如何创建“未找到记录”错误消息
如何创建“未找到记录”错误消息
提问于 2018-10-13 15:34:48
我正在尝试添加一些代码到下面的示例中，这将允许生成"Record Not Found“错误-如果找不到记录：
<?php
header('Content-type=application/json; charset=utf-8');
//database constants
define('DB_HOST', 'localhost');
define('DB_USER', 'root');
define('DB_PASS', '');
define('DB_NAME', 'test');
//connecting to database and getting the connection object
$conn = new mysqli(DB_HOST, DB_USER, DB_PASS, DB_NAME);
//Checking if any error occured while connecting
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
die();
}
$stmt = $conn->prepare('SELECT * FROM donor WHERE city =? AND gender=? AND
bloodgroup=? AND age=?;');
$stmt->bind_param('ssss',$_GET['city'],$_GET['gender'],$_GET['bloodgroup'],$_GET['age']);
$stmt->execute();
//binding results to the query
$stmt->bind_result($id, $name, $gender, $city, $contact, $bloodgroup,$age);
$donors = array();
//traversing through all the result
while($stmt->fetch()){
$temp = array();
$temp['id'] = $id;
$temp['name'] = $name;
$temp['gender'] = $gender;
$temp['city'] = $city;
$temp['contact'] = $contact;
$temp['bloodgroup'] = $bloodgroup;
$temp['age'] = $age;
array_push($donors, $temp);
}
//displaying the result in json format
echo json_encode($donors);
?>您建议我将启用“找不到错误”错误的代码放在哪里？
回答 1
Stack Overflow用户
发布于 2018-10-13 15:55:10
您可以使用$stmt->num_rows，也可以使用以下代码进行检查
if (empty($donors)) {
echo json_encode(array('msg' => 'Record Not Found'));
}else{
echo json_encode($donors);
}页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/52790607
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