Writing in database via JSON php issue

Question

Here is the code is written is RadarsMySql.php

 <?php
 $con = mysql_connect("localhost","root","dacian");


 $db = mysql_select_db("radars");

 $sql = mysql_query("SELECT latitude,longitude,description FROM radars WHERE id > '0'");

 while($row=mysql_fetch_assoc($sql))
 $output[]=$row;
 print(json_encode($output));
 foreach($out["radars"] as $radars) { 
 $latitude = addslashes($radars[latitude]); 
 $longitude= addslashes($radars[longitude]); 
 $description = addslashes($radars[description]);

 mysql_query("INSERT INTO radars (latitude, longitude, description) VALUES('$name', '$ingredients', '$ingredients2')") or die (mysql_error()); 
 }

  mysql_close(); ?>

It gives me this error :

Notice: Undefined variable: out in C:\xampp\htdocs\RadarsMySql.php on line 13

Warning: Invalid argument supplied for foreach() in C:\xampp\htdocs\RadarsMySql.php on line 13

I need this because i want to write data from android app using JSON. Could anyone tell me what is wrong or to give me some tips ?

The error messages PHP gives you are helpful, clear, and informative. Did you try reading them? — TRiG
– TRiG, Commented May 11, 2012 at 17:35
From what I see, $out is not defined as an array anywhere. What should $out be? — EmmanuelG
– EmmanuelG, Commented May 11, 2012 at 17:35

TRiG · Accepted Answer · 2012-05-14 20:00:57Z

2

foreach ($out["radars"] as $radars) {}

If $out["radars"] is an array, this is fine. If it isn't, you'll get a bug: Invalid argument supplied for foreach.

In your case $out["radars"] doesn't exist at all. In fact $out doesn't exist at all. So you get another bug: Undefined variable out.

You're initialising a variable $output but then trying to use it as $out. That won't work.

To pull data out of the database, encode it as JSON, and output it:

$sql = 'SELECT latitude,longitude,description FROM radars WHERE id>0'
$result = mysql_query($sql);

$rows = array();
while ($row = mysql_fetch_assoc($sql)) $rows[] = $row;
echo json_encode($rows);

And to receive JSON posted to the server, process it, and add it to the database:

// It's being posted as application/json, not as application/x-www-form-urlencoded, so it won't populate the $_POST array.
if ($json = file_get_contents('php://input')) {
    // Never mind. We'll do it ourselves.
    $a = json_decode($json, true); // Now we have a nice PHP array.
    $insert = '';
    foreach ($a as $v) {
        if ($insert) $insert .= ',';
        $insert .= ' ("' . $d->escape($v['lattitude']) . '", "' . $d->escape($v['longitude']) . '", "' . $d->escape($v['description']) . '")';
    }
    $sql = 'INSERT INTO `radars` (`latitude`, `longitude`, `description`) VALUES' . $insert;
    $d->exec($sql);
}

// I'm assuming you have a database class here, $d.
// $d->exec() could be simply mysql_query() or mysqli_query().
// $d->escape() should be mysql_real_escape_string() or mysqli_real_escape_string(), but both of those functions will fail if a database connection is not currently open.
// So $d->escape() should (a) check whether a connection is currently open, (b) open one if not, (c) perform the escape and return the result.

edited May 14, 2012 at 20:00

answered May 11, 2012 at 17:34

TRiG

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9 Comments

stanga bogdan Over a year ago

Notice: Undefined index: radars in C:\xampp\htdocs\RadarsMySql.php on line 13 Warning: Invalid argument supplied for foreach() in C:\xampp\htdocs\RadarsMySql.php on line 13

stanga bogdan Over a year ago

that's what i get after i modified out with ouput. It's ok ?

TRiG Over a year ago

@stanga. I don't know what $output is. Try print_r($output); exit; and add that to the question.

TRiG Over a year ago

Actually, scrap that. I'm going to completely rewrite my answer.

stanga bogdan Over a year ago

Ok. I don't really understand what you say but I wonder if you know a tutorial or you could say me what should I do / write there the enstablish a connection between android application and mysql database ?

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Ryan Kempt · Accepted Answer · 2012-05-11 17:32:47Z

0

Where are you defining $out? ;)

foreach($out["radars"] as $radars) {

answered May 11, 2012 at 17:32

Ryan Kempt

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1 Comment

stanga bogdan Over a year ago

... nowhere. I'm a newbie in mysql/php programming. Could you help me with this ? I have 3 columns latitude , longitude , description . How could i write the php to allow me posting from android app ?

Chintan Raghwani · Accepted Answer · 2012-05-11 17:36:46Z

0

use

     foreach($output["radars"] as $radars) {

which you have created above print statement

answered May 11, 2012 at 17:36

Chintan Raghwani

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1 Comment

stanga bogdan Over a year ago

it's ok if I only receive a notice and a warning ? no errors ?

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