PHP - Query Empty When Trying To Insert Into Database (odbc Connection)

I have this code:
Code: [Select]
$sql = "SELECT adpat, adptnm, addate, adtime, adddsc, disdt, distm, drname, adptyp, ptprvtyp FROM hma711.zadpatnmf
JOIN hma711.addocrmf on adamdr = drno WHERE addate BETWEEN '20111115' AND '20111115' AND adpat BETWEEN '2000000' AND '2999999'";

$execute = odbc_exec($conn, $sql);

$num = odbc_num_rows($execute);
echo $num;

while($row = odbc_fetch_array($execute)){
    echo $row['adpat'];
}
It gets an error on this line: echo $row['adpat'];

The error is Notice: Undefined index: adpat in C:\xampp\htdocs\erboard\test.php on line 29

The query works directly on the AS400 I'm pulling from and the $num is being populated with the correct number of records.

This is the first time I have ever used php with odbc so I'm not sure what I'm missing. Any help would be greatly appreciated. Thanks!

CAN ANYONE TELL ME WHAT i AM DOING WRONG. I WANT TO RETRIEVE THE VERY FIRST RECORD IN MY DATABASE WHEN THE CODE EXECUTE IT ONLY SHOWS THE LAST RECORD IN THE DATABASE

<?PHP
$thisMonth = date('M');
$thisDay = date('j');
$eventMonth = array();
$eventDay = array();
$eventTime = array();
$eventName = array();
$eventLocation = array();
$dbMonth="";
$dbDay="";

$i=0;

$conn = odbc_connect('eventsDB','','');

$sql= "SELECT month,day, time, event,location FROM Events";
$rs="$conn,$sql";
if (!$conn)
{
   exit("Connection Failed: " . $conn);
}
 $rs=odbc_exec($conn,$sql);

if(!$rs)
{
   exit("Error in SQL");
}
 echo "DATABASE OPEN";
 

while($i<3)
{
   $dbMonth= odbc_result($rs,"month");
   echo $eventMonth[$i]=odbc_result($rs,"month")."\n"; 
   
   
   if($dbMonth<>$thisMonth)
   {
      odbc_fetch_row($rs);   
   
   }
                echo $eventMonth[$i]=odbc_result($rs,"month")."\n"; 
               echo $eventDay[$i]=odbc_result($rs,"day")."\n";
               echo $eventTime[$i]=odbc_result($rs,"time")."\n";
               echo $eventDay[$i]=odbc_result($rs,"event")."\n";
               echo $eventLocation[$i]=odbc_result($rs,"location")."\n";
               $i++;
               odbc_fetch_row($rs);
            
                  
         
   

echo $i;
} //ends while loop
odbc_close($conn);
?>

Okay guys I have finished my db and I want to upload it to a live server but don't know exactly what to change in the code to get the ODBC_connect to open the path at the new server location please help.

my database  is located in a subfolder called Databases

i.e. "www.mydomain.com/Databases/myEvents.mdb"

what do i change in the connection string below to get it to open the file on the live server? Thanks in advance.


$conn = odbc_connect('myEvents','','');

Hey guys,

I`m having problems finding information on how to insert a image on a blob column trough odbc on a oracle database. Does anyone knows where can i find it, or can anyone help me with sample code? I really need to use odbc functions (cant use oci8 )

Tanks for your time reading this and for the possible input you may add on this.

Sincerely

Arestas

after cloasing connection of database i still got the values form database.

Code: [Select]
<?php
session_start();
/* 
 * To change this template, choose Tools | Templates
 * and open the template in the editor.
 */
require_once '../database/db_connecting.php';
$dbname="sahansevena";//set database name
 $con=  setConnections();//make connections use implemented methode in db_connectiong.php
 mysql_select_db($dbname, $con);
//update the time and date of the admin table
 $update_time="update admin set last_logged_date =CURDATE(), last_log_time=CURTIME() where username='$uname'limit 3,4";
 //my admin table contain 5 colums they are id, username,password, last_logged_date, last_log_time
 $link= mysql_query($update_time);
 // mysql_select_db($dbname, $link);
//$con=mysql_connect('localhost', 'root','ijts');

       $result="select * from admin where username='a'";

