Php代码结果与mySQL查询不匹配

mySQL表

ID |名称

1 | A

2 | B

3 | C

4 | D

5 | E

mySQL查询

$query = "SELECT * FROM ego_work WHERE 1;
$result = mysql_query($query);

$rows = array();
while ($row = mysql_fetch_array($result)) {
    $rows[] = $row;
}

php代码

<?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; ?>

, 

, 

结果

1,2,3,4,5

E、 A、B、C、D

我做错了什么？我正在尝试将第二个结果设为A、B、C、D、E

您可以尝试下面的代码吗

<?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; ?>

, 

, 

你能试试下面的代码吗

<?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; ?>

, 

, 

能否将

变量转储的结果发布到
$rows
上？我怀疑您表中的数据与您想象的不一样。您能否将
var\u dump
的结果发布到
$rows
？我怀疑您表中的数据与您认为的不一样。