2011-12-17 85 views
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

+1

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

回答

1

你可以試試下面的代碼

<?php foreach ($rows as $work): ?> 
<span> <?php echo $work['id']; ?>, </span> 
<?php endforeach; ?> 

<?php reset($rows);?> 
<br \> 
<?php foreach ($rows as $work): ?> 
<span> <?php echo $work['name']; ?>, </span> 
<?php endforeach; ?>