        $result=mysql_query($result);
mysql_close($con);
//here i just check after closing data baseconnection whether i do get reselts but i do, why?
echo "after the cnnection was closed";
if(!$result){
echo "cont fetch data";

}else{

   $row= mysql_fetch_array($result);
   echo "id".$row[0]."usrname".$row[1]."passwped".$row[2]."date".$row[3]."time".$row[4];
}
//  echo "<html>";
//echo "<table border='1' cellspacing='1' cellpadding='2' align='center'>";
//      echo "<thead>";
//           echo"<tr>";
//                   echo "<th>";
//                   echo ID;
//                   echo"</th>";
//                   echo" <th>";echo Username; echo"</th>";
//                    echo"<th>";echo Password; echo"</th>";
//                    echo"<th>";echo Last_logged_date; echo "</th>";
//                   echo "<th>";echo Last_logged_time;  echo "</th>";
//               echo" </tr>";
//           echo" </thead>";
//           echo" <tbody>";
//while($row= mysql_fetch_array($result,MYSQL_BOTH)){
//             echo "<tr>";
//              echo "<td>";
//              echo $row[0];
//               echo "</td>";
//                   echo "<td>";
//              echo $row[1];
//              echo "</td>";
//                  echo "<td>";
//              echo $row[2];
//               echo "</td>";
//                   echo "<td>";
//                echo $row[3];
//              echo "</td>";
//                  echo "<td>";
//              echo $row[4];
//               echo "</td>";
//            echo "</tr>";
//     }
//         echo" </tbody>";
//         echo "</table>";
//         echo "</html>";
     session_destroy();
session_commit();

echo "session and database are closed but i still get values from doatabase session is destroyed".$_SESSION['admin'];
?>

session is destroyed but database connection is not closed.

thanks

What I am looking for is that if any field in a database is empty, the form comes back and puts in the word "empty" into the web page, but not the database.


I got this right now, it shows the first part fine, but the second part it doesn't show the "Empty" comment... I don't understand why... as far as I can tell the code is fine...

Code: [Select]
<?php
            //Nonconformity, Disposition, Comments and Comments & Additional Details
            echo '<div id="box3">';
                if (!empty($row['Nonconformity']) || !empty($row['Disposition']) || !empty($row['Comments']) || !empty($row['CommentsAdditional_Details'])) {
                    echo '<div id="non"><span class="b">Nonconformity:&nbsp;&nbsp;</span><br />' . $row['Nonconformity'] . '</div>';
                    echo '<div id="dis"><span class="b">Disposition:&nbsp;&nbsp;</span><br />' . $row['Disposition'] . '</div>';
                    echo '<div id="comm"><span class="b">Comments:&nbsp;&nbsp;</span><br />' . $row['Comments'] . '</div>';
                    echo '<div id="comma"><span class="b">Comments and/or Additional Details:&nbsp;&nbsp;</span><br />' . $row['CommentsAdditional_Details'] . '</div>';}
                        else
                            if (empty($row['Nonconformity']) || empty($row['Disposition']) || empty($row['Comments']) || empty($row['CommentsAdditional_Details'])) {
                                echo '<div id="non"><span class="b">Nonconformity:&nbsp;&nbsp;</span><br />Empty</div>';
                                echo '<div id="dis"><span class="b">Disposition:&nbsp;&nbsp;<br /></span>Empty</div>';
                                echo '<div id="comm"><span class="b">Comments:&nbsp;&nbsp;<br /></span>Empty</div>';
                                echo '<div id="comma"><span class="b">Comments and/or Additional Details:&nbsp;&nbsp;</span><br />Empty</div>';}
            echo '</div>';
?>

Hello guys,

I've installed the mongoDB PHP driver using "sudo pecl install mongodb"

Testing shows driver is installed correctly and it appears nicely in phpinfo():

ludo@aegir:~$ php7.2 -i | grep -i mongodb
mongodb
MongoDB support => enabled
MongoDB extension version => 1.5.3
MongoDB extension stability => stable
mongodb.debug => no value => no value 

For connecting I've got:

   $db = new MongoDB\Driver\Manager("mongodb://localhost:27017"); 

Now I want to define a collection and insert a record, i've been using code from here  https://www.php.net/manual/en/set.mongodb.php:

$collection = $db->mydatabase->mytable;

$result = $collection->insertOne( [ 'name' => 'test', 'item' => 'testitem' ] );

But this fails  Notice: Undefined property: MongoDB\Driver\Manager::$mydatabase in /var/www....

