Question

我有以下PHP代码，用于从表中返回用户详细信息

<?php

$con=mysqli_connect("localhost","root","");
if(!$con)
    {
        die('Could not connect'.mysqli_error());
    }
mysqli_select_db($con,"mysql");
$username=$_POST["username"];
$password=$_POST["password"];
$statement=mysqli_prepare($con,"Select * from bbau_login where username= ? and password= ? ");
mysqli_stmt_bind_param($statement,"ss",$username,$pasword);
mysqli_stmt_execute($statement);
mysqli_stmt_store_result($statement);
mysqli_stmt_bind_result($statement,$id,$name,$username,$password);

$user=array();

while(mysqli_stmt_fetch($statement))
    {
        $user['name']=$name;
        $user['username']=$username;
        $user['password']=$password;
    }
echo json_encode($user);
mysqli_stmt_close($statement);
mysqli_close($con);

?>

但是当我从应用程序中获取此代码时，它返回null结果。如果我运行带有硬编码值的SQL

Select * from bbau_login where username= 'aqsdfg' and password= 'adjbf'

然后我得到了所需的结果，但没有使用php脚本中指定的sql

此外，我检查了我在$ username和$ password中获取了正确的值。我想我需要在引号中传递$ username和$ password。请有人帮忙用引号撰写正确的查询。

Answer 1

传递密码时，您写的密码拼写错误

变化

 mysqli_stmt_bind_param($statement,"ss",$username,$pasword);

到

mysqli_stmt_bind_param($statement,"ss",$username,$password);

SQL返回空行

1 个答案: