PHP - I Would Like To Grab The Id Of An Edited Record And Insert Into Another Table. Is This Possible?

Not sure if this is right.  I can't get it to insert my record.  Can someone please tell me if I'm doing it right?   cus_functions.php

function dbRowInsert($table_name, $form_data)
{
global $conn;
   
    $fields = array_keys($form_data);

    $sql = "INSERT INTO ".$table_name."
    (`".implode('`,`', $fields)."`)
    VALUES('".implode("','", $form_data)."')";

       return mysqli_query($sql);
}

process.php

 include("new_db.php");
 include("cus_functions.php");
 
 $do=$_GET['do'];
 
 if($do=='addpro'){
 
if(isset($_POST['submit'])){

$input = $_POST['title'];
$comp = '0';

$form_data = array('title' => $input, 'completed' => $comp);
 dbRowInsert('projects', $form_data);

}}

I have a form that submits a record and saves it to the database, I've got that working already, I'm trying to figure out if there is a cleaner way to delete a record with a button that gets a value from the unique key, in this case 'id'
here is how the delete button looks like:

   // get info from table
$query  = "SELECT id, Title, Message FROM table_name";
$result = mysql_query($query);

//displaying all data
while($row = mysql_fetch_assoc($result))
{
echo"
<div class='status'><h3>{$row['Title']} <br></h3>" . "
         <h5>{$row['Message']} <br><br></h5>"; 
 $id = $row['id']; //trying to get unique key from database
//delete button
 echo "</div>
 <form action='delete.php' method='post' />
 <input type='hidden' name='delete' value='yes' /> 
 <input type='hidden' name='id' value='$id' />
 <input type='submit' value='remove' /></form>";
}

As you can see, i'm trying to give $id a unique value but for some reason i'm not getting it.
The delete.php code looks like this:

if (isset($_POST['delete']))// check if delete was clicked
        {
         $query = "DELETE FROM table_name WHERE id='$id'";
echo "$id Deleted successfully";
}
elseif(!mysql_query($query, $db_server))
{
echo "DELETE failed: $query<br />" .
myql_error() . "<br /><br />";
         }
mysql_close();

I'm new to php and still learning so, if you think there are other ways to do this, please let me know, the button won't do anything to the data.

Thanks in advance

I am trying to get this query correct.  I want to insert a record into the database upon form submission but only if the record does not already exist.  If the record exists, then I want it to be updated in the database.    What is happening:  Upon form submit, a new record is entered into the database every time.  Note:  The contact_id column is both primary key and unique in my database.  Here is my code:

if($_POST['submit']){
  $con=mysqli_connect("localhost","username","password","database_name");
  // Check connection
  if (mysqli_connect_errno())
    {
      echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }
	
	$org = mysql_real_escape_string($_POST['organization']);
  	$namefirst = mysql_real_escape_string($_POST['firstName']);
        $namelast = mysql_real_escape_string($_POST['lastName']);
	$emailaddy = mysql_real_escape_string($_POST['email']);
	$phonenum = mysql_real_escape_string($_POST['phone']);
	$appquestion = mysql_real_escape_string($_POST['appquestion']);
	$banner = mysql_real_escape_string($_POST['banner']);
	$bulletin = mysql_real_escape_string($_POST['bulletin']);
	$giveaway = mysql_real_escape_string($_POST['giveaway']);
	$app = mysql_real_escape_string($_POST['app']);
	$tshirt = mysql_real_escape_string($_POST['tshirt']);
	$tshirtp = mysql_real_escape_string($_POST['tshirtp']);
	$print = mysql_real_escape_string($_POST['print']);
	$party = mysql_real_escape_string($_POST['party']);
	$orgnotes = mysql_real_escape_string($_POST['notes']);
	
 
$sql="INSERT INTO database_name (contact_id, first_name, last_name, email_address, phone_number, org, appquestion, banner, bulletin, giveaway, app, tshirt, promised_tee, print, party, org_notes)
      VALUES
      ('','$namefirst','$namelast','$emailaddy','$phonenum','$churchorg','$appquestion','$banner','$bulletin','$giveaway','$app','$tshirt','$tshirtp','$print','$party','$orgnotes')	
	    
	  ON DUPLICATE KEY UPDATE first_name = '$namefirst', last_name = '$namelast', email_address = '$emailaddy', phone_number = '$phonenum', org = '$org', appquestion = '$appquestion', banner = '$banner', bulletin = '$bulletin', giveaway = '$giveaway', app = '$app', tshirt = '$tshirt', promised_tee = '$tshirtp', print = '$print', party = '$party', org_notes = '$orgnotes'" ;

						

    if (!mysqli_query($con,$sql))
    {
      die('Error: ' . mysqli_error($con));
    }
      echo "1 record added";



  mysqli_close($con);
}

From everything I have read, I need to use ON DUPLICATE KEY UPDATE to replace the old information with new information in the database upon form submission.  While the insert part of my code is working, the portion with ON DUPLICATE KEY UPDATE is not working.   Why might this portion of the code not be working?  Is there a better way to insert else update the information?   Thank you for any help or guidance you can give me!  I've been working on this concept for three days and have read a ton of information about it, but am still not able to get it to work.

