我正在尝试使用PHP脚本从服务器获取Status列的值。我收到以下错误:
<b>Catchable fatal error</b>: Object of class mysqli_result could not be converted to string in <b>C:\xampp\htdocs\loginstatuscheck.php</b> on line <b>21</b><br />
插入工作正常,在我尝试执行提取查询的同一行上。我目前的PHP脚本如下所示:
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "iFocusBlogs";
$obtaineduserName = urldecode($_POST['enteredusername']);
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$query_get_status = "SELECT Status FROM LoginTable WHERE UserName ='" .$obtaineduserName. "'";
// perform the query and store the result
$result = $conn->query($query_get_status);
echo $result;
//if ($conn->query($query_get_status) === TRUE) {
// echo $result;
//} else {
echo "Error: " . $result . "<br>" . $conn->error;
//}
mysql_close();
$conn->close();
?>
我的查询是否正确?
发布于 2018-06-05 19:53:19
如果你想查看对象的内容,可以使用var_dump($var);
下面是如何转储与请求匹配的所有SQL行:
while($obj = $result->fetch_object()){
var_dump($obj);
}
由于你似乎是在请求一个独特的行,所以你可以简单地使用:
$query_get_status = "SELECT Status FROM LoginTable WHERE UserName ='" .$obtaineduserName. "'";
// perform the query and store the result
$result = $conn->query($query_get_status);
var_dump($result->fetch_object());
使用fetch_object()
将MySQL结果转换为一个对象,这样就可以访问它的属性了。
$result = $conn->query($query_get_status)->fetch_object();
$status = $result->Status;
https://stackoverflow.com/questions/-100004754
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