我在Android应用程序中从我的数据库获取错误 [英] Am getting errors from my db in android app

问题描述

Hello Guys我想选择值并显示在文本框中，但我点击按钮点击我的edittext

这里是错误

 < br /> 警告：mysqli_num_rows（）要求参数1为mysqli_result，在 /h7/ae/public_html/fashion360/Com.php   11 中给出布尔值>< br /> [] 

请帮帮我，请告诉我这是怎么回事，我怎么能解决这个错误

谢谢



我尝试过：

<？php 
包括'Db.php'; 
 $ con = mysqli_connect（$ HostName，$ HostUser，$ HostPass，$ DatabaseName）或死亡（无法连接）; 
 mysqli_select_db（$ con，$ DatabaseName）或die（不能选择db）; 
 mysqli_set_charset（$ con，'utf8'）; 
 $ sql =SELECT TOP 1抱怨抱怨; 
 $ result = mysqli_query（$ con，$ sql）; 
 $ json = array（）; 
 if（mysqli_num_rows（$ result））{
 while（$ row = mysqli_fetch_assoc（$ result））{
 $ json ['complainno'] [] = $ row; 
} 
} 
 mysqli_close（$ con）; 
 echo json_encode（$ json）; 
？> 

解决方案

con = mysqli_connect（

HostName，

HostUser，

Hello Guys i want to select values and show in textbox but i am getting on my edittext on button click
here is the error

<br /> Warning:  mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in /h7/ae/public_html/fashion360/Com.php on line 11<br /> []

please help me please tell me whats the matter and how can i resolved this error
thanks

What I have tried:

<?php
include 'Db.php';
$con=mysqli_connect($HostName,$HostUser,$HostPass,$DatabaseName)or die("Cannot Connect");
mysqli_select_db($con,$DatabaseName)or die("Cannot select db");
mysqli_set_charset($con,'utf8');
$sql = "SELECT TOP 1 complainno FROM complain";
$result  =mysqli_query($con,$sql);
$json = array();
if(mysqli_num_rows($result)){
while ($row=mysqli_fetch_assoc($result)){
$json['complainno'][]=$row;
}
}
mysqli_close($con);
echo json_encode($json);
?>

解决方案

con=mysqli_connect(

HostName,

HostUser,

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