Not sure how to update a record in my table here. I got my insert and delete statement working. Just can't get the update to work correctly. I'm not really sure how to do it.



Here is my code for my page, and my update and delete page



Code: [Select]
<?php

ini_set ("display_errors", "1");
error_reporting(E_ALL);

require_once('database.php');
session_start();




if (isset($_POST['add_grade']))
{



$query = "INSERT INTO grades (grade_id, student_id, grade_type, grade_name, grade_points) ";
$query .= "VALUES (:grade_id, :student_id, :grade_type, :grade_name, :grade_points) ";

$statement = $db->prepare($query);
$statement->bindValue (':student_id', $_SESSION['student_id']);
$statement->bindValue (':grade_id', $_SESSION['grade_id']);
$statement->bindValue (':grade_type', $_POST['grade_type']);
$statement->bindValue (':grade_name', $_POST['grade_name']);
$statement->bindValue (':grade_points', $_POST['grade_point']);

$statement->execute();

$statement->closeCursor();





$grade_type = $_POST['grade_type'];

if ($grade_type == 'Lab') {


$final *= .60;
}

echo $final;


$grade_type = $_POST['grade_type'];

if ($grade_type == 'Mid-Term') {


$final *= .20;
}

echo $final;



$grade_type = $_POST['grade_type'];

if ($grade_type == 'Final') {


$final *= .20;
}

echo $final;


}

?>
<!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">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>View Course Grades</title>

<link rel="stylesheet" type="text/css" href="main.css" />
</head>

<body>


   



<?php

$student_name = $_SESSION['student_name'];
$student_id = $_SESSION['student_id'];
$query = "SELECT * FROM grades WHERE student_id = :student_id ";

$statement = $db->prepare($query);
$statement->bindValue (':student_id', $student_id);
$statement->execute();
$grades = $statement->fetchAll();
$statement->closeCursor();


echo "<h1>Show Grades for $student_name </h1>";

foreach ($grades as $grade)
{

echo $grade['grade_type'] . " " . $grade['grade_name']. " " . $grade['grade_points'] . "<br />";
}





?>

<div id="content">
            <!-- display a table of products -->
           
            <table>
                <tr>
                    <th>Grade Type</th>
                    <th>Grade Name</th>
                    <th>Grade Points</th>
                    <th>Remove</th>
                </tr>
                <?php foreach ($grades as $grade) : ?>
                <tr>
                    <td><?php echo $grade['grade_type'];   ?></td>
                    <td><?php echo $grade['grade_name'];   ?></td>
                    <td><?php echo $grade['grade_points']; ?></td>
                    <td><form action="delete_grade.php" method="post">
                       
                        <input type="submit" name="remove" value="Delete" />




                    </form></td>
                </tr>
                <?php endforeach; ?>
            </table>
           
        </div>
    </div>

    <div id="footer">
       
    </div>






<form name="grades" method="post" action="grades.php">

<p>Grade Type<SELECT NAME="grade_type">
<OPTION VALUE="Mid-Term">Mid-Term
<OPTION VALUE="Final">Final
<OPTION VALUE="Lab">Lab
</SELECT>
<br>







Grade Name:<input type="text" name="grade_name" value=""><br />
Grade Points:<input type="text" name="grade_point" value="">

<input type="submit" name="add_grade" value="Add Grade">


</form>

</table>

</body>

</html>





My delete and update page




Code: [Select]
<html>
<head>
<title>Delete Grade</title>
</head>
<body>
<form method="post" action="delete_grade.php">

<?php


ini_set ("display_errors", "1");
error_reporting(E_ALL);

$dbc = mysqli_connect('localhost', 'se266_user', 'pwd', 'se266') 
or die(mysql_error());

//delete grades

if (isset($_POST['remove']))
{          
foreach($_POST['delete'] as $delete_id)
{             
$query = "DELETE FROM grades WHERE grade_id = $delete_id";             
mysqli_query($dbc, $query) or die ('can\'t delete user');          
}        
echo 'user has been deleted.<br />';   
}

if (isset($_POST['update']))
{          
foreach($_POST['update'] as $update_id)
{             
$query = "UPDATE grades SET grade_id = $grade_id";             
mysqli_query($dbc, $query) or die ('can\'t update user');    
       
}
}

//Display grade info with checkbox to delete  
$query = "SELECT * FROM grades";  
$result = mysqli_query($dbc, $query);  
while($row = mysqli_fetch_array($result))
{      
echo '<input type="checkbox" value="' .$row['grade_id'] . '" name="delete[]" />'; 
echo ' ' .$row['grade_type'] .' '. $row['grade_name'];
echo '<br />';  
}
mysqli_close($dbc);

?>

<input type="submit" name="remove" value="Remove" />
<input type="submit" name="update" value="Update" />

</form>
</body>
</html>

I have the following code:

Code: [Select]
$result = mysql_query("SELECT * FROM ESSAY_QUESTIONS WHERE SUBJECT = 'ENGLISH'")
or die(mysql_error());

while($essay_data=mysql_fetch_array($result)){

$question = $essay_data['QUESTION'];
$id = $essay_data['ID'];

echo "<a href=\"englishessays.php?id=$id>";
echo "$question </a>";
echo "<br><br>";
}
It seems to be almost working but it just displays one link and the address is all the rest of the code including </a>"; echo "<br><br>"; and other questions.

I presume there's an error with how I've written the echo statements so can anyone see it?  Or can anyone suggest a better way to do this?

On the next page, the php will read the ID from the address and display the information form that record.

Hi, I just want to know how to check if a row/record already exists for a given table, but this code gives me message in browser: Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in myTest.php

This is the the snippet I am running:
Code: [Select]
<?php
$connxn=mysql_connect("localhost","root");
$db=mysql_select_db("siliconvalley");
$toyNodePath="#document/siliconValleyHELLO";
$queryA=mysql_query("SElECT FROM pathexpress (path_express) WHERE path_express='$toyNodePath'");//make sure AT MOST ONE ROW EXISTS!!
if(mysql_num_rows($queryA)==1)
print "ALREADY IN table pathexpress!";
else {
mysql_query("INSERT INTO pathexpress (path_express) VALUES ('$toyNodePath')");
print "ADDED $toyNodePath to table pathexpress!";
}

?>
Please, any help is greatly appreciated!

I have some code that will update a record and is generic, meaning any POST variables can be used - whatever you have on the form.  See below:
Code: [Select]
$set = array();
foreach($_POST as $field => $value){
    $field = mysql_real_escape_string($field);
    $value = mysql_real_escape_string($value);
    $set[] = "`{$field}` = '{$value}'";
}


$query .= implode(", ",$set) . " WHERE $id_name = '".$id."' LIMIT 1";

mysql_query($query) or die(mysql_error());
My question is, how would I modify this to insert a NEW record (not update an existing one).  I'm not sure how to order this within a foreach statement because the add query has a different form: insert into tabel (all the fieldnames here) VALUES (all the values here)

This topic has been moved to MySQL Help.

http://www.phpfreaks.com/forums/index.php?topic=348130.0

I posted a topic earlier and there was confusion as to what I could not work out.   I am able to display the checkbox, either checked or unchecked depending of the value or 1 or 0 in the obcDisplay field.

<input type="checkbox" name="obcDisplay" value="<?=$r['obcDisplay'] ?>" <?=($r['obcDisplay']) ? 'checked="checked"' : ''; ?>/>

I am trying to update the DB table by the checkbox.  If the checkbox is checked (obcDisplay = 1) and I uncheck the box and submit, I want the table updated so that obcDisplay now equals 0.  When the form is displayed, the checkbox is now uncheck (obcDisplay = 0).  Now I check the box and submit and the update changes the field from 0 to 1.   I can display the checkbox as checked or unchecked by the DB field obcDisplay being either 0 or 1.  It is changing the value where I am having the problem.

Hello, Here is my current code: index.php

<html>
<body>
<form id="hey" name="hey" method="post" onsubmit="return false">
Name:<input type="text" name="name">
<input type="submit" name="submit" value="click">
</form>
<table class="table table-bordered" id="update">
<thead>
<th>Name</th>
</thead>
<tbody>
<script type="text/javascript">
      $("#hey").submit(function() {
          
        $.ajax({
          type: 'GET',
          url: 'response.php',
          data: {   username: $('#name').val()},
          success: function (data) {
            $("#update").prepend(data);
          },
          error: function (xhr, ajaxOptions, thrownError) {
            alert(thrownError);
          }
        });
      });
      
    </script>
</tbody>
</table>

response.php

<?php
$username = $_GET['username'];
echo "<tr>";
echo "<td>$username</td>";
echo "</tr>";
?>

This code works fine, it prints out the rows as what the user enters. Now what I want to do is, log all these entries to a mySQL database and also, display these rows over the site. i.e., any user who is online, should be seeing this without having to refresh the page too.. Real time updates in a way.   How can I achieve that? Thanks!
 

Hi I am very new to PHP & Mysql. I am trying to insert values into two tables at the same time. One table will insert a single row and the other table will insert multiple records based on user insertion. Everything is working well, but in my second table, 1st Table ID simply insert one time and rest of the values are inserting from 2nd table itself. Now I want to insert the first table's ID Field value (auto-incrementing) to a specific column in the second table (only all last inserted rows).   Ripon.   Below is my Code:

<?php



$con = mysql_connect("localhost","root","aaa");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }


mysql_select_db("ccc", $con);


$PI_No = $_POST['PI_No'];
$PO_No = $_POST['PO_No'];


$qry = "INSERT INTO wm_order_entry (
Order_No,
PI_No,
PO_No) 


VALUES(
NULL,
  '$PI_No',
 '$PO_No')";
$result = @mysql_query($qry);




$val1=$_POST['Size'];
$val2=$_POST['Style'];
$val3=$_POST['Colour'];
$val4=$_POST['Season_Code'];
$val5=$_POST['Dept'];
$val6=$_POST['Sub_Item'];
$val7=$_POST['Item_Desc'];
$val8=$_POST['UPC'];
$val9=$_POST['Qty'];
$N = count($val1);
for($i=0; $i < $N; $i++)
{


$profile_query = "INSERT INTO order_entry(Size,
Style,
Colour,
Season_Code,
Dept,
Sub_Item,
Item_Desc,
UPC,
Qty,
Order_No
) VALUES( '$val1[$i]','$val2[$i]','$val3[$i]','$val4[$i]','$val5[$i]','$val6[$i]','$val7[$i]','$val8[$i]','$val9[$i]',LAST_INSERT_ID())";
$t_query=mysql_query($profile_query);
}
header("location: WMView.php");
mysql_close($con);
?>

Output is attached.

Hello,

I'm editing an XML file using PHP, the XML file contains a playlist of songs for a flash mp3 player.
Now I'm able to edit the XML file, but I'm not able to save it  .

Here's my code:

    $random_speed  = mt_rand(1,15); // Discspeed
    $song_title    = /*mysql_real_escape_string($_GET['song_title'])*/ "My First XML Edit";
    $song_artist   = /*mysql_real_escape_string($_GET['song_artits'])*/ "A great author";
    $song_filename = /*basename(mysql_real_escape_string($_GET['song_filename']))*/ "myfirstxmledit.mp3";
    
    /* Open image directory and get a random image name */
    $dir          = "./img";
    /* Scan the directory and put everything in an array */
    $files        = scandir($dir);
    /* Count the elements in the array */
    $count_array  = count($files);
    /* Pick a random index from the array */
    $rand_array   = array_rand( $files, 1 );
    /* Pick the value of the random index */
    $random_image = $files[$rand_array];
    /* Loop through array and if element is not equal to "." or ".." then set the random_image variable */
    if( ($random_image == "." ) || ($random_image == ".." ) ){
        while( ($random_image == "." ) || ($random_image == ".." ) ){
            $random_image = $files[$rand_array];
        }
    }elseif( ($random_image != "." ) || ($random_image != ".." ) ){
        echo $random_image;
    }
    /* Open the music.xml file */
    $xml = new DOMDocument('1.0', 'utf-8');
    $xml->formatOutput = true;
    $xml->preserveWhiteSpace = false;
    $xml->load('music.xml');
    /* Begin settin new element to music.xml file */
    $newItem = $xml->createElement('song');
    $newItem->appendChild($xml->createElement('title', "'$song_title'"));
    $newItem->appendChild($xml->createElement('artist', "'$song_artist'"));
    $newItem->appendChild($xml->createElement('url', "songs/'$song_filename'"));
    $newItem->appendChild($xml->createElement('image', "img/'$random_image'"));
    $newItem->appendChild($xml->createElement('discspeed', $random_speed));
    /* Add the new elements to the xml tag 'song' */
    $xml->getElementsByTagName('song')->item(0)->appendChild($newItem);
    /* Save the xml file */
    $xml->save();

I've tried to enter the filename in the last $xml->save() function, didn't work, added the location + filename, didn't work ...
Anyone knows what or how do I save the file?

Thanks!

Hey all!
In the code in question I echo out individual records of data from MySQL successfully. For each record there is a number which is used as a var in the javascript that does the count-down-timer part.
However when I view the resulting page the timer works dynamically only with the first record. With the rest, the timer is static.

Code: [Select]
<?
$result0 = mysql_query("SELECT * FROM table WHERE field='$value'");
while ($riw0 = mysql_fetch_assoc($result0)) {
$seconds1 = $riw0['seconds'] ;
//// echo out data and set variable for the number of seconds to count down
?>

<script language="JavaScript">
var countDownInterval=<?=$seconds1?>;
var c_reloadwidth=200
</script>
<ilayer id="c_reload" width=&{c_reloadwidth}; ><layer id="c_reload2" width=&{c_reloadwidth}; left=0 top=0></