PHP - Insert Or Update Function Ajax

Hi Guys,

I want my UPDATE function from the code below to only update those fields that are NOT empty - if they are empty on submit then do not update or change any values from that field. I know that sounds odd, but with my file fields, when I submit to change something else at a later date it will overwrite the photo, download1,2 & 3 values and as a result I loose my files from the mysql table.

Cheers, S

<?php 

include('config.php'); 
if (isset($_GET['Ter']) ) { 
$ter = (int) $_GET['Ter']; 
if (isset($_POST['submitted'])) {

//Photo & Document Upload Upload
$timestamp_photo = time();
$timestamp_download1 = time();
$timestamp_download2 = time();
$timestamp_download3 = time();
//This is the directory where the files will be saved
//Photos
$photo_target = "images/"; 
$photo_target = $photo_target .$timestamp_photo. basename( $_FILES['photo']['name']);
//Documents
$download_target = "documents/"; 
$download_target1 = $download_target .$timestamp_download1. basename( $_FILES['download1']['name']);
$download_target2 = $download_target .$timestamp_download2. basename( $_FILES['download2']['name']);
$download_target3 = $download_target .$timestamp_download3. basename( $_FILES['download3']['name']);

//This gets all the other information from the form 
$photo = ($_FILES['photo']['name']);
$download1 = ($_FILES['download1']['name']);
$download2 = ($_FILES['download2']['name']);
$download3 = ($_FILES['download3']['name']);
//Pause Photo/Document Upload


foreach($_POST AS $key => $value) { $_POST[$key] = mysql_real_escape_string($value); } 
$sql= "UPDATE `ter` SET `Ter` =  '{$_POST['Ter']}' , `BranchName` = '{$_POST['BranchName']}' ,  `BranchAddress` =  '{$_POST['BranchAddress']}' ,  `BranchTel` =  '{$_POST['BranchTel']}' , `BranchEmail` =  '{$_POST['BranchEmail']}' , `BranchLink` =  '{$_POST['BranchLink']}' , `Theme` =  '{$_POST['Theme']}' , `LocalInfo` =  '{$_POST['LocalInfo']}' , `BranchInfo` =  '{$_POST['BranchInfo']}' , `photo` =  '$timestamp_photo{$_FILES['photo']['name']}' , `download1` =  '$timestamp_download1{$_FILES['download1']['name']}' , `download1name` =  '{$_POST['download1name']}' , `download2` =  '$timestamp_download2{$_FILES['download2']['name']}' , `download2name` =  '{$_POST['download2name']}' , `download3` =  '$timestamp_download3{$_FILES['download3']['name']}' , `download3name` =  '{$_POST['download3name']}' WHERE `Ter` = '$ter' "; 

mysql_query($sql) or die(mysql_error()); 


//Unpause Photo/Document Upload
//Writes the photo to the server 
if(move_uploaded_file($_FILES['photo']['tmp_name'], $photo_target))
{

echo "<br />The file ".$timestamp_photo. basename( $_FILES['photo']['name']). " has been uploaded. <br />";
}

else {
echo "";
} //End of Photo/Document Upload

//Writes the photo to the server 
if(move_uploaded_file($_FILES['download1']['tmp_name'], $download_target1))
{

echo "<br />The file ".$timestamp_download1. basename( $_FILES['download1']['name']). " has been uploaded. <br />";
}

else {
echo "";
} //End of Photo/Document Upload

//Writes the photo to the server 
if(move_uploaded_file($_FILES['download2']['tmp_name'], $download_target2))
{

echo "<br />The file ".$timestamp_download2. basename( $_FILES['download2']['name']). " has been uploaded. <br />";
}

else {
echo "";
} //End of Photo/Document Upload

//Writes the photo to the server 
if(move_uploaded_file($_FILES['download3']['tmp_name'], $download_target3))
{

echo "<br />The file ".$timestamp_download3. basename( $_FILES['download3']['name']). " has been uploaded. <br />";
}

else {
echo "";
} //End of Photo/Document Upload


echo (mysql_affected_rows()) ? "<br />Edited Branch.<br />" : "<br />Nothing changed. <br />";  
} 
$row = mysql_fetch_array ( mysql_query("SELECT * FROM `ter` WHERE `Ter` = '$ter' "));
?>

I am attempting something similar to the following:

Code: [Select]
mysql_query("UPDATE table SET name='$name' WHERE id=$id");
if (mysql_affected_rows()==0) {
mysql_query("INSERT INTO table (id, name) VALUES ('$name',$id);
}

If the $id row does not exists in the table, mysql_affected_rows() returns 0 and a new $id row gets inserted
but if the $id row already exists and UPDATE changes nothing, mysql_affected_rows() still returns 0
and gives an 'duplicate id' error as expected.

