I'm a rookie in MYSQL PDO, and I'm trying to convert from SQLi to PDO.
I have solved most of the line but one of the line I'm still getting errors after I changed to PDO.
please see code....
<?php session_start(); include_once("../cs/db.php");
//echo 'lll'.$_SESSION['user']; if(isset($_SESSION['user_id'])){
if(isset($_SESSION['cart_id'])){
$cart_id_session =
$_SESSION['cart_id'];
}else{
$q_fetch_cart_id = $conn->query("select cart_id from cart where user_id =
'".$_SESSION['user_id']."'");
$cart_id_set = $q_fetch_cart_id ->fetch(PDO::FETCH_ASSOC);
$cart_id_session = $cart_id_set[0];
$_SESSION['cart_id'] = $cart_id_session;}}
?>
As you can see on last three lines, I changed to (PDO::FETCH_ASSOC); and that line have solved but then next line where it said
$cart_id_session = $cart_id_set[0];
I'm getting errors said
Undefined offset: 0
Is this has to do with PDO! if so what is correct code is!
Many thanks for your help.
AM
Related
I am kind of new one for mysql and php. a week ago this code worked perfectly and when now I am trying it shows this error message
Error : You have an error in your SQL syntax; check the manual that
corresponds to your MariaDB server version for the right syntax to use
near 's product portfolio has diversified to encompass a highly
successful multi-brand' at line 1
I search how to solve that after spending a whole day, but couldn't figure it out.
I have tried similar questions here in stackoverflow, Yet I am stucked here.
A help would be really admired
Given below is my code
<?php
if(isset($_POST['upload']))
{ $company_name =$_POST['company_name'];
$service =$_POST['service'];
$email =$_POST['email'];
$password =$_POST['password'];
$details =$_POST['details'];
$fileName = $_FILES['Filename']['name'];
$fileName1 = $_FILES['Filename1']['name'];
$fileName2 = $_FILES['Filename2']['name'];
$fileName3 = $_FILES['Filename3']['name'];
$fileName4 = $_FILES['Filename4']['name'];
$target = "company_images/";
$fileTarget = $target.$fileName;
$fileTarget1 = $target.$fileName1;
$fileTarget2 = $target.$fileName2;
$fileTarget3 = $target.$fileName3;
$fileTarget4 = $target.$fileName4;
$tempFileName = $_FILES["Filename"]["tmp_name"];
$tempFileName1 = $_FILES["Filename1"]["tmp_name"];
$tempFileName2 = $_FILES["Filename2"]["tmp_name"];
$tempFileName3 = $_FILES["Filename3"]["tmp_name"];
$tempFileName4 = $_FILES["Filename4"]["tmp_name"];
$result = move_uploaded_file($tempFileName,$fileTarget);
$result1 = move_uploaded_file($tempFileName1,$fileTarget1);
$result2 = move_uploaded_file($tempFileName2,$fileTarget2);
$result3 = move_uploaded_file($tempFileName3,$fileTarget3);
$result4 = move_uploaded_file($tempFileName4,$fileTarget4);
$file = rand(1000,100000)."-".$_FILES['file']['name'];
$file_loc = $_FILES['file']['tmp_name'];
$file_size = $_FILES['file']['size'];
$file_type = $_FILES['file']['type'];
$folder="pdf_uploads/";
// new file size in KB
$new_size = $file_size/1024;
// new file size in KB
// make file name in lower case
$new_file_name = strtolower($file);
// make file name in lower case
$final_file=str_replace(' ','-',$new_file_name);//anthima
if(move_uploaded_file($file_loc,$folder.$final_file))
{
$query = "INSERT INTO company_details( company_name,service, email, password, details,image_path,file_name,image_path1,file_name1,image_path2,file_name2,image_path3,file_name3,file,type,size,image_path4,file_name4) VALUES ('$company_name','$service','$email','$password','$details','$fileTarget','$fileName','$fileTarget1','$fileName1','$fileTarget2','$fileName2','$fileTarget3','$fileName3','$final_file','$file_type','$new_size','$fileTarget4','$fileName4')";
$con->query($query) or die("Error : ".mysqli_error($con));
mysqli_close($con);
}
}
?>
<?php
Given below is the test data error
VALUES ('singer','Hardware','singer#gmail.com','singer','Singer has been in Sr' at line 1
Because you never sanitize anything and put the data straight into your query,
$company_name =$_POST['company_name'];
$service =$_POST['service'];
$email =$_POST['email'];
$password =$_POST['password'];
$details =$_POST['details'];
...
