PHP - Transactions Problem: Inserting Data Into Two Tables.

No idea what I am doing here.  I have a joined table which is:

$sql = "SELECT itemnum, image1, title, close, quantity, bidquantity, ".
          "bidprice, buyitprice, seller, biditem, buser, bidder " .
          "FROM Items LEFT OUTER JOIN Bids " .
          "ON Items.itemnum = Bids.biditem " .
          "WHERE buser != '' and close > '$time' and outbid = '0'".
          #"GROUP BY buser " .
          #"HAVING bidder=$buser or buser=\'$busername\' ".
          "ORDER by buser";
$result = mysqli_query($con, $sql) or die ( mysqli_error());
$row = mysqli_fetch_assoc($result);

I need to loop through the table on buser and insert select data into two other tables.  One of the tables is the control table which will only have one row per buser.  The other table I guess you would call the data table of the two and may have more than one row depending.  The catch is these will be linked together by an invoice number.  Table A, the control table has a field for invoice number, table B does not.  Of course once the data is inserted into the new tables they will be joined and from there on it should be easy.  I just don't know now how to get there.  I've thought about using an array to loop through, but I don't know enough about them to make any sense of this.  Not even sure the group by may help me in doing this.

Anyone have any ideas on this?  I sure do need it and thanks in advance!

  $username = $_POST['uid'];
  $email = $_POST['mail'];
  $password = $_POST['pwd'];
  $passwordRepeat = $_POST['pwd-repeat'];
  $date = $_POST['date2'];
  $stream = $_POST['relationship'];

		$sql1 = "INSERT INTO users (uidUsers, emailUsers, pwdUsers, relationship) VALUES (?, ?, ?, ?);";
		$sql2 = "INSERT INTO Family1 (username, application_filed, relationship) VALUES (?, ?, ?);";
		$sql3 = "INSERT INTO Family2 (username, application_filed, relationship) VALUES (?, ?, ?);";
		mysqli_query($sql1, $conn);
		mysqli_query($sql2, $conn);
		mysqli_query($sql3, $conn);
		
   $stmt = mysqli_stmt_init($conn);
		if (!mysqli_stmt_prepare($stmt, $sql2)) {
          header("Location: ../signup.php?error=sqlerror");
          exit();
        }
		        else {

          mysqli_stmt_bind_param($stmt, "sss", $username, $date, $stream);
          $result = mysqli_stmt_get_result($stmt);
		  if ($row = mysqli_fetch_assoc($result))
		  ($username==$_SESSION['uid'] and $stream =='nursing');
			mysqli_stmt_execute($stmt);
		}
		
		if (!mysqli_stmt_prepare($stmt, $sql3)) {
          header("Location: ../signup.php?error=sqlerror");
          exit();
        }
		        else {

          mysqli_stmt_bind_param($stmt, "sss", $username, $date, $stream);
          $result = mysqli_stmt_get_result($stmt);
		  if ($row = mysqli_fetch_assoc($result))
		  ($username==$_SESSION['uid'] and $stream =='doctoral');
			mysqli_stmt_execute($stmt);
		}
		
        if (!mysqli_stmt_prepare($stmt, $sql1)) {
          header("Location: ../signup.php?error=sqlerror");
          exit();
        }

 if (!mysqli_stmt_prepare($stmt, $sql1)) {
          header("Location: ../signup.php?error=sqlerror");
          exit();
        }
        else {
          $hashedPwd = password_hash($password, PASSWORD_DEFAULT);
          mysqli_stmt_bind_param($stmt, "ssss", $username, $email, $hashedPwd, $stream);
          mysqli_stmt_execute($stmt);
          header("Location: ../signup.php?signup=success");
          exit();

I was wondering if someone could point me in the right direction. I have this code. They idea I had behind it is to insert values into different tables depending on variables being passed. 

So when user fills out a form and selects $stream="nursing" I want results to go to table 'users' and 'Family1', but not 'Family2' table. and if user selects $stream='doctoral' results should go to table 'users' and  'Family2', and not go to 'Family1'

But with my query I get results go to both table and also users table. And there is no restriction to what users selects, variable $stream being passed no matter what it is.

