[英]codeigniter - trying to get property of non-object error because i was trying to get the value of the user's input
[英]I'm trying to get a property value of a non-object
这听起来像是重复的,但是我在这里搜索了与此相关的所有主题,但还是没有运气... :(
我有以下代码:
$result = array();
include 'conn.php';
$sql = "SELECT * FROM editlistm WHERE checkvou IN (SELECT checkvou FROM editlist)";
$result = $conn->query($sql);
//$numRows = $result->count($rs);
$numRows = $result->num_rows;
if ($numRows > 0) {
// output data of each row
while($row = $result->fetch_object()) {
array_push($items, $row);
}
echo json_encode($items);
}
conn.php:
$servername = "localhost";
$username = "root";
$password = "";
$dbase = "cadacctg";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbase);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
我没有发现任何错误,但一直在说:
'试图在第11行的C:...中获取非对象的属性'($ numRows = $ result-> num_rows;)
为什么它不被视为对象? 我该怎么办? 提前致谢 :)
使用fetch_object
或fetch_array
获取字段值。
var_dump($result) before $numRows = $result->num_rows;
如果为假,则查询中存在问题
问题是,此代码:
$result = $conn->query($sql);
可能会失败。
http://php.net/manual/en/mysqli.query.php#refsect1-mysqli.query-returnvalues
失败时返回FALSE。 对于成功的SELECT,SHOW,DESCRIBE或EXPLAIN查询,mysqli_query()将返回mysqli_result对象。 对于其他成功的查询,mysqli_query()将返回TRUE。
因此,您需要对其进行检查,例如:
$result = $conn->query($sql);
// For SELECT query, check if it's false
if (!$result) {
exit('Query has failed.');
}
检查您的值正确无误:
if ($result = $conn->query($sql)) {
///value returns not-false,
$numRows = $result->num_rows;
...
}
else {
//there is a problem with your SQL
}
如果->query
正常运行,这只会为$numRows
返回非布尔值-负值。
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