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Android将数据插入mysql json_Android_Mysql_Json - Fatal编程技术网

Android将数据插入mysql json

Android将数据插入mysql json,android,mysql,json,Android,Mysql,Json,我正在尝试使用json将数据插入mysql这是我的php代码,没有问题,但它插入查询无法工作。json输出始终显示:{“code”:0}它必须是{“code”:1}帮助 <?php header('Content-type=application/json; charset=utf-8'); $con = mysql_connect("xxxx","xxx","xxx"); $db = mysql_select_db("android_data");

我正在尝试使用json将数据插入mysql这是我的php代码,没有问题,但它插入查询无法工作。json输出始终显示:
{“code”:0}
它必须是
{“code”:1}
帮助

<?php

    header('Content-type=application/json; charset=utf-8');

    $con = mysql_connect("xxxx","xxx","xxx");
    $db = mysql_select_db("android_data");

    $us = $_POST["uname"];
    $pa = $_POST["upass"];
    $pe = $_POST["uper"];

    $flag["code"] = 0;

    $sql = "insert into administrator (username,password,permission) values('".$us."','".$pa."','".$pe."')";
    $query = mysql_query($sql);

    if ($query)
    {
        $flag["code"] = 1;
        echo json_encode($flag);
        //echo $flag['StatusID'];
    }

    mysql_close($con);
    echo json_encode($flag);
?>

试试这个:

$query = mysql_query("INSERT INTO administrator (username, password, permission) VALUES( '$us', '$pa', '$pe' )");
还有一件事需要注意,看起来您要为一个成功的查询回显两次输出

您可能希望执行以下操作:

if ($query)
{
    $flag["code"] = 1;
    $flag["message"] = "success";
    echo json_encode($flag);


} else {
    // failed to insert row
   $flag["code"] = 0;
   $flag["message"] = "Oops! An error occurred.";
   echo json_encode($flag);
}

 mysql_close($con);