Is this the wrong way to go here? Did I completely mess up the logic?

This UPDATE script works:

$query = $pdo->prepare("UPDATE posts SET headline = ? WHERE id = ?");
    $query->bindValue("1", $_POST['headline']);
    $query->bindValue("2", "2");
    $query->execute();

Though, this INSERT script does not work:

$query = $pdo->prepare("INSERT INTO posts (headline, post) VALUES (?, ?)");
    $query->bindValue("1", $_POST['headline']);
    $query->bindValue("2", $_POST['post']);
    $query->execute();

Nothing gets added, when executing this script.   The connection:

<?php

$dsn = "mysql:dbname=phpblog;host=localhost";
$user = "root";
$password = "";
try{
    $pdo = new PDO($dsn, $user, $password);
}catch(PDOException $e){
    echo "Connection failed: " . $e->getMessage();
}

?>

Any suggestions why the INSERT script is not working?
Edited by glassfish, 25 November 2014 - 09:26 AM.

Can someone please help me figure out how to insert "image" from table users, that contains the .jpg data associated with "screen_name", into table friends "friendimg"

Here are the tables:
friends: id, member, friendwith, friendimg
users: id, screen_name, image

$_GET["id"] only contains "screen_name" and"image" is not included

add.php
Code: [Select]
<?php
 if(isset($_GET["id"]))
 {
echo "<br/><a href=\"member?add=".$_GET["id"]."\">Add ".$_GET["id"]." to your list of friends</a>";
 }
?>member.php
Code: [Select]
<?php
 if(isset($_GET["add"]))
 {
$username = $_SESSION["screen_name"];
$friend = $_GET["add"];

$query = "SELECT * FROM friends WHERE member='$username' AND friendwith='$friend'";
$result = mysql_query($query);
$exist = mysql_num_rows($result); 
if($exist=='0')
{
$query = "INSERT INTO friends(member,friendwith) VALUES('$username','$friend')";
mysql_query($query);
echo "<br/>You are now friends with $friend";
}
else 
{
echo "<br/>$friend is already in your list of friends!";
}
 }
?>

i want insert data from text box in html form.there are many text boxes gave name using 2 loops..please tell how can insert data to table?please reply

Hello,
I have three scripts below that I was wondering if someone can look at as a second opinion.

What they all have in common is they are supposed to append data to some tables in my database.  They match the code on all my other scripts but for some reason these particular scripts will not append the data to the files.

This first one...
is supposed to insert the name of the doctor this person uses, and then gets the last ID created and inserts it into another table...no matter what I try, it will not do it.

Code: [Select]
<?php

session_start();
$userid=$_SESSION['userID'];
$firstName=$_SESSION['fname'];
$lastName=$_SESSION['lname'];
$camperID=$_POST['camperID'];
$physicianFName=$_POST['physicianFName'];
$physicianLName=$_POST['physicianLName'];
$physicianSuffix=$_POST['physicianSuffix'];
$practiceName=$_POST['practiceName'];
$physicianAddress1=$_POST['physicianAddress1'];
$physicianAddress2=$_POST['physicianAddress2'];
$physicianAddress3=$_POST['physicianAddress3'];
$physicianCity=$_POST['physicianCity'];
$physicianState=$_POST['physicianState'];
$physicianZip=$_POST['physicianZip'];
$physicianOfficeP=$_POST['physicianOfficeP'];
$physicianOfficeF=$_POST['physicianOfficeF'];
$physicianSpecialty=$_POST['physicianSpecialty'];


echo $camperID. ' ' .$physicianFName. ' ' .$physicianLName;

//htc information and physician

$htcName=$_POST['htcInfo'];
$htcPhysician=$_POST['htcPhysician'];

$con=mysql_connect("xxx.xxx.xxx.xx", "xxx", "xxxx") or die ("cannot connect");
mysql_select_db("testcamp") or die ("cannot select database");

