PHP - Cannot Update Mysqli Unless I Upload Image

Hi all,

As you will see this is my first post in I really hope somebody can nail this for me, as I have been working on this for some time and I cannot get it to work.

Basically this is the final part of a CMS I'm building, and all was fine until I needed to build in an option of a multiple image uploader to accompany all other stock details.

I have managed to nail the 'insert' part of the multiple image uploader in this case up to 4 images, and the code is below for this part, any help with improvements though will be greatly appreciated.

Code: [Select]
if(isset($_POST['btnsubmit']))
{

$flag=$_POST['flag'];
if ($flag==0)
{
 
$name=$_POST['txtname'];
$desc1=$_POST['e1m1'];
 
$meta=$_POST['txtmeta'];
$sr=$_POST['srno1'];
$name=$_POST['txtname'];
$ref=$_POST['Ref'];
$desc=$_POST['e1m1'];
$maker=$_POST['Maker'];
$date=$_POST['Date'];
$weight=$_POST['Weight'];
$height=$_POST['Height'];
$depth=$_POST['Depth'];
$width=$_POST['Width'];
$price=$_POST['txtprice'];
$sold=$_POST['txtsold'];
$active=$_POST['active'];
$pcats=$_POST['pcats'];
$subcats=$_POST['subcats'];

$str_str='';
    $p='';
    $j=0;
    for($i=0;$i<=3;$i++){
    $j++;
    $p=$_REQUEST['p$j'];
    $file=$_FILES['pic']['name'][$i];
   
    if(!empty($file) ){
              $str_str.=",pic$j='$file'";
     }
     else if(!empty($p))$str_str.=",pic$j='$p'";
     
     $path1="imgdata/stock/".$file;
     copy($_FILES['pic']['tmp_name'][$i], $path1);     
     
    }

$q24=mysql_query("update stock set stock_Name='$name', stock_MetaTitle='$meta', parent_Category='$pcats', sub_Category='$subcats', stock_Image='imgdata/stock/$pic[0]', stock_Image2='imgdata/stock/$pic[1]', stock_Image3='imgdata/stock/$pic[2]', stock_Image4='imgdata/stock/$pic[3]', stock_Ref='$ref', stock_Description='$desc1', stock_Maker='$maker', stock_Date='$date', stock_Weight='$weight', stock_Height='$height', stock_Depth='$depth',stock_Width='$width', stock_Price='$price', stock_Sold='$sold', stock_Active='$active', stock_DateTime='$dt2' where stock_Id=$sr") or die (mysql_error());
$flag=1;
$conf="Data Updated Successfully - Click <a href='http://www.accendsandbox.co.uk/adminSallam/admin_categories.php'>here</a> to continue";
$update="1";


Code: [Select]
<tr>
<td bgcolor="#A0B050" width="161">
<div style="font-family:Verdana, Arial, Helvetica, sans-serif; position:relative; font-size:12px; color:#293334; font-weight:bold; position:relative; float:left; left:1px;">Stock Image 1 (Main):</div>
</td>
<td bgcolor="#888888">
<input type="file" name="pic1[]" id="pic1[]" size="50" />
</td>
</tr>
<tr>
<td bgcolor="#A0B050" width="161">
<div style="font-family:Verdana, Arial, Helvetica, sans-serif; position:relative; font-size:12px; color:#293334; font-weight:bold; position:relative; float:left; left:1px;">Stock Image 2:</div>
</td>
<td bgcolor="#888888">
<input type="file" name="pic1[]" id="pic1[]" size="50" />
</td>
</tr>
<tr>
<td bgcolor="#A0B050" width="161">
<div style="font-family:Verdana, Arial, Helvetica, sans-serif; position:relative; font-size:12px; color:#293334; font-weight:bold; position:relative; float:left; left:1px;">Stock Image 3:</div>
</td>
<td bgcolor="#888888">
<input type="file" name="pic1[]" id="pic1[]" size="50" />
</td>
</tr>
<tr>
<td bgcolor="#A0B050" width="161">
<div style="font-family:Verdana, Arial, Helvetica, sans-serif; position:relative; font-size:12px; color:#293334; font-weight:bold; position:relative; float:left; left:1px;">Stock Image 4:</div>
</td>
<td bgcolor="#888888">
<input type="file" name="pic1[]" id="pic1[]" size="50" />
</td>
</tr>
<input type="submit" name="btnsubmit" value="Submit">
<input type="submit" name="btndelete" value="Delete" DEFANGED_OnClick="return check();">
<input type="hidden" name="srno1" value="<?= $rows["stock_Id"];?>">
<input type="hidden" name="action" value="Upload">

