【发布时间】:2019-04-24 02:02:31
【问题描述】:
php 代码:
<?php
header("Content-type:application/json");
require "db_conn.php";
$user_name = $_POST["user_name"];
$user_pass = $_POST["pwd"];
$mysql_qry = "select * from user_accounts where user_name like '$user_name' and pwd like '$user_pass';";
$result = mysqli_query($conn,$mysql_qry);
if( $num_rows = mysqli_num_rows($result) > 0 ){
while( $row = mysqli_fetch_array($result)){
array_push($response,array("user_name"=>$row['user_name'],"first_name"=>$row['first_name'] , "last_name"=>$row['last_na$}
echo json_encode(array("server_response"=>$response));
//echo json_encode(array("result"=>123));
}else{
echo "login fail";
}
?>
我的回复是: {"server_response":null}
但如果我取消注释示例输出并注释另一部分,我将得到输出: {“结果”:123}
如果我直接初始化变量而不是 $_POST 也出于某种原因,我会得到正确的输出
【问题讨论】:
-
您是如何访问该页面的?您是否正在发送 HTTP POST 请求并提供 user_name 和 pwd 参数?
-
POST 请求然后给出这两个参数