Update with 3 condition

Question

0.00/5 (No votes)

See more:

hey Guys I am trying to create Update Mysql command but it is not working can you tell me what i did wrong?

This is how it looks when place GEt one form

tickettitle=sssss&
issuedetails=ssssssssssss&
privatenotes=sssssssssss&
status=1&
uploadfile=image.jpg&
save=&
category=1&
subcat=1&
option=1&
priority=1&
companyname=2&
compid=112224&
trackid=CS20220202-67&
assign=1&
calername=&
caleremail=&
calernumber=%2B38766520452#

What I have tried:

<?php
 if (isset($_POST["save"])){

   if (($_POST["status"]==1)){
      $tickettitle= $_POST["tickettitle"];
      $issuedetails=$_POST["issuedetails"];
      $privatenotes=$_POST["privatenotes"];
      $priority= $_POST["priority"];     
      $category= $_POST["category"];
      $subcat=$_POST["subcat"];
      $option=$_POST["option"];
      $status= $_POST["status"];
      $companyname=$_POST["companyname"];
      $compid=$_POST["compid"];
      $calername= $_POST["calername"];
      $caleremail=$_POST["caleremail"];
      $calernumber=$_POST["calernumber"];
      $asign=$_POST["assign"];
   
      $ran_id = rand(time(), 100000000);
      $time = time();
      $date=date('Ymd');
      $cs="CS";
      $dubleo="-";
      $traid = $cs.$date.$dubleo.$tid;
      $date2=date('Y-m-d H:i:s');
    
      if(isset($_FILES['uploadfile'])){
   $img_name = $_FILES['uploadfile']['name'];
   $img_type = $_FILES['uploadfile']['type'];
   $tmp_name = $_FILES['uploadfile']['tmp_name'];
   $img_explode = explode('.',$img_name);
   $img_ext = end($img_explode);
   $extensions = ["jpeg", "png", "jpg", "PNG"];
   if(in_array($img_ext, $extensions) === true){
       $types = ["image/jpeg", "image/jpg", "image/png", "image/PNG"];
       if(in_array($img_type, $types) === true){
           $time = time();
           $new_img_name = $time.$img_name;
           if(move_uploaded_file($tmp_name,"../php/images/".$new_img_name)){

  
   $sql="UPDATE tickets SET  calername='$calername', caleremail='$caleremail', category='$category',priority='$priority',tickettitle='$tickettitle',
   issuedetails='$issuedetails',message_html='$issuedetails',lastchange='$date2',firstreply='$date2',privatenotes='$privatenotes',ip='$ip',status='$status',
   openedby='$flname',firstreplyby='$flname',`closedby`='$flname',assignto='$asign',assignedby='$flname',lastreplier='$flname',`archive`='$new_img_name',attachments='$new_img_name',
   subcat=' $subcat',issue='$option',companyname='$companyname',compid='$compid' WHERE	trackid='$traid' ";
   mysqli_query($dbc, $sql);
   header("Location: knwo.php");
   exit();
}
}
   }
}
 }
}

   ?>

Posted 2-Feb-22 10:05am

Member 13084733

Updated 3-Feb-22 20:47pm

v2

Add a Solution

Comments

Dave Kreskowiak 2-Feb-22 16:12pm

"Not working" is the most common problem description and it's equally useless.

What error messages do you get? What isn't working? What do you expect this code to do and what isn't it doing?

Member 13084733 2-Feb-22 16:14pm

I am not getting any message and I am trying to update trackid and when I click save button it is not updating table

Richard Deeming 3-Feb-22 4:33am

Your code is vulnerable to SQL Injection[^]. NEVER use string concatenation/interpolation to build a SQL query. ALWAYS use a parameterized query.
PHP: SQL Injection - Manual[^]
PHP: Prepared statements and stored procedures - Manual[^]

2 solutions

Solution 1

Hmm.

1. What is $tid? $dbc?

2. Inspect the value of $traid !

3. An ID field (such as $traid) should be just that, and not contain meaningful information such as a date.

4. system calls always have an error mechanism. Your code does not care about the return value of PHP: mysqli::query - Manual[^]. Read how it is done.

5. BTW: you should always use parameterized queries for safety and robustness.

Posted 2-Feb-22 11:24am

Luc Pattyn

Comments

Member 13084733 2-Feb-22 17:33pm

$findid = mysqli_query($dbc, "SELECT id FROM tickets ORDER BY id DESC LIMIT 1 ");
if($findedid = mysqli_fetch_array($findid))
{
$dbc= mysqli_connect($dbHost, $dbUsername, $dbPassword, $dbName);
$tid = $findedid['id'];
$date=date('Ymd');
$cs="CS";
$dubleo="-";
$traid = $cs.$date.$dubleo.$tid;
}
?>

Member 13084733 2-Feb-22 17:34pm

$dbc= mysqli_connect($dbHost, $dbUsername, $dbPassword, $dbName);

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

_Asif_ · Accepted Answer · 2022-02-03T20:47:00

Your UPDATE SQL seems to have problems. Please check the following fields.

`closedby`='$flname',
`archive`='$new_img_name'

should be

closedby='$flname',
archive='$new_img_name'

Secondly you need to use parameterization to avoid SQL injection attacks.

Last but not least one way to isolate the problem is to reduce the scope of update till it gets success and then gradually increase update scope. What I mean here is comment out all updates except one and check if it gets success. If it does success then remove commented columns update one by one and you will find the field that was failing the update.

Hopefully this should solve your problem