All seems to be fine above, I can upload 1,2,3 or 4 images and the image goes into the server and the path to the database.

But I then needed the option for my client to be able to click to edit a certain stock and then aswel as edit the other details, if he wanted to change one pic, no pics or all 4 he could and when he clicked submit, if there was a new image it would change if not it would stay as it is.

So here is my attempt and at the moment it doesnt work, so I'm looking for help of any kid and anything can change.

Code: [Select]
$name=$_POST['txtname'];
$ref=$_POST['Ref'];
$desc=$_POST['e1m1'];
$maker=$_POST['Maker'];
$date=$_POST['Date'];
$weight=$_POST['Weight'];
$height=$_POST['Height'];
$depth=$_POST['Depth'];
$width=$_POST['Width'];
$price=$_POST['txtprice'];
$sold=$_POST['txtsold'];
$meta=$_POST['txtmeta'];
$active=$_POST['active'];
$pcats=$_POST['pcats'];
$subcats=$_POST['subcats'];

$pic1='';
for($i=0;$i<4;$i++){
if(isset($_FILES['pic1']['name'][$i]))$pic1[$i]=$_FILES['pic1']['name'][$i];
else $pic1[$i]='';
}
for($i=0;$i<4;$i++){
if(isset($_FILES['pic1']['name'][$i]))$path1= "./imgdata/stock/".$_FILES['pic1']['name'][$i];
//echo $_FILES['pic1']['tmp_name'][$i]." :". $path1;
if(!empty($_FILES['pic1']['name'][$i])&&isset($_FILES['pic1']['name'][$i]))copy($_FILES['pic1']['tmp_name'][$i], $path1);
}
       
$q=mysql_query("insert into stock (stock_Name, stock_MetaTitle, parent_Category, sub_Category, stock_Ref, stock_Description, stock_Maker, stock_Date, stock_Weight, stock_Height, stock_Depth, stock_Width, stock_Price, stock_Sold, stock_Image, stock_Image2, stock_Image3, stock_Image4, stock_Active, stock_DateTime) values('$name','$meta','$pcats','$subcats','$ref','$desc','$maker','$date','$weight','$height','$depth','$width','$price','$sold','imgdata/stock/$pic1[0]','imgdata/stock/$pic1[1]','imgdata/stock/$pic1[2]','imgdata/stock/$pic1[3]','$active','$dt2')") or die (mysql_error());
$conf="Data Inserted Successfully - Click <a href='http://www.accendsandbox.co.uk/adminSallam/admin_stock.php'>here</a> to continue";
$update=1;