I know I could use a SELECT to test for the existance of the $id row.
In Perl there is an '0E0', 'zero but true', condition to handle this. Is there an equivilant in PHP?
 

I need to insert data into a table if it doesn't exist or update if it does.

Can someone give me an example of this code.

im using mySQL 5.1.42
im using phpMyAdmin 3.1.o

quick question  i have located in my code

$r = mysql_query("insert into points values ('${v[0]}', ${v[1]}, '$At') on duplicate key update Points = Points + ${v[1]}");

-it is updateing the player points but i need it to do a new entry  how would i go about that i tried just taking out the on duplicate and it gives me a tstring insert error on next line. if i leave it in i get no error

I am trying to create a script that takes information from a form and puts in a database.
In the action page, I decided to post a URL that shows the user there story that they posted. This is where I ran into the problem. . I realized that for the optional fields I could not just use a seperate insert statement, because this creates a new row. So I desided to use update statments, but this STILL does not work, they values or simply not getting inserted.  Here is the code:
Code: [Select]
<?php

if(isset($_POST['hidden']))
{

die('SPAM BOT!');
}
if (
!isset($_POST['title']) 
&& !isset($_POST['summary'])
&& !isset($_POST['story'])
&& !isset($_POST['rating'])
&& !isset($_POST['cat'])
)
{
die("<div id='impor'>You forgot to enter one(or more) of the following fields <br /> 
1. Title <br />
2. Summary <br />
3. Story<br />
 </div>
 ");

}


mysqlConnect();


//take data from form an\ put them in variable
$title_form = bb(mysql_real_escape_string($_POST['title'])); //required 
$summ_form = bb(mysql_real_escape_string($_POST['summary']));// required
$story_form = bb(mysql_real_escape_string($_POST['story'])); 
$cat_form = $_POST['cat'];
$rating_form = $_POST['rating'];
$username = $_SESSION['user'];
// Make the other var into a list of links




mysql_query("
INSERT INTO story_info (title, sum, story, user, cat, rating)
VALUES('$title_form','$summ_form', '$story_form,', '$username', '$cat_form','$rating_form')
");
if(isset($_POST['notes']))
{
$notes_form = mysql_real_escape_string($_POST['notes']);
$notes_final = bb($notes_form);
mysql_query("
UPDATE story_info
SET notes = '$notes_final'
WHERE story = '$story_form' AND user = '$username' AND sum = '$summ_form' AND title = '$title_form'
");
}
//put other in array. Use while loop to put link code. Then but it back into one non array variable
if(isset($_POST['u_id']))
{
$uid = mysql_real_escape_string($_POST['u_id']);
$uid_db = str_replace(' ','_', $uid);
$blerg = "
UPDATE story_info
    SET series_id = '$uid_db'
WHERE story = '$story_form' AND user = '$username' AND sum = '$summ_form' AND title = '$title_form'
";
mysql_query($blerg);

}
echo "<h1> Your Story Has Been Posted! Thanks for posting $username .   </h1>";
echo "Please review the post below <br />";
echo "<h2> $title_form </h2>";
echo "<strong> <h2> Summary: </h2> </strong> $summ_form";
echo "<h4> Story: </h4>";
echo "$story_form";
if(isset($notes))
{
echo "<h4> Author's Notes: </h4> ";
echo "$notes_final";


}
if (isset($uid_db))
{
echo '<h3> Unique Series ID </h3>';
echo '<p> Make sure to write down this! <br />' .$uid_db .'</p> ';
}

$db = mysql_query("
SELECT story_id
FROM story_info
WHERE story='$story_form' AND user='$username'
")or die(mysql_error());
$rows = mysql_fetch_assoc($db);

$id = $rows['story_id'];





echo "Catagory: $cat_form <br />
Rating: $rating_form  <br />
";
echo "<a href='?p=page&id=$id'> Click here to view your story! </a>'";
?>


Please help!

how can i make this code to insert a single, multiple and update rows in the database. The code only insert new rows in the database.

