为foreach（）提供了无效的参数
[英]Invalid argument supplied for foreach()
问题描述
i'm coding a project and have some troubles when the system inform error: Invalid argument supplied for foreach() in: foreach($dbh->query($q1) as $row) 
我正在编码一个项目，并且在系统通知错误时遇到一些麻烦：为foreach（）提供了无效的参数：foreach（$ dbh-> query（$ q1）as $ row）
and can't get data from the database. 并且无法从数据库中获取数据。 How can i fix it, im a newbie, so if i don't understand, please teach me! 我是新手，我该如何解决，如果我不明白，请教我！ thanks! 谢谢！ thank for your helps but i still can't fix it 感谢您的帮助，但我仍然无法解决
<?php include("top.html"); ?>
<body>
<div id="main">
<h1>Results for <?php echo $_GET['firstname'] . " " . $_GET['lastname'] ?></h1> <br/><br/>
<div id="text">All Films</div><br/>
<table border="1">
<tr>
<td class="index">#</td>
<td class="title">Title</td>
<td class="year">Year</td>
</tr>
<?php
$dbh = new PDO('mysql:host=localhost;dbname=imdb_small', 'root', '');
$q1 = "SELECT id
FROM actors
WHERE first_name = '".$_GET['firstname']."' AND last_name = '".$_GET['lastname']."'
AND film_count >= all(SELECT film_count
FROM actors
WHERE (first_name LIKE'".$_GET['firstname']." %' OR first_name = '".$_GET['firstname']."')
AND last_name = '".$_GET['lastname']."')";
$id = null;
foreach($dbh->query($q1) as $row){
$id = $row['id'] ;
}
if($id == null){
echo "Actor ".$_GET['firstname']." ".$_GET['lastname']."not found.";
}
`
$sql2 = "SELECT m.name, m.year
FROM movies m
JOIN roles r ON r.movie_id = m.id
JOIN actors a ON r.actor_id = a.id
WHERE (r.actor_id='".$id."')
ORDER BY m.year DESC, m.name ASC";
$i = 0;
foreach($dbh->query($sql2) as $row){
echo "<tr><td class=\"index\">";
echo $i+1;
echo "</td><td class=\"title\">";
echo $row['name'];
echo "</td><td class=\"year\">";
echo $row['year'];
echo "</td></tr>";
$i++;
}
$dbh = null;
?>
</table>
</div>
</div>
<?php include("bottom.html"); ?>
</body>
</html>
3 个解决方案
解决方案1
0 2016-05-13 18:29:34
You could do this 你可以这样做
$st = $dbh->query($q1);
if ( $st ) {
while ( $row = $st->fetch() ) {}
}
Try to make your code more readable
解决方案2
0 已采纳 2016-05-13 18:37:06
Check that your query succeeded before iterating on the result: 在迭代结果之前，请检查查询是否成功：
if (false !== ($result = $dbh->query($d1))) {
foreach($result as $row){
$id = $row['id'] ;
}
}
By the way, I don't understand what you are trying to do with this pointless loop. 顺便说一句，我不明白您在尝试这种无意义的循环。
解决方案3
-1 2016-05-13 18:28:23
$result = $dbh->query($q1);
foreach($result as $row){$id = $row['id'];}
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