php - 第 9 行 E : mysqli_fetch_object() expects parameter 1 to be mysqli_result, php 中给出的警告 :\xampp\htdocs\getSubCatList. 字符串

Question

请帮我找出代码中的错误。

它显示以下警告:

mysqli_fetch_object() expects parameter 1 to be mysqli_result, string given in E:\xampp\htdocs\getSubCatList.php on line 9

代码:

<?php
$con = mysqli_connect("localhost","root","","dbname");
$parentname = $_POST['$parentname'];
$con->query("SET NAMES 'utf8'");
$sql1 ="SELECT `fld_category_id` FROM `tbl_category_master` WHERE `fld_category_name` = '$parentname' limit 1";
//$result = mysql_query($sql1);
try{
$value = mysqli_fetch_object($sql1);
$cat_id= $value;}
catch (Exception $e) { 

echo $e->errorMessage(); }

$sql="SELECT `fld_category_name` FROM `tbl_category_master` WHERE `fld_parent_category_id` = '$cat_id'";
 $result1=$con->query($sql);
        while($e=mysqli_fetch_assoc($result1)){
    $output[]=$e; **strong text**
    }   

    $con->close();
?>

Answer 1

我想你在 $value = mysqli_fetch_object($sql1);

之前错过了一行

这是关于执行您的查询。

$data = mysqli_query($con,$sql1);

然后是你的代码

$value = mysqli_fetch_object($data);