$query = "INSERT INTO
company_details( company_name,service, email, password, details,image_path,file_name,image_path1,file_name1,image_path2,file_name2,image_path3,file_name3,file,type,size,image_path4,file_name4)
VALUES (
'$company_name','$service','$email','$password','$details','$fileTarget','$fileName','$fileTarget1','$fileName1','$fileTarget2','$fileName2','$fileTarget3','$fileName3','$final_file','$file_type','$new_size','$fileTarget4','$fileName4'
)";
your problem is most likely in the data
's product portfolio has diversified to encompass a highly successful multi-brand
Maybe you have unscaped apostrophes in your data, so you're kinda SQL-injecting yourself. The query ends before the string shown in the error.
The solution is to escape special chars before inserting like in this question: How do I escape only single quotes?
In your case, start with the details
$details = addcslashes($_POST['details'], "'");
or
$details = addslashes($_POST['details']);
But keep adding test scenarios for your code. E.g. what happens if company name gets something like Mc'Donaldson? What is the set of chars you want to accept for each field? Then you will know how to validate those fields and create your functions (or reuse something)
This is my code:
<?php
ob_start();
session_start();
include("index.php");
if (isset($_POST['user'],$_POST['pass'])):
$con=mysqli_connect("connect info");
if ( !$con ){
die('Could not connect to Database: '.mysqli_error());}
$pass=($_POST['pass']);
$query0 = "SELECT * FROM user WHERE username = '" . $_POST['user'] ."' AND password = '" . $pass . "'";
$resource0 = mysqli_query($con,$query0);
if(!$resource0):
die("Error conducting query. ".mysqli_error($con));
endif;
if(mysqli_num_rows($resource0) == 0){
echo "Username not find";
header("Location: /login.php");}
$result0 = mysqli_fetch_row($resource0);
$_SESSION['ID'] = $result0[0];
$_SESSION['userType'] = $result0[3];
if(!isset($_SESSION['ID'])):
//header("Location: ...");
else:
header("Location: ....");
endif;
else:
if(isset($_POST['from'])):
$_SESSION['from'] = $_POST['from'];
endif;
?>
<?php endif; ?>
This code takes a user's username and password information and stores and searching the database for the userID. Once is finds the information it stores in the session variable 'ID'.
PROBLEM: When the session 'ID' variable is passed to the next page it is not set. Surprisingly, this code without the ob_start function was working yesterday morning but wasn't working by the afternoon. I know it's not setting because two things: When I echo on the next page, nothing appears and because when I try to run a query with the session 'ID' variable I get a mysql error saying the query could not be conducted.
This is the code on the next page that is not working. At first I thought, it might be something wrong with my query because I was getting this error:
"Error conducting query. 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 '' at line 1"
But when I tried printing the session 'ID' variable nothing is printed.
<?php session_start;
// Create connection
$con=mysqli_connect(connect info....);
// Check connection
if (mysqli_connect_errno($con))
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$query0 = "SELECT * FROM studentProfile WHERE sID = ".$_SESSION['ID'];
$resource0 = mysqli_query($con,$query0);
if(!$resource0):
echo $_SESSION['ID'];
die("Error conducting query. ".mysqli_error($con));
endif;
.......(rest of code)
Solutions Already Tried: I have tried them all...session_write_close(), session_regenerate_id(true), session_commit(), ob_end_flush(). I have tried initially setting the session variable on the home page but once the functions are performed in the session 'ID' variable is no longer set at all.
Please help! I have read all the forums I could find on this problem but nothing seems to work.
what you have written on the next page that your code is not working. Please provide me with detail.
try this on the page which you have directed by header location. May be your session can work.
<?php
include('config.php');
if(!isset($_SESSION['ID']))
{
header("Location: ../index.html");
}
else{
$user_id=$_SESSION['ID'][0];
$user_name=$_SESSION['ID'][1];
?>
and end the else part at the last of the page
Perhaps I am just being a complete idiot but I am trying to insert a record into a MySQL table but it doesn't seem to be working. When I test it (i.e. get the script to echo the values so I can check that they are being posted by the form), they are being sent but the query isn't posting to the database. Like I said, perhaps I am being a complete idiot but I felt that perhaps a fresh set of eyes might speed up my troubleshooting because I have been fighting with this issue for the past 2 hours!
