我有以下PHP代码从表
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($con);
?>
但是当我从应用程序中点击这段代码时,它返回null结果。如果我用硬编码值
运行sqlSelect * from bbau_login where username= 'aqsdfg' and password= 'adjbf'
然后我得到所需的结果,但不是与php脚本中指定的sql
我还检查了我在$username和$password中获得了正确的值。我想我需要在引号中传递$用户名和$密码。请有人帮忙写正确的查询引号
嗯,我认为pasword/password
的一致拼写将大有帮助:
# vv
$password=$_POST["password"];
:
mysqli_stmt_bind_param($statement,"ss",$username,$pasword);
# ^
你可能很好,正如你所说的,"在$username
和$password
中获得适当的值",但如果你实际上不使用 $password
中的内容:-)
您在传递密码时拼写错误
改变 mysqli_stmt_bind_param($statement,"ss",$username,$pasword);
mysqli_stmt_bind_param($statement,"ss",$username,$password);