PHP代码的结果不匹配的MySQL查询

ID |名称

1 | A

2 | B

3 | C

4 | D

5 | È

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; ?>

RESULT

1,2,3,4,5

E，A，B ，C，D

我做错了什么？我试图让第二个结果为A，B，C，d，E

来源

2011-12-17 ber2g

你能张贴在`$ rows`一个`var_dump`的结果？我怀疑你的桌子上的数据不是你认为的那样。 – Shad 2011-12-17 05:00:14

你可以试试下面的代码

<?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; ?>

来源

2011-12-18 10:50:35

PHP代码的结果不匹配的MySQL查询

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