数组值，"$arr['key']" 不是正确的两种方式之一。

要么是

"$arr[key]"

或

"{$arr['key']}"

Answer 4

您应该在第24行使用此代码

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

Answer 5

您也可以使用：

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

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

}

这个简单代码中的错误在哪里？

5 个答案: