inserting data in mysql data base in php

Question

i am sending data from iphone to server using php it is sending data from iphone but not inserting in mysql i am using following php code.

 <?php
  $con =     

  mysql_connect("surveyipad.db.6420177.hostedresource.com","tom","ben");
   if (!$con)
    {
   die('Could not connect: ' . mysql_error());
   }

   mysql_select_db("surveyipad", $con);



   $device_Id=$_POST['device_Id'];
   $R1=$_POST['R1'];
   $R2=$_POST['R2'];
   $R3=$_POST['R3'];
   $comment=$_POST['comment'];
   $update_date_time=$_POST['update_date_time'];

    $query=("INSERT INTO survey_responsese_pfizer (device_Id,R1,R2,R3,comment,update_date_time)

  VALUES ('$device_Id','$R1','$R2','$R3','$comment','$update_date_time')");

   mysql_query($query,$con);
   printf("Records inserted: %d\n", mysql_affected_rows());

    echo($device_Id)
   ?>

What debugging have you done? Wheres the ending semi-colon for echo($device_Id)? you also have sql-injection, you should escape inputs else adding ' to one of your parameters will break the query... — Lawrence Cherone
– Lawrence Cherone, Commented Jan 17, 2013 at 8:03
try printing out mysql_error(), to see if there was some sort of error and post it here — Darvex
– Darvex, Commented Jan 17, 2013 at 8:04
keep your strings outside of the quotes. Use sprintf for example. Also, be on your guard for user input, it may be false. Mysql_real_escape_string will be needed on your $_POST values and - as @zan said - go have a look at PDO. Sidenote: the use of die() is something not needed here, use propper error handling in stead — Tikkes
– Tikkes, Commented Jan 17, 2013 at 8:06
@Darvex Records inserted: -1 My Sql Error: 0 this is the result i am getting after printing — user1808433
– user1808433, Commented Jan 17, 2013 at 8:08
Hmmm...perhaps the value is getting inserted then? Did you check your db? — Darvex
– Darvex, Commented Jan 17, 2013 at 8:14

Lawrence Cherone · Accepted Answer · 2013-01-17 08:37:11Z

Ok one only learns by example, stop using the mysql_functions(), They are no longer maintained and are officially deprecated. And in PHP 5.6 they will most likely be removed, rendering you code broken.

Move over to PDO with prepared querys. A Port of your current code using PDO:

<?php
// SQL Config
$config['sql_host']='surveyipad.db.6420177.hostedresource.com';
$config['sql_db']  ='surveyipad';
$config['sql_user']='tom';
$config['sql_pass']='ben';

// SQL Connect
try {
        $db = new PDO("mysql:host=".$config['sql_host'].";dbname=".$config['sql_db'], $config['sql_user'], $config['sql_pass']);
        $db->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
        $db->setAttribute(PDO::ATTR_EMULATE_PREPARES, false);
}catch (Exception $e){
        die('Cannot connect to mySQL server.');
}

// Check for POST, add isset($_POST['device_Id']) ect to add validations
if($_SERVER['REQUEST_METHOD']=='POST'){
   // Build your query with placeholders
   $sql = "INSERT INTO survey_responsese_pfizer 
                  (device_Id,R1,R2,R3,comment,update_date_time)
                   VALUES
                  (:device_Id, :R1, :R2, :R3, :comment, :update_date)";

    // Prepare it
    $statement = $db->prepare($sql);
    // Assign your vairables to the placeholders
    $statement->bindParam(':device_Id', $_POST['device_Id']);
    $statement->bindParam(':R1', $_POST['R1']);
    $statement->bindParam(':R2', $_POST['R2']);
    $statement->bindParam(':R3', $_POST['R3']);
    $statement->bindParam(':comment', $_POST['comment']);
    $statement->bindParam(':update_date', $_POST['update_date_time']);
    // Execute the query
    $statement->execute();

    echo htmlspecialchars($device_Id);
}
?>

Untested tho, hope it helps.

Why use die()? You catch an exception so why kill it? Otherwise, good answer! +1

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