Code: [Select]
<tr>
<td bgcolor="#A0B050" width="161">
<div style="font-family:Verdana, Arial, Helvetica, sans-serif; position:relative; font-size:12px; color:#293334; font-weight:bold; position:relative; float:left; left:1px;">Stock Image 1 (Main):</div>
</td>
<td bgcolor="#888888">
<input type="file" name="pic[]" size="50" /> <input type="hidden" name="p1" value="<?php echo $pic1;?>" /> &nbsp;<img src="<?php echo $pic1;?>" height="100px" />
</td>
</tr>
<tr>
<td bgcolor="#A0B050" width="161">
<div style="font-family:Verdana, Arial, Helvetica, sans-serif; position:relative; font-size:12px; color:#293334; font-weight:bold; position:relative; float:left; left:1px;">Stock Image 2:</div>
</td>
<td bgcolor="#888888">
<input type="file" name="pic[]" size="50" /> <input type="hidden" name="p2" value="<?php echo $pic2;?>" /> &nbsp;<img src="<?php echo $pic2;?>" height="100px" />
</td>
</tr>
<tr>
<td bgcolor="#A0B050" width="161">
<div style="font-family:Verdana, Arial, Helvetica, sans-serif; position:relative; font-size:12px; color:#293334; font-weight:bold; position:relative; float:left; left:1px;">Stock Image 3:</div>
</td>
<td bgcolor="#888888">
<input type="file" name="pic[]" size="50" /> <input type="hidden" name="p3" value="<?php echo $pic3;?>" /> &nbsp;<img src="<?php echo $pic3;?>" height="100px" />
</td>
</tr>
<tr>
<td bgcolor="#A0B050" width="161">
<div style="font-family:Verdana, Arial, Helvetica, sans-serif; position:relative; font-size:12px; color:#293334; font-weight:bold; position:relative; float:left; left:1px;">Stock Image 4:</div>
</td>
<td bgcolor="#888888">
<input type="file" name="pic[]" size="50" /> <input type="hidden" name="p4" value="<?php echo $pic4;?>" /> &nbsp;<img src="<?php echo $pic4;?>" height="100px" />
</td>
</tr>

What happens above as its all on the same page, is when a stock item is selected to be edited the form to update the images changes to the image upload options above, instead of the original ones for a new stock item at the top of this post.

I can provide anything you need to help me with this, so please can somebody take a look and see if it can be got working, as its been a long problem for me this.

Cheers

Dear Sir/Madame

I am making a website where user can insert data and wait for the admin to approve/reject the form. Now i am stuck with the update status where an admin can submit with a click pending to approval or reject with comments. I am new to PHP programming. Can somebody help me with the issue. Part 1 is inserting the data and part two is fetching the data but i am unable to solve the status approve/reject and comment at the same time on the view.php? page. Kindly help. Thank you. 

<?php


$host="localhost";
$username="root";
$pass="";
$db="ems1";

$conn=mysqli_connect($host,$username,$pass,$db);
if(!$conn){
	die("Database connection error");
}

// insert query for register page
if(isset($_REQUEST['proposal']))
{

$details=$_POST['details'];
$location=$_POST['location'];
$date=$_POST['date'];
$time=$_POST['time'];

$status="Pending";
$comment=$_POST['comment'];
	
	
	 $query="INSERT INTO `proposal` (`details`,`location`,`date`,`time`,`status`,`comment`) VALUES ('$details','$location','$date','$time','$status','$comment')";
	
	$res=mysqli_query($conn,$query);
	if($res){
		$_SESSION['success']="Not Inserted successfully!";
		header('Location:');
	}else{
		echo "Leave not Applied, please try again!";
	}
	}
?>

<div class="col-xs-6 col-xs-push-3 well">
<form class="form-horizontal" method="post" action="" >
<input type="hidden" name="proposal" value="">
  <fieldset>
    <legend>New Proposals  </legend>
	        <!----left box----------->
		<!----right box----------->
	<div class="col-xs-9">
<div class="form-group">
      <label for="inputEmail" class="col-lg-3"><b>Details:</b></label>
      <div class="col-lg-9">
        <input type="text" name="details" class="form-control">
      </div>
    </div>
   <div class="form-group">
      <label for="inputEmail" class="col-lg-3"><b>Location:</b></label>
      <div class="col-lg-9">
        <input type="text" name="location" class="form-control" >
      </div>
    </div>
	<div class="form-group">
      <label for="inputEmail" class="col-lg-3"><b>Date:</b></label>
      <div class="col-lg-9">
        <input type="date" name="date"  class="form-control">
      </div>
    </div>
	<div class="form-group">
      <label for="inputEmail" class="col-lg-3"><b>Time:</b></label>
      <div class="col-lg-9">
        <input type="time" name="time"  class="form-control" >
      </div>
    </div>
	