Code: [Select]

if (($handle = fopen('inventorylist.csv', "r")) !== FALSE) {
    while (($data = fgetcsv($handle, 100000, ",")) !== FALSE) {
        $num = count($data);
     

                  $sql="INSERT into inventory(itemNumber,itemDesc,quantityHand,category,Whse) values('$data[0]','$data[1]','$data[2]','$data[3]','$data[4]')";

                    mysql_query($sql) or die(mysql_error());

       
    }
    fclose($handle);

        }

Hi i need to count the number items in an album now on the album i have create a field called count which counts the number items in that album

but now i dnt knw how to update those as soon insert a item in the table item

so i thought woul be possible to create a script that when insert data into item tables also update album table row count by adding one like update count with A now A equal row count ++1  any help on start of it if possible???/

I have a form right now that Updates a table on my database.

I was wondering if I could possibly still have that working but also Insert the same information into another table on the database when the submit button is pressed?

Hello:

Can someone give me some insight on how to be able to not just add (INSERT) a photo and photo description with the following code, but also how to add the ability to UPDATE (if the user wants to replace a photo or adjust the photo description):

Code: [Select]
<?php
include("config.inc.php");

// initialization
$result_final = "";
$counter = 0;

// List of our known photo types
$known_photo_types = array( 
'image/pjpeg' => 'jpg',
'image/jpeg' => 'jpg',
'image/gif' => 'gif',
'image/bmp' => 'bmp',
'image/x-png' => 'png'
);

// GD Function List
$gd_function_suffix = array( 
'image/pjpeg' => 'JPEG',
'image/jpeg' => 'JPEG',
'image/gif' => 'GIF',
'image/bmp' => 'WBMP',
'image/x-png' => 'PNG'
);

// Fetch the photo array sent by preupload.php
$photos_uploaded = $_FILES['photo_filename'];

// Fetch the photo caption array
$photo_caption = $_POST['photo_caption'];

while( $counter <= count($photos_uploaded) )
{
if($photos_uploaded['size'][$counter] > 0)
{
if(!array_key_exists($photos_uploaded['type'][$counter], $known_photo_types))
{
$result_final .= "File ".($counter+1)." is not a photo<br />";
}
else
{
mysql_query( "INSERT INTO gallery_photos(`photo_filename`, `photo_caption`, `photo_category`) VALUES('0', '".addslashes($photo_caption[$counter])."', '".addslashes($_POST['category'])."')" );
$new_id = mysql_insert_id();
$filetype = $photos_uploaded['type'][$counter];
$extention = $known_photo_types[$filetype];
$filename = $new_id.".".$extention;

mysql_query( "UPDATE gallery_photos SET photo_filename='".addslashes($filename)."' WHERE photo_id='".addslashes($new_id)."'" );

// Store the orignal file
copy($photos_uploaded['tmp_name'][$counter], $images_dir."/".$filename);

// Let's get the Thumbnail size
$size = GetImageSize( $images_dir."/".$filename );
if($size[0] > $size[1])
{
$thumbnail_width = 100;
$thumbnail_height = (int)(100 * $size[1] / $size[0]);
}
else
{
$thumbnail_width = (int)(100 * $size[0] / $size[1]);
$thumbnail_height = 100;
}

// Build Thumbnail with GD 1.x.x, you can use the other described methods too
$function_suffix = $gd_function_suffix[$filetype];
$function_to_read = "ImageCreateFrom".$function_suffix;
$function_to_write = "Image".$function_suffix;

// Read the source file
$source_handle = $function_to_read ( $images_dir."/".$filename ); 

if($source_handle)
{
// Let's create an blank image for the thumbnail
      $destination_handle = ImageCreate ( $thumbnail_width, $thumbnail_height );

// Now we resize it
       ImageCopyResized( $destination_handle, $source_handle, 0, 0, 0, 0, $thumbnail_width, $thumbnail_height, $size[0], $size[1] );
}