$sql="INSERT INTO Physicians(physicianFName, physicianLName, physicianSuffix, practiceName, physicianAddress1, physicianAddress2, physicianAddress3, physicianCity,
physicianState, physicianZip, physicianOfficeP, physicianOfficeF, physicianSpecialty) VALUES ('$physicianFName', '$physicianLName', '$physicianSuffix', '$physicianAddress1',
'$physicianAddress2', '$physicianAddress3', '$physicianCity', '$physicianState', '$physicianZip', '$physicianOfficeP', '$physicianOfficeF', '$physicianSpecialty')";

$result=mysql_query($sql,$con);
$physicianID=mysql_insert_id();

$sql2="INSERT INTO CamperPhysicianInfo(physicianID, camperID) VALUES ('$physicianID', '$camperID')";
$result2=mysql_query($sql2,$con);

$sql3="INSERT INTO CamperHTCInfo(camperID, htcID) VALUES ('$camperID', '$htcName')";
$result3=mysql_query($sql3,$con);

$sql4="INSERT INTO camperHTCPhysicianInfo(camperID, htcPhysicianID) VALUES ('$camperID', '$htcPhysician')";
$result4=mysql_query($sql4,$con);

?>

This next one...is supposed to enter data into the CamperRegistrations tables and also enter data into another table for a questionaire that was completed on the form previous to this page...same thing no matter what I try will not append data.

Code: [Select]
<?php

$question12=$_POST['question12'];
$question12a=$_POST['question12a'];





session_start();
$userid=$_SESSION['userID'];
$firstName=$_SESSION['fname'];
$lastName=$_SESSION['lname'];

$camperID=$_SESSION['camperID'];


$con=mysql_connect("xxx.xxx.xxx.xx", "xxx", "xxx") or die ("cannot connect");
mysql_select_db("testcamp") or die ("cannot select database");

$sql="INSERT INTO Question12(camperID, question12Answer, teaching) VALUES ('$camperID', '$question12', '$question12a')";
$result=mysql_query($sql,$con);

$date=date("Y-m-d");
$defaultStatus=1;

$sql2="INSERT INTO CamperRegistrations(userID, camperID, registrationDate, registrationStatus) VALUES ('$userid', '$camperID', '$date', '$defaultStatus')";
$result2=mysql_query($sql2,$con);

?>

and then on this same page with this script...
is it possible to carry session information when using HTML links instead of a submit form?

I need the userid to carry to the next page with a link, but it won't do that either...

Any help would be appreciated and thanks in advance for looking.

~jcjst21

Stuck again on simple code.  I am trying to insert some fields extracted from one table into another.  I'm using code that worked elsewhere.  The SQL statement flies, the script runs, the input array is printed back

 I get an echo back from the end of the script but nothing is added to the table.  Even aded an echo  print_r in the conditional and I know the data is getting to the execute command.    The script follows with a sample of the input array.  I have attached am image of the table I am trying to insert the data into.

--Kenoli

The script:
<?php
	
require '__classes/DB.php';

$sql = "SELECT name, table_id, image_name, description, medium  FROM tbl_person_data ";

$stmt = $pdo->query($sql);
$array1 = $stmt->fetchall(PDO::FETCH_ASSOC);
	
	
$stmt = $pdo->prepare("INSERT INTO Images (name, person_id, filename, description, medium) VALUES (?,?,?,?,?)");

    //$pdo->beginTransaction();
    foreach ($array1 as $row)
    {
        $stmt->execute($row);
    }
    
echo "<pre>";
print_r ($row);
echo "</pre>";
    
		
echo '<h4>Got to end of file</h4>';
	
?>

$array1: The input array

[0] => Array
        (
            [name] => Carol Lettko
            [table_id] => 21
            [image_name] => Carol_Lettko-DSC_3022.jpg
            [description] => Baby Herons/Brickyard
            [medium] => photo
        )