Has anyone got a workingg example of defining and inserting into mongoDB. Seems the driver was updated fairly recently and so stuff on the internet is not always relevant. 

Edited May 17, 2019 by ludo1960

I have a form with a series of text boxes where the User enters specific data, be it text, numeric, or date/time.  It is not mandatory to fill in every box, but when I submit the form to be put into a mysql table, I get errors such as

"Incorrect time value: '' for column 'starttime' at row 1"
"Incorrect decimal value: '' for column 'workkj' at row 1"


How can I setup the mysql table to accept blank answers?  Or is there something I need to do in php to fix this?

Sorry if this is simple... I'm still learning.
Thanks for the help!

Hi guys

I have the code below and have 4 text fields, the issue is i'd like to prevent insert query if textfield is empty, do u know how?




if (isset($_POST['register']) && $_POST['register'])
    {

$reference = rand(11111111,99999999);

$name1 = addslashes(strip_tags($_POST['name1']));
$name2 = addslashes(strip_tags($_POST['name2']));
$name3 = addslashes(strip_tags($_POST['name3']));
$name4 = addslashes(strip_tags($_POST['name4']));

$addname4=mysql_query("INSERT INTO nearest (stref, name) VALUES('$reference','$name4')");
$addname3=mysql_query("INSERT INTO nearest (stref, name) VALUES('$reference','$name3')");
$addname2=mysql_query("INSERT INTO nearest (stref, name) VALUES('$reference','$name2')");
$addname1=mysql_query("INSERT INTO nearest (stref, name) VALUES('$reference','$name1')");

}




thanks

If you also have any feedback on my code, please do tell me. I wish to improve my coding base.

Basically when you fill out the register form, it will check for data, then execute the insert query. But for some reason, the query will NOT insert into the database. In the following code below, I left out the field ID. Doesn't work with it anyways, and I'm not sure it makes a difference.

Code:

  Code: [Select]
mysql_query("INSERT INTO servers (username, password, name, type, description, ip, votes, beta) VALUES ($username, $password, $name, $server_type, $description, $ip, 0, 1)");
Full code:

Code: [Select]
<?php
include_once("includes/config.php");
 
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<title><? $title; ?></title>
<meta http-equiv="Content-Language" content="English" />
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<link rel="stylesheet" type="text/css" href="style.css" media="screen" />
</head>
<body>
 
<div id="wrap">
 
<div id="header">
<h1><? $title; ?></h1>
<h2><? $description; ?></h2>
</div>
 
<? include_once("includes/navigation.php"); ?>
 
<div id="content">
<div id="right">
<h2>Create</h2>
<div id="artlicles">
<?php
 
if(!$_SESSION['user'])
{
 
        $username = mysql_real_escape_string($_POST['username']);
        $password = mysql_real_escape_string($_POST['password']);
        $name = mysql_real_escape_string($_POST['name']);
        $server_type = mysql_real_escape_string($_POST['type']);
        $description = mysql_real_escape_string($_POST['description']);
 
        if(!$username || !$password || !$server_type || !$description || !$name)
        {

        echo "Note: Descriptions allow HTML. Any abuse of this will result in an IP and account ban. No warnings!<br/>All forms are required to be filled out.<br><form action='create.php' method='POST'><table><tr><td>Username</td><td><input type='text' name='username'></td></tr><tr><td>Password</td><td><input type='password' name='password'></td></tr>";
        echo "<tr><td>Sever Name</td><td><input type='text' name='name' maxlength='35'></td></tr><tr><td>Type of Server</td><td><select name='type'>
       
        <option value='Any'>Any</option>
        <option value='PvP'>PvP</option>
        <option value='Creative'>Creative</option>
        <option value='Survival'>Survival</option>
        <option value='Roleplay'>RolePlay</option>
       