// Let's save the thumbnail
$function_to_write( $destination_handle, $images_dir."/tb_".$filename );
ImageDestroy($destination_handle );
//

$result_final .= "<img src='".$images_dir. "/tb_".$filename."' /> File ".($counter+1)." Added<br />";
}
}
$counter++;
}

// Print Result
echo <<<__HTML_END

<html>
<head>
<title>Photos uploaded</title>
</head>
<body>
$result_final
</body>
</html>

__HTML_END;
?>

The tutorial says to edit a photo, simply add:
Code: [Select]
function edit_photo( $photo_id, $new_caption, $new_category )        <br>
{        <br>
  mysql_query( "UPDATE gallery_photo SET photo_caption='".addslashes( $new_caption )."', photo_category='".addslashes( $new_category )."' WHERE photo_id='".addslashes( $photo_id )."'"  );        <br>
}


But I can not figure out how to call that function.

Anyone help me out?

Thanks.

I have created a registration page to access my website. After the user registrate himself should appear an alert saying that the registration was OK and a redirect to main.php page... however for some reason if I create an insert statement the alert and the redirect don't appear... If I remove the insert the alert and the redirect works... why?

This is part of the code of my 3 files:

registration.php (ajax call)

$('#submit').click(function() {
    var username2 = $('#uname2').val();
    var password2 = $('#psw2').val();
      $.ajax({
        url: 'ajax/response.php',
        type: 'GET',
        data: {username2 : username2, password2: password2},
            success: function(data) {
                if(data === 'correct') {
                    alert("Username and Password have been created!"); //don' work with the insert
                    location.replace("main.php"); //don' work with the insert
                    } else {
                         alert("Username or password are not correct... please register yourself!");
                   }
              }
       });
});

response.php (answer to ajax call)

if(isset($_GET['username2']) && isset($_GET['password2'])) {
    $username2 = $_GET['username2'];
    $password2 = $_GET['password2'];
      if (checkUser($pdo, $username2) === true) {
        echo 'duplicate';
    } else {
        insertUserPwd($pdo, $username2, $password2); //including this line the redirect and the alert doesn't work... the insert is OK
        echo 'correct';
    }
}

data_access.php (the function works but doesn't permit alert and redirect to appear)

function insertUserPwd(PDO $pdo, $usr, $pwd)
{
    $data = [
    'id' => '',
    'user' => $usr,
    'password' => $pwd
    ];
    $sql = "INSERT INTO users (id, user, password) VALUES (:id, :user, :password)";
    $stmt= $pdo->prepare($sql);
    $stmt->execute($data);
}

Can someone help me to fix the code? 

 

I can't find the error, someone help me please.                   $conn = mysql_connect("localhost","root","Pass");
                $err_db = mysql_select_db('bd_amics');
                $sql = ("INSERT INTO `'".$_SESSION["use"][14]."'` (ID,Amic,PubID) VALUES ('".$_SESSION["person"][14]."', "1", '".$_SESSION["person"][15]."')");
                mysql_query("SET NAMES utf8");
                mysql_query($sql, $conn);
                mysql_close();                   $conn3 = mysql_connect("localhost","root","Pass");
                $err_db3 = mysql_select_db('bd_amics');
                $sql3 = ("UPDATE `'".$_SESSION["person"][14]."'` SET Amic="2" WHERE ID='".$_SESSION["use"][14]);
                mysql_query("SET NAMES utf8");
                mysql_query($sql3, $conn3);
                mysql_close();

Hi I'm currently experiencing a problem with my query. I've used the `INSERT IGNORE` option in my query and it works pretty well not to add duplicates, but the problem is that if my unique field match it doesn't update any other info in the row, is there maybe another option to update a row if an existing field exists but insert new row if it doesn't exists?   my current query is as follow  

$query = mysqli_query($con,"INSERT IGNORE INTO mxit (ip,time,user_agent,contact,userid,id,login,nick,location,profile) 
VALUES ('$ip','$post_time','$mxitua','$mxitcont','$mxituid','$mxitid','$mxitlogin','$mxitnick','$mxitloc','$mxitprof')") or die(mysqli_error($con)); 

I put a snippet up yesterday, and thought it had been solved. It had a typo error, but there is still something wrong. The whole page follows, because it might be easier to see what I am doing. The form successfully inserts a new record OK, but it will not update a record that already exists as is intended. Instead, it always creates another new record. Where can I be wrong?


