mysql_field_name(resource $result, int $field_offset) 返回指定索引位置的欄位名稱字串,失敗返回 false。
<?php $sql = "SELECT id, username, email, created_at FROM users"; $result = mysql_query($sql); $name = mysql_field_name($result, 0); echo "第 0 欄位名: $name\n"; // 輸出: id $name = mysql_field_name($result, 2); echo "第 2 欄位名: $name\n"; // 輸出: email ?>
<?php
$num_fields = mysql_num_fields($result);
for ($i = 0; $i < $num_fields; $i++) {
echo "Column $i: " . mysql_field_name($result, $i) . "\n";
}
?>
<?php
function export_csv($table, $link) {
$result = mysql_query("SELECT * FROM `$table`", $link);
$fp = fopen('php://output', 'w');
// 寫入表頭
$headers = [];
for ($i = 0; $i < mysql_num_fields($result); $i++) {
$headers[] = mysql_field_name($result, $i);
}
fputcsv($fp, $headers);
// 寫入資料
while ($row = mysql_fetch_row($result)) {
fputcsv($fp, $row);
}
fclose($fp);
}
?>
| 擴展 | 方法 |
|---|---|
| mysqli | mysqli_fetch_field_direct($result, $offset)->name |
| PDO | $stmt->getColumnMeta($i)['name'] 或 array_keys($stmt->fetch(PDO::FETCH_ASSOC)) |
SELECT id AS user_id 返回 user_id