        </select></td></tr>
       
        <tr><td>Description</td><td><textarea maxlength='1500' rows='18' cols='40' name='description'></textarea></td></tr>";
        echo "<tr><td>Submit</td><td><input type='submit'></td></tr></table></form>";
        }
        elseif(strlen($password) < 8)
        {
        echo "Password needs to be higher than 8 characters!";
        }
        elseif(strlen($username) > 13)
        {
                echo "Username can't be greater than 13 characters!";
        }
        else
        {
                $check1 = mysql_query("SELECT username,name FROM servers WHERE username = '$username' OR name = '$name' LIMIT 1");
               
                if(mysql_num_rows($check1) < 0)
                {
                        echo "Sorry, there is already an account with this username and/or server name!";
                }
                else
                {
                        $ip = $_SERVER['REMOTE_ADDR'];

                        mysql_query("INSERT INTO servers (username, password, name, type, description, ip, votes, beta) VALUES ($username, $password, $name, $server_type, $description, $ip, 0, 1)");
                        echo "Server has been succesfully created!";
                }
        }
       
}
else
{
        echo "You are currently logged in!";
}
 
?>
</div>
</div>
 
<div style="clear: both;"> </div>
</div>
 
<div id="footer">
<a href="http://www.templatesold.com/" target="_blank">Website Templates</a> by <a href="http://www.free-css-templates.com/" target="_blank">Free CSS Templates</a> - Site Copyright MCTop
</div>
</div>
 
</body>
</html>

hey i need help im tryig to get information from my user and then process it in my database so i can use it to log them to a different web site im trying to use this method to get the information from the user but need help to get it please help me.
  Code: [Select]
  //make the database connection.
        $conn = mysql_connect("localhost", "Black Jack");
        mysql_select_db("chaper7", $conn);
       
        //create a query
       
        $sql = "SELECT * FROM hero";
        $result = mysql_query($sql, $conn);

On some occasions I need to connect to a second and third database in the same script (maybe 5% of scripts have at least a second connection).

Usually I would just select the new database. However, my host requires different users to be created for each database.

What is the best way to do this?

Close current connection (say db1) and open new (say db2) OR keep all open, creating 2nd and 3rd connections.

I am happy with the design of my database, and don't want to merge all these tables into one db.

Overall I am still happy with my host, so I'd rather not change.

<?php class UserQuery { public function Adduser($id,$username,$email,$password)

{

$conn = new Config();

$sql =("INSERT INTO test.user (id, username, email, password) VALUES ('$id', '$username', '$email',$password)");

$conn->exec($sql); }

}

 

getting an "exec doesnt exist "    error,  saying exec doesnt exist in my db file.

it doesnt need to exist does it ?

anyone any idea why ? 

 

I'm not sure what i have done wrong here...
all it says is "Query was empty"


<?php
SESSION_START();

?>

//some html

<?php

if($_SESSION['email']){

$mysql = "SELECT id,firstname FROM users WHERE 'id'='".$_SESSION['email']."'";
$res = mysql_query($sql) or die(mysql_error());

if(mysql_num_rows($res) == 0){
session_destroy();
echo"Please <a href=\"/index.html\">Login</a> or <a href=\"/register.php\">Join Us.</a>";
}else{
$row = mysql_fetch_assoc($res);
echo "welcome back, <a href=\"./index.php?act=profile&id=".$row['id']."\">".$row['firstname']."</a>\n";

}
}else {
echo"Please <a href=\"/index.html\">Login</a> or <a href=\"/register.php\">Join Us.</a>";

}
?>

The Script: pdo_connect.php:

<?php
$dsn = "mysql:host=localhost;dbname=2354";

$db = new PDO($dsn, "root", "");

?>

pdo_test.php:

<?php
    try{
        require_once("pdo_connect.php");
    }catch(Exception $e){
        $error = $e->getMessage();
    }

?>
<!DOCTYPE html>
<html>
<head>
    <title>Database Connection with PDO</title>

</head>
<body>
<h1>Connection with PDO</h1>
<?php
    if($db){
        echo "<p>Connection successful.</p>";
    }elseif(isset($error)){
        echo "<p>$error</p>";
    }

?>
</body>
</html>

I have just watched a tutorial and tried out this script. The issue is that I am getting the following notice alongside with the error message:

Notice: Undefined variable: db in C:\xampp\htdocs\oophp\pdo_test.php on line 18

SQLSTATE[HY000] [1049] Unknown database '2354'

By the way this notice does happen in the tutorial as well.   My question is: How to have this in ways, where the notice does not occur?