<?php
session_start();

// See if logged in member by checking session data
if (isset($_SESSION['recid'])) {
    // Put stored session variables into local php variable
    $userid      = $_SESSION['recid'];
    $username    = $_SESSION['name'];
    $school      = $_SESSION['college'];
    $payer_email = $_SESSION['payer_email'];
    $group1      = $_SESSION['group1'];
    $group2      = $_SESSION['group2'];
    $group3      = $_SESSION['group3'];
    $group4      = $_SESSION['group4'];
    $group5      = $_SESSION['group5'];
    $group6      = $_SESSION['group6'];

    
} else { ?>
<!-- make login box -->
<html>
<tr>
  <td>&nbsp;
    <p>&nbsp;
    <p>&nbsp;
    <p></td>
</tr>
<table style= border align="center" border="1" width="400" height="200">
    <td align="center" valign="center">You need to<a href="login_allocate_quiz.php">log in</a>as an administrator to<br />
      allocate quizzes to student groups.</td>
</table>
</html>
<?php exit(); 
}

    // Set error message as blank upon arrival to page
    $errorMsg = "";
    // Check to see if the form has been submitted 
    if (isset($_POST['equip'])){
    //Connect to the database through include 
    include_once "demo_conn.php";
    //set variables    
    $equip     = $_POST['equip'];
    $egroup1   = $_POST['egroup1'];
    $egroup2   = $_POST['egroup2'];
    $egroup3   = $_POST['egroup3'];
    $egroup4   = $_POST['egroup4'];
    $egroup5   = $_POST['egroup5'];
    $egroup6   = $_POST['egroup6'];
    
    //create variables to manipulate filenames in database
    $lowquip = strtolower($equip);
    $wordquip = str_replace (" ", "", $lowquip);
    $bigpic    = $wordquip.'big';
    $smallpic  = $wordquip.'small';
    $bigpic    = $bigpic.'.jpg';
    $smallpic = $smallpic.'.jpg';
    $errorMsg = '<font size="-1"><u>ERROR:</u>&nbsp;&nbsp;&nbsp;You have not entered the following required information.';
    $nogroup = ((!$egroup1) && (!$egroup2) && (!$egroup3) && (!$egroup4) && (!$egroup5) && (!$egroup6));
    $noequip = (!$equip);
    if ($nogroup)
        {$errorMsg .= "--- Student Group. (No data was forwarded to the database.)</font>";}
    else if($noequip)
        {$errorMsg .= "--- Quiz Name. (No data was forwarded to the database.)</font>";}
    else 
        {$query = mysql_query("SELECT equip FROM topics WHERE equip = '$equip' AND managerId = '$userid'");
              if (mysql_num_rows($query) > 0)
                mysql_query("UPDATE topics SET egroup1 = '$egroup1', egroup2 = '$egroup2', egroup3 = '$egroup3', egroup4 = '$egroup4', egroup5 = '$egroup5', egroup6 = '$egroup6' WHERE equip = '$equip' AND managerId = '$userid'");
            else
                mysql_query("INSERT INTO topics (managerId, equip, title, url_big, url_small, egroup1, egroup2, egroup3, egroup4, egroup5, egroup6) VALUES ('$userid','$wordquip', '$equip', '$bigpic', '$smallpic', '$egroup1', '$egroup2', '$egroup3', '$egroup4', '$egroup5', '$egroup6')") or die (mysql_error("error 1321"));            
        $URL='make_quiz_list_test4.php';
        header ('Location: '.$URL);
        }    
    }
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml" dir="ltr" lang="en-US" xml:lang="en">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<meta http-equiv="X-UA-Compatible" content="IE=EmulateIE7" />
<meta http-equiv="X-UA-Compatible" content="IE=4">
<!-- IE5 mode -->
<meta http-equiv="X-UA-Compatible" content="IE=7.5">
<!-- IE7 mode -->
<meta http-equiv="X-UA-Compatible" content="IE=100">
<!-- IE8 mode -->
<meta http-equiv="X-UA-Compatible" content="IE=a">
<!-- IE5 mode -->
<title>Manager selected quiz list</title>
<link rel="stylesheet" href="style.css" type="text/css" media="screen" />
<!--[if IE 6]><link rel="stylesheet" href="style.ie6.css" type="text/css" media="screen" /><![endif]-->
<!--[if IE 7]><link rel="stylesheet" href="style.ie7.css" type="text/css" media="screen" /><![endif]-->
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Hi All,

I have an insert/update that i am using in several places across my site, it works fine everywhere apart from on one page.