    [1] => Array
        (
            [name] => 
            [table_id] => 22
            [image_name] => Carol_Lettko-DSC_0164.JPG
            [description] => Heron/Brickyard
            [medium] => photo
        )

    [2] => Array
        (
            [name] => 
            [table_id] => 23
            [image_name] => Carol_Lettko-IMG_5723.jpg
            [description] => Kayaker/Brickyard
            [medium] => photo
        )

 

I am at a loss why my query is not inserting values into db. Even if I do echo $query or var_dump($query) there is nothing printed at all. All values are being passed successfully just not being inserted. I am getting 'Could not connect' but I do not know why. All connections are established and as a test I took this code and ran it on it's own with dummy data and it inserted the data fine. I can only think it has something to do with the $response_array.  Where am I going wrong. and would appreciate any help. Thanks

Code: [Select]
<?php require_once('Connections/sample.php'); ?>
<?php

session_start();

$new = 1;
$activity = 'General Contact Enquiry';
$mobile = 'Submitted from mobile';
$name = mysql_real_escape_string($_POST['GC_name']);
$department = mysql_real_escape_string($_POST['GC_department']);
$message = mysql_real_escape_string($_POST['GC_message']);
$email = mysql_real_escape_string($_POST['GC_email']);
$company = mysql_real_escape_string($_POST['GC_company']);
$position = mysql_real_escape_string($_POST['GC_position']);

//response array with status code and message
$response_array = array();

//validate the post form

//check the name field
if(empty($name)){

    //set the response
    $response_array['status'] = 'error';
    $response_array['message'] = 'Name cannot be blank';

//check the name field
} elseif(empty($company)) {

    //set the response
    $response_array['status'] = 'error';
    $response_array['message'] = 'You must enter a company name';

//check the position field
}elseif(empty($position)) {

    //set the response
    $response_array['status'] = 'error';
    $response_array['message'] = 'You must enter a position';

//check the email field
} elseif(empty($email)) {

    //set the response
    $response_array['status'] = 'error';
    $response_array['message'] = 'You must enter a valid email address';

//check the dept field
}elseif($department=="Choose Department") {

    //set the response
    $response_array['status'] = 'error';
    $response_array['message'] = 'You must select a department';

//check the message field
}elseif(empty($message)) {

    //set the response
    $response_array['status'] = 'error';
    $response_array['message'] = 'You must enter a message';

//check the dept field
}
else {

    //set the response
    $response_array['status'] = 'success';
    $response_array['message'] = 'Your enquiry has been sent succesfully';
    $flag=1;
 }

//send the response back
echo json_encode($response_array);

if($flag == 1) {

    mysql_select_db($database_sample, $sample);
    $query = 'INSERT INTO feedback (company, department, name,  email, position, feedback, date, new) VALUES (\''.$company.'\', \''.$department.'\', \''.$name.'\', \''.$email.'\', \''.$position.'\', \''.$message.'\', NOW() , \''.$new.'\')';
    mysql_query($query) or die("Could not connect");
    }
    

?>

I am trying to insert product descriptions into a mysql database, however they are failing because of apostrophes in the text. How do I fix this? Everything is working fine except $productdescription

$ProductDescription contains,

"Record In Stereo Sound, Then Play Back Your Videos Instantly On You Computer. The Sx130 Is Is All About Creativity, With Fun New Scene Modes Like Miniature Effect For Movies, And Fisheye Effect For Artistically Distorted Stills. While You're Having Fun Creating, The Camera's Smart Flash Exposure And Advanced Smart Auto Systems Are Ensuring That Every Image Is The Best It Can Be. Add The Digic 4 Image Processor, 12.1 Megapixels And Canon's Optical Image Stabilizer, And You've Got The Ideal Camera For Making The Good Times Last."

SQL Error on insert,

"You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 're Having Fun Creating, The Camera's Smart Flash Exposure And Advanced Smart Aut' at line 1 "

Code: [Select]
mysql_select_db("testdb") or die(mysql_error());

mysql_query ("INSERT INTO product (merchantname, producttitle, productdescription, gtin, availability, price) VALUES ('$merchantname[1]','$producttitle[1]','$productdescription[1]','$gtin[1]',''$availability[1]','$price[1]')");

i want to  insert only whats checked to db:)

but everything get posted

for now i came upwith a for loop but thats for inserting all data not only  whats checked any one know a  good way to do  this?