      <div class="col-lg-9">
          <input type="hidden" name="status" class="form-control" >
      </div>
    </div>
	<div class="form-group">
      <label for="inputEmail" class="col-lg-3"><b></b></label>
      <div class="col-lg-9">
          <input type="hidden" name="comment"  class="form-control">
      </div>
    </div>
	</div>
	
    
    <div class="form-group">
      <div class="col-lg-12">
        <button type="reset" class="btn btn-default">Cancel</button>
        <button type="submit" class="btn btn-primary">Submit</button>
      </div>
    </div>
  </fieldset>
</form>
</div>
</div>	

<body>
    
<h2 style="text-align:center; color:orangered;">
DASHBOARD
</h2>
<table>   
<h3>
                    <tr style="background-color:#E4EBC5; color:orangered;">

                    <th>ID</th>
                    <th>Details</th>
                    <th>Location</th>
                    <th>Status</th>
                    <th>Comment</th>
                    

                    </tr>
                    </h3>
</table>
    
  
<?Php
////////////////////////////////////////////
require "dbconfig.php"; // MySQL connection string

$count="SELECT  id,details,location,time,status,comment FROM proposal";

    
if($stmt = $connection->query($count)){


while ($nt = $stmt->fetch_assoc()) {
echo "
  

<body>
<table>

<tr>   
   
  <td><a href=view.php?id=$nt[id]>$nt[id]</a></td>
  <td>$nt[details]</td>
  <td>$nt[location]</td>
  <td>$nt[status]</td>
  <td>$nt[comment]</td>
  

     </tr>
     
      </table>
  </body>
  ";
   
}

}else{
echo $connection->error;
}
?>

 

Hi, I am having trouble coding for mysqli update. Please, somebody
tell me the correct way. I'm trying to update the "lastused" (current date)
field in "emailtbl". Somebody please tell the best way to code this.
Below is the message and following, the current code:
  Fatal error: Call to undefined function curdate() in C:\xampp\htdocs\home\lastused.php on line 14  

 
$db = new mysqli('localhost', 'root', 'pass', 'mydb');
if($db->connect_errno > 0)
{die('Unable to connect to database [' . $db->connect_error . ']');}
$sql = <<<SQL
    SELECT *
    FROM `emailtbl`
    WHERE `id` = '$id'
SQL;

if(!$result = $db->query($sql))
{die('There was an error running the query [' . $db->error . ']');}
$lastused = $_POST['lastused'];
$lastused = curdate();
echo "last date accessed is ".$data['lastused'];
$result->free();
$db->escape_string('This is an unescape "string"');
$db->close();
?>
$update 

= mysqli_query($dbconnect, "UPDATE emailtbl SET
   lastused = curdate() WHERE id ='$id'");
    if($update == false)
    { die("UPDATE FAILED: ".mysqli_error($dbconnect)); }
    echo "$lastused is the last date this account was accessed";
 

Hi, i have a user update function in my code so I can easily change user fields int he database
        public function updateUser ($username, $value, $what) {
$q = "UPDATE `users` SET ? = ? WHERE username = ?";
if ($stmt = $this->db_connection->prepare($q)) {
$stmt->bind_param("sis", $what, $value, $username);
$stmt->execute();
}
}

The database is connected successfully, but say i run a updateUser('blaine0002', 'blah', 'password'); nothing would get updated and no errors are thrown.
Am i doing something wrong? Thank you!

I have several registration systems that were designed some time ago running mysql. How long do I have until I have to competely move to mysqli? I am working on recoding them, but it is taking some time. Does anyone know when it will become an issue?