我正在尝试从数据库中检索一些值。我检索它们后,我想将它们存储到一个数组中,这个数组通过json发送它。不幸的是,我陷入了错误,我无法找到任何解决方案。 你能帮助我吗?
<?php
include_once 'database.php';
$unique_id = 1;//$_GET['id'];
try {
$results = $db->prepare("
SELECT image_id,image_url,offer_id
FROM `Images`
WHERE offer_id = ?");
$results->bindValue(1, $unique_id);
$results->execute();
} catch (Exception $e) {
echo "Data could not be retrieved from the database.";
exit;
}
$product = $results->fetch(PDO::FETCH_ASSOC);
if($product['image_url'] != ''){
$arr="".$product['image_url']."";
//print_r($arr);
foreach($arr as $file){ //get an array which has the names of all the files and loop through it
$obj['name'] = $file;//get the filename in array
$obj['size'] = filesize("uploads/".$file); //get the flesize in array
$result[] = $obj; // copy it to another array
}
header('Content-Type: application/json');
echo json_encode($result); // now you have a json response which you can use in client side
}
?>
渲染
<br />
<b>Warning</b>: Invalid argument supplied for foreach() in <b>/Applications/XAMPP/xamppfiles/htdocs/projects/uploads/upload_get.php</b> on line <b>20</b><br />
<br />
<b>Notice</b>: Undefined variable: result in <b>/Applications/XAMPP/xamppfiles/htdocs/projects/uploads/upload_get.php</b> on line <b>26</b><br />
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