Hello All,

First, I'm new to php and I have spent a lot of time searching forums and google for the answer to my question with no luck, so if it's a topic already covered, PLEASE reply with a link or point me in the right direction.

OK, I have a form that let's users upload multiple images. The upload portion works fine. What I can't figure out how to do is put the image name of each file that gets uploaded all into the same table record. I also need each file name to be put into it's own column in the table. For example, file1, file2 and file3 will get dumped into the same record and into it's respective column (uploaded_image1, uploaded_image2, uploaded_image3). I've included the working upload code and the code that inserts some of the information into the database (just not the image names).

Code: [Select]
echo "<textarea name=challenge rows=15 cols=60> </textarea><br/><br/>"
echo "<textarea name=insight rows=15 cols=60> </textarea><br/><br/>"
for($i=1; $i<=$max_no_img; $i++){
echo "Image $i
<input type=file name='image[]' ><br/>";
}
echo "<input type=submit value=Submit>";

Code: [Select]
// Connect to DB
$host = 'xxx';
$user = 'xxx';
$pass = 'xxx';
$db = 'xxx';
mysql_connect($host,$user,$pass) or die(mysql_error());
mysql_select_db($db) or die(mysql_error());

// Post Variables
$challenge = $_POST['challenge'];
$insight = $_POST['insight'];
$filename=serialize($_POST['filename']);

// Upload Images
while(list($key,$value) = each($_FILES[image][name]))
{
    if(!empty($value))
    {
        $filename = $value;
        $add = "upimage/$filename";
        copy($_FILES[image][tmp_name][$key], $add);
        chmod("$add",0777);
    }
}

// Insert into database
$query = "INSERT INTO table (
                        challenge,
                        insight,
                        uploaded_image1,
                        uploaded_image2,
                        uploaded_image3,)".
                       
                "values (
                        '$challenge',
                        '$insight',
                        '$filename'
                        '$filename2'
                        '$filename3')";
                       
mysql_query($query) or die('Error, query failed : ' . mysql_error());

I know just enough about php that the reason the above wont work, is because $filename2 and $filename3 haven't been given a value. Unfortunately, I have no idea how to separate that information in the array. If anyone can point me in the right direction I'd be forever grateful!

Thanks,
Gary

Hey all
I really need help with a project I am doing.
I need to create a website with a registration form that checks if the user exists if not to add the user details to my local database MySQL.
My webpage looks like it should, it connects with the database but when I enter a new user it does nothing when it should save the new user to the database!
I am guessing my problem is within the if...else section.
Please help my code is:

Code: [Select]

<?php

include('connect.php');  //connection details to database in a connect.php page

$name = "";
$surname = "";
$username = "";
$password = ""; 
$confirmp =  "";
$errorMessage = "";
$num_rows = 0;

//if form was submitted
if ($_SERVER['REQUEST_METHOD'] == 'POST'){

//get values from fields
$submit = $_POST['Submit'];
$title = $_POST['title'];
$name = $_POST['name'];
$surname = $_POST['surname'];
$username = $_POST['username'];
$password = $_POST['password']; 
$confirmp =  $_POST['confirmp'];

//getting string lengths
$nameLength = strlen($name);
$surnameLength = strlen($surname);
$usernameLength = strlen($username);
$passwordLength = strlen($password);
$confirmpLength = strlen($confirmp);

