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我在访问此链接时出现以下错误: (是的,我知道以后我会使用json
,因为它更好,但我是尝试此) http://justedhak.comlu.com/insert.php?username=m&password=mmysqli_fetch_array()期望参数1是mysqli_result,布尔错误,但查询的结果是varchar
Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in /home/a6901827/public_html/insert.php on line 18
但查询的结果是VARCHAR不boealan资源为varchar和查询将给出结果“是”。为什么然后我有这样的错误?
<?php
$host='mysql12.000webhost.com';
$uname='z';
$pwd='6';
$db="a6901827";
$con = mysqli_connect($host,$uname,$pwd) or die("connection failed");
if (mysqli_connect_errno($con))
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$username = $_POST['username'];
$password = $_POST['password'];
$result = mysqli_query($con,"SELECT res FROM samle where
name='$username' and password='$password'");
$row = mysqli_fetch_array($result);
$data = $row[0];
if($data){
echo $data;
}
mysqli_close($con);
?>
是的,我发誓我只是阅读关于后期的教程,并得到,我是新来的PHP,我用它与android @ chris85 – Moudiz