duanbi9202
duanbi9202
2015-08-07 04:54

从android php mysql数据插入应用程序获取json

This is my android json response for php mysql data insert application. please help me to get the original json from this return response or how to skip return web (HTML) table from return response (data success fully inserting to the data base table)

<br />
<font size='1'><table class='xdebug-error xe-deprecated' dir='ltr' border='1' cellspacing='0' cellpadding='1'>
<tr><th align='left' bgcolor='#f57900' colspan="5"><span style='background-color: #cc0000; color: #fce94f; font-size: x-large;'>( ! )</span> Deprecated: mysql_connect(): The mysql extension is deprecated and will be removed in the future: use mysqli or PDO instead in C:\wamp\www\AndroidFileUpload\db_connect.php on line <i>28</i></th></tr>
<tr><th align='left' bgcolor='#e9b96e' colspan='5'>Call Stack</th></tr>
<tr><th align='center' bgcolor='#eeeeec'>#</th><th align='left' bgcolor='#eeeeec'>Time</th><th align='left' bgcolor='#eeeeec'>Memory</th><th align='left' bgcolor='#eeeeec'>Function</th><th align='left' bgcolor='#eeeeec'>Location</th></tr>
<tr><td bgcolor='#eeeeec' align='center'>1</td><td bgcolor='#eeeeec' align='center'>0.0010</td><td bgcolor='#eeeeec' align='right'>252784</td><td bgcolor='#eeeeec'>{main}(  )</td><td title='C:\wamp\www\AndroidFileUpload\upload_news.php' bgcolor='#eeeeec'>..\upload_news.php<b>:</b>0</td></tr>
<tr><td bgcolor='#eeeeec' align='center'>2</td><td bgcolor='#eeeeec' align='center'>0.0020</td><td bgcolor='#eeeeec' align='right'>263888</td><td bgcolor='#eeeeec'>DB_CONNECT->__construct(  )</td><td title='C:\wamp\www\AndroidFileUpload\upload_news.php' bgcolor='#eeeeec'>..\upload_news.php<b>:</b>26</td></tr>
<tr><td bgcolor='#eeeeec' align='center'>3</td><td bgcolor='#eeeeec' align='center'>0.0020</td><td bgcolor='#eeeeec' align='right'>263976</td><td bgcolor='#eeeeec'>DB_CONNECT->connect(  )</td><td title='C:\wamp\www\AndroidFileUpload\db_connect.php' bgcolor='#eeeeec'>..\db_connect.php<b>:</b>11</td></tr>
<tr><td bgcolor='#eeeeec' align='center'>4</td><td bgcolor='#eeeeec' align='center'>0.0020</td><td bgcolor='#eeeeec' align='right'>264792</td><td bgcolor='#eeeeec'><a href='http://www.php.net/function.mysql-connect' target='_new'>mysql_connect</a>
(  )</td><td title='C:\wamp\www\AndroidFileUpload\db_connect.php' bgcolor='#eeeeec'>..\db_connect.php<b>:</b>28</td></tr>
</table></font>

{"success":1,"message":"Product successfully created."}

at the last line of the response have needed json result. any help appreciate. thanks in advance.

here is my php code

<?php


// array for JSON response
$response = array();

// check for required fields
if (isset($_POST['news_Title']) && isset($_POST['news_Details']) && isset($_POST['news_Image'])&& isset($_POST['news_Video']) && isset($_POST['reporter_Name']) && isset($_POST['reporter_Email']) && isset($_POST['reporter_Phone'])) {

    $news_Title = $_POST['news_Title'];
    $news_Details = $_POST['news_Details'];
    $news_Image = $_POST['news_Image'];
    $news_Video = $_POST['news_Video'];
    $reporter_Name = $_POST['reporter_Name'];
    $reporter_Email = $_POST['reporter_Email'];
    $reporter_Phone = $_POST['reporter_Phone'];

    // include db connect class
    require_once __DIR__ . '/db_connect.php';

    // connecting to db
    $db = new DB_CONNECT();

    // mysql inserting a new row
    $result = mysql_query("INSERT INTO news(news_Title, news_Details, news_Image, news_Video, reporter_Name, reporter_Email , reporter_Phone) VALUES('$news_Title', '$news_Details', '$news_Image', '$news_Video', '$reporter_Name', '$reporter_Email', '$reporter_Phone')");

    // check if row inserted or not
    if ($result) {
        // successfully inserted into database
        $response["success"] = 1;
        $response["message"] = "Product successfully created.";

        // echoing JSON response
        echo json_encode($response);

    } else {
        // failed to insert row
        $response["success"] = 0;
        $response["message"] = "Oops! An error occurred.";

        // echoing JSON response
        echo json_encode($response);
    }
} else {
    // required field is missing
    $response["success"] = 0;
    $response["message"] = "Required field(s) is missing";

    // echoing JSON response
    echo json_encode($response);
}

?>
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2条回答

  • du4010 du4010 6年前

    Simply i solved my problem by using tricky way. but this is not the exact correct way to solve this.

    String jsonStrObj = strReturn.substring(strReturn.lastIndexOf('{') + 1);
    try {
        jObj = new JSONObject(jsonStrObj);
    } catch (JSONException e) {
        Log.e("JSON Parser", "Error parsing data " + e.toString());
    }
    

    Waiting for the correct answer.. thank you

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  • douzhuan0309 douzhuan0309 6年前

    There is no problem in response. Your mysql_connect() function will deprecated in future version of php Use msyqli.

    <?php
    $connection = mysqli_connect('localhost', 'username', 'password', 'database');
    ?>
    

    or turn off all deprecated warnings including them from mysql_*:

    <?php
    error_reporting(E_ALL ^ E_DEPRECATED);
    ?>
    
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