//testing if strings are between certain  numbers
 if ($nameLength > 1 && $nameLength <= 20) {
$errorMessage = "";
}
else {
$errorMessage = $errorMessage . "Name must be between 2 and 20 characters" . "<br>";
}
 if ($surnameLength >= 2 && $surnameLength <= 50) {
$errorMessage = "";
}
else {
$errorMessage = $errorMessage . "Surname must be between 2 and 50 characters" . "<br>";
}
  if ($usernameLength = 6) {
$errorMessage = "";
}
else {
$errorMessage = $errorMessage . "Username must be 6 characters long" . "<br>";
}

if ($passwordLength = 6) {
$errorMessage = "";
}                                                                              
else {
$errorMessage = $errorMessage . "Password must be 6 characters long" . "<br>";
}
  if ($confirmpLength = 6) {
$errorMessage = "";
}                                                                              
else {
$errorMessage = $errorMessage . "Password must be 6 characters long" . "<br>";
}

  if ($errorMessage == "")  {
  
$query = "SELECT * FROM user WHERE username = '$username' AND password = '$password'";
$result = mysql_query($query);
$num_rows = mysql_num_rows($result);

//check to see if the $result is true
    if ($num_rows = 1){
      $errorMessage = "Username already exists";
      }
    else {
        if($password == $confirmp){
        
        $query = "INSERT INTO user (title, name, surname, username, password) VALUES ('$title', '$name', '$surname', '$username', '$password')";
        $result = mysql_query($query);
        
        session_start();
   $_SESSION['login'] = "1";
        header ("Location: login.php");
        }
        else {
         $errorMessage = "Passwords do not match!";
        }
        }
        }
        else {
      $errorMessage = "Error Registering";
      }   
      }
else {
      $errorMessage = "Please enter your details";  
      }           
  
?>


<html>
<head>
<title>Mia's Beauty Products</title>
</head>
<body>
<p><img src = "banner1.jpg" width = "975" height = "95" alt = "Mia's Beauty Product" /></p>
<br>
<p align= "center"><a href="register.php">Register</a> | <a href="login.php">Login</a> | <a href="insert.php">Insert</a> | <a href="list.php">List</a></p>
<form method = "post" action = "register.php">
<table>
<tr><td>Title:</td><td><select name = "title">
                  <option>Miss</option>
                  <option>Mrs</option>
                  <option>Mr</option>
                  </select></td></tr>
<tr><td>Name:</td><td><input name = "name" type = "text" value ="<?php print $name;?>"></td></tr>
<tr><td>Surname:</td><td><input name = "surname" type = "text" value ="<?php print $surname;?>"></td></tr>
<tr><td>Username:</td><td><input name = "username" type = "text" value ="<?php print $username;?>"></td></tr>
<tr><td>Password:</td><td><input name = "password" type = "password" value ="<?php print $password;?>"></td></tr>
<tr><td>Confirm Password:</td><td><input name = "confirmp" type = "password" value ="<?php print $confirmp;?>"></td></tr>
<tr><td><input type = "submit" name = "Submit" value = "Submit"></td></tr>
</table>
</form>
<p align= "center"><a href="code.txt">Code</a></p>
<br>
<?php
      print $errorMessage;
?>
<p><img src = "banner2.jpg" width = "975" height = "95" alt = "Mia's Beauty Product" /></p>
</body>
</html>
 

Thank you
 Amanda

hi guys

Ive written this php to take in two variables from the http POST, the idea is that I can multiple devices submit temperature readings to the php script, the script then append the unix time stamp and then the the 3 variables - device id, temp and unix time are then stored in a mysql DB.

I can get the variables to present on a php page for debugging but I cant get the variables to be stored in the mysql DB.

See the code:

Code: [Select]
<?php

$unixtime = time();
// get device variables
$device_id=$_GET['device'];
$device_temp=$_GET['temp'];

/* 
//for testing purposes
echo "unixtime: " . $unixtime . "<br />";
echo "device id: " . $device_id . "<br />";
echo "device_temp: " . $device_temp . "<br />";
 */

// Make a MySQL Connection
mysql_connect("localhost", "username", "password") or die(mysql_error());
mysql_select_db("test") or die(mysql_error());

//mysql query
$query = "INSERT INTO temperature VALUES ('',$device_id,$temp,$unixtime)";

// Insert a row of information into the relevant device table
mysql_query($query);
or die(mysql_error());  

mysql_close();
echo "Data Inserted!";


?>

I cant see where Im going wrong to correct this, but as nothing is displayed on the page i believe I am not forming the query correctly?
- any ideas would be much appreciated.

