1
MySQL表PHP代碼的結果不匹配的MySQL查詢
ID |名稱
1 | A
2 | B
3 | C
4 | D
5 | È
MySQL查詢
$query = "SELECT * FROM ego_work WHERE 1;
$result = mysql_query($query);
$rows = array();
while ($row = mysql_fetch_array($result)) {
$rows[] = $row;
}
PHP代碼
<?php foreach ($rows as $work): ?>
<span> <?php echo $work['id']; ?>, </span>
<?php endforeach; ?>
<br \>
<?php foreach ($rows as $work): ?>
<span> <?php echo $work['name']; ?>, </span>
<?php endforeach; ?>
RESULT
1,2,3,4,5
E,A,B ,C,D
我做錯了什麼?我試圖讓第二個結果爲A,B,C,d,E
你能張貼在`$ rows`一個`var_dump`的結果?我懷疑你的桌子上的數據不是你認爲的那樣。 – Shad 2011-12-17 05:00:14