[PHP]一个简易的返回JSON的PHP连接MySQL中间件

一个简易的php连接MySQL中间件,帮一个客户写的,最后不用了,就发出来了

实测易语言,C++,JavaScript都能够post成功,只需post一个参数名为sql的sql语句

<?php
$MySql_ServerIP = "127.0.0.1";
$MySql_UserName = "root";
$MySql_Password = "root";
$MySql_DBName ="";
$MySQL_Conn = new Mysqli($MySql_ServerIP,$MySql_UserName,$MySql_Password,$MySql_DBName);
if ($MySQL_Conn->connect_error){
    die("连接失败,失败原因: ".$MySQL_Conn->connect_error);
}
if($_SERVER['REQUEST_METHOD'] == 'POST')
{
    if(empty($_POST["sql"]))
    {
        echo "sql错误";
        exit();
    }
    else
    {
        $sql = $_POST["sql"];
        $result = $MySQL_Conn->query($sql);
        $JsonArr = array();
        while($row = $result->fetch_assoc())
        {
            $JsonArr[] = $row;
        }
        echo json_encode($JsonArr);
        $MySQL_Conn->close();
    }
}
else
{
    echo "禁止访问!";
}
?>

 

posted @ 2018-03-21 22:55  leeli73  阅读(314)  评论(1编辑  收藏  举报