Thank you
Mathew

hello dear experts   good day

dear friend - i need your help.

import urllib
import urlparse
import re
# import peewee
import json
from peewee import *



#from peewee import MySQLDatabase ('cpan', user='root',passwd='rimbaud')


db = MySQLDatabase('cpan', user='root',passwd='rimbaud')

class User(Model):
    name = TextField()
    cname = TextField()
    email = TextField()
    url = TextField()

    class Meta:
        database = db # this model uses the cpan database


User.create_table() #ensure table is created


url = "http://search.cpan.org/author/?W"
html = urllib.urlopen(url).read()
for lk, capname, name in re.findall('<a href="(/~.*?/)"><b>(.*?)</b></a><br/><small>(.*?)</small>', html):
    alk = urlparse.urljoin(url, lk)

    data = { 'url':alk, 'name':name, 'cname':capname }

    phtml = urllib.urlopen(alk).read()
    memail = re.search('<a href="mailto:(.*?)">', phtml)
    if memail:
        data['email'] = memail.group(1)


data = json.load('email') #your json data file here

for entry in data: #assuming your data is an array of JSON objects
    user = User.create(name=entry["name"], cname=entry["cname"],
        email=entry["email"], url=entry["url"])
    user.save()

guess that there a data-file must exist: one that have been created by the script during the parsing... is this right?

)
martin@linux-70ce:~/perl> python cpan_100.py
Traceback (most recent call last):
  File "cpan_100.py", line 47, in <module>
    data = json.load('email') #your json data file here
  File "/usr/lib/python2.7/json/__init__.py", line 286, in load
    return loads(fp.read(),
AttributeError: 'str' object has no attribute 'read'
martin@linux-70ce:~/perl>

this script does not work



 

)
martin@linux-70ce:~/perl> python cpan_100.py
Traceback (most recent call last):
  File "cpan_100.py", line 47, in <module>
    data = json.load('email') #your json data file here
  File "/usr/lib/python2.7/json/__init__.py", line 286, in load
    return loads(fp.read(),
AttributeError: 'str' object has no attribute 'read'
martin@linux-70ce:~/perl>

well i suppose that in the first part of the script - the parser - part we parse data and - therefore we should create a file.
I guess that this file is taken up within the second part of the script...

load data .

well why the script does not work!?
 

Hello,

I'm having trouble inserting data into a MySQL table.

The user has a form which is HEIGHT and WIDTH and NUMBER_OF_OBSERVATIONS. All these values are stored in the same table. NUMBER_OF_OBSERVATIONS does what it needs to and inserts its calculated results into the database, as does DATE_TIME.

I've been trying for a couple of days to get this done, and I am having no luck in getting it sorted.

Any help is greatly appreciated!

Generator.php
Code: [Select]
$WIDTH = $_POST['WIDTH'];
$HEIGHT = $_POST['HEIGHT'];
$NUMBER_OF_OBSERVATIONS = $_POST['NUMBER_OF_OBSERVATIONS'];

$db1 = new Number_Information();
$db1->openDB();
$OBSERVATION_ID = $db1->insert_Observation();
echo "<br /><br />Success. ID: <strong>$OBSERVATION_ID<strong>";
$db1->closeDB();
Number_Information.php
Code: [Select]
function insert_Observation() {

        //$Date_Now = datetime();
        $sql = "INSERT INTO Observations (DATE_TIME, HEIGHT, WIDTH)
VALUES (NOW(), '{$esc_HEIGHT}','{$esc_WIDTH}' )";
        $result = mysql_query($sql, $this->conn);
        if (!$result) {
            die("SQL Insertion error: " . mysql_error());
        } else {
            return mysql_insert_id($this->conn);
        }
The table name is Observations

Again - Any help is greatly appreciated!