I am clearly missing something but cant for the life of me work it out.

My php:

if ($_SERVER['REQUEST_METHOD']=='POST') {
    $jobId = $_SESSION['current_job_id'];
    $qty = $_POST['drinkItemQty'];
    // prepare insert query
    $stmt = $conn->prepare("INSERT INTO ssm_drink_order (drink_qty, job_id, drink_id) 
                            VALUES (?,?,?) 
                            ON DUPLICATE KEY UPDATE 
                            drink_qty = VALUES(drink_qty)"
                        ); 
    
    
    foreach ($_POST['drinkItemId'] as $k => $diid) {

        if ($qty[$k] > 0) {

            $stmt->bind_param("iii", $qty[$k], $jobId, $diid);

            $stmt->execute();
        }

        if ($qty[$k] < 1) {

            $stmt1 =$conn->prepare("DELETE FROM ssm_drink_order WHERE job_id = ?");
            $stmt1->bind_param('i', $jobId);
            $stmt1->execute();
        }
    }   
}

the page html

<tbody>
    <tr>
        <th style="width:70%;" class="text-center">Drink Item</th>
        <th class="text-center">Quantity</th>
    </tr>

    <tr>
        <td>
            House Gin
            <input name="drinkItemId[]" type="hidden" value="2">
        </td>
        <td class="text-center">
            <input name="drinkItemQty[]" type="number" value="999" class="text-center">
        </td>
    </tr>
    <tr>
        <td>
            House Brandy
            <input name="drinkItemId[]" type="hidden" value="4">
        </td>
        <td class="text-center">
            <input name="drinkItemQty[]" type="number" value="" class="text-center">
        </td>
    </tr>
    <tr>
        <td>
            House Vodka
            <input name="drinkItemId[]" type="hidden" value="1">
        </td>
        <td class="text-center">
            <input name="drinkItemQty[]" type="number" value="" class="text-center">
        </td>
    </tr>
    <tr>
        <td>
            House Whiskey
            <input name="drinkItemId[]" type="hidden" value="3">
        </td>
        <td class="text-center">
            <input name="drinkItemQty[]" type="number" value="" class="text-center">
        </td>
    </tr>
</tbody>

Any help is as ever greatly appreciated

Well, I made this yesterday then realised, I need to check is it exists...
I got this but when I go to accept the application and the member exists in the table it enters it anyway...

Code: [Select]
<?php
$member=$_POST['memberid'];
$status=$_POST['Status'];

$con = mysql_connect("host","user","pass");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("a2186214_hbclan",$con);
$sql="UPDATE application SET Status = '$status' WHERE ID = '$member'";
$sql1="INSERT INTO table_members(name) SELECT application.Name FROM application WHERE application.ID = '$member'";
if ($status == 'ACCEPTED')
{
if(mysql_num_rows(mysql_query("SELECT name FROM table_members WHERE name = '$member'")))
{
if(mysql_query($sql, $con) or die(mysql_error()))
{
echo 'Status Changed.<br /><a href="../applications.php">Return To Members List</a>';
}
else
{
die('Could not submit: ' . mysql_error());
}
}
else
{
if(mysql_query($sql, $con) or die(mysql_error()))
{
if(mysql_query($sql1, $con) or die(mysql_error()))
{
echo 'Status Changed.<br /><a href="../applications.php">Return To Members List</a>';
}
}
else
{
die('Could not submit: ' . mysql_error());
}
}
}
else
{
if(mysql_query($sql, $con) or die(mysql_error()))
{
echo 'Status Changed.<br /><a href="../applications.php">Return To Members List</a>';
}
else
{
die('Could not submit: ' . mysql_error());
}
}
mysql_close($con);

?>