Here is the code:
// Connects to your Database
mysql_connect("localhost", "dbuser", "dbpword") or die(mysql_error());
mysql_select_db("dbname") or die(mysql_error());
// Get Variables
$sectorid = $_POST['sectorid'];
$parentid = $_POST['parentid'];
$sectorname = $_POST['sectorname'];
$status = $_POST['status'];
$creon = $_POST['creon'];
$creby = $_POST['creby'];
$modon = $_POST['modon'];
$modby = $_POST['modby'];
//Insert Record
mysql_query("INSERT INTO cand_emp_sector (sectorid, parentid, sectorname, status, creon, creby, modon, modby)
VALUES ('$sectorid', '$parentid', '$sectorname', '$status', '$creon', '$creby', '$modon', '$modby)");
//On completion, redirect to next page
header("Location: canddb.new.7i.php");
Any assistance would be greatly appreciated.
Thanks
you are missing a quote at the end
, '$modby')");
^---------here
Check the result for errors:
$result = mysql_query("INSERT INTO cand_emp_sector (sectorid, parentid, sectorname, status, creon, creby, modon, modby)
VALUES ('$sectorid', '$parentid', '$sectorname', '$status', '$creon', '$creby', '$modon', '$modby)");
if($result === false) die('query failed..');
I know I am simply missing the simplest thing here but cant seem to figure it out.
so this works with this code but changes all rows of the database as opposed to just the one with the page id...
<? $pageid= $_GET["id"];
$sql = "SELECT id, first_name, last_name, email, bio, job, job2, job3 FROM `".weapons."` WHERE id = $pageid";
if(isset($_POST['Update']))
{
$first_name = $_POST['first_name'];
$last_name = $_POST['last_name'];
$job = $_POST['job'];
$job2 = $_POST['job2'];
$job3 = $_POST['job3'];
$bio = $_POST['bio'];
$email = $_POST['email'];
$sql = "UPDATE weapons SET first_name='$first_name', email='$email' , job='$job', job2='$job2', job3='$job3', bio='$bio', last_name='$last_name'";
if (#mysql_query($sql)) {
echo('<p>Update Complete</p>');
} else {
echo('<p>Error updating: ' . mysql_error() . '</p>');
}
}else{ ...
however when adding the WHERE clause, like as follows
$sql = "UPDATE weapons SET first_name='$first_name', email='$email' , job='$job', job2='$job2', job3='$job3', bio='$bio', last_name='$last_name' WHERE id = $pageid";
I get an error
Error updating: 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 '' at line 1
Any help would be great, thanks
EDIT
I actually missed a super easy thing, which is what I initially assumed, I had at first in my form had <form method='post' enctype='multipart/form-data' action='submit.php'> however that obviously messed up the get id because there isnt an idea, so even if that page was submit.php?id=4 when you hit submit it wouldnt run because the id would be gone.
Switching the code to <form method='post' enctype='multipart/form-data' action='#'> did just the trick.
Thanks for the help guys and I am looking into the sql injection now and working on how to better secure my site.
Please escape your strings before you create your SQL statement. Various characters in your input values will both break your query and open a HUGE security hole. That may very well be your problem. Look at this post for more info How can I prevent SQL injection in PHP?
In short, you assignments would look like this:
$first_name = mysql_real_escape_string($_POST['first_name']);
echo $sql; before you run it and post what that outputs.
Been staring at this all day and can't seem to figure out why my update statement fails to update the field 'image_filename':
$fileName = $_FILES['image_filename'];
if($fileName["name"] <> ""){
$imageFile = $fileName['name'];
$destination = "../../../../assets/resources/images/".$fileName['name'];
move_uploaded_file($fileName['name'], $destination);
}
$updateSQL = sprintf("UPDATE content SET image_filename='$imageFile' WHERE id=%s",
GetSQLValueString($_POST['resource_id'], "int"));
mysql_select_db($database_conn_talent, $conn_talent);
$Result1 = mysql_query($updateSQL, $conn_talent) or die(mysql_error());
Can a SQL pro tell me what I"m missing? Much thanks in advance for your feedback!
You appear to be building a query, but never executing it. Also, Drupal'll handle all the sprintfing for you, if you let it.
$query = "UPDATE content SET image_filename='$imageFile' WHERE id=%i";
db_query($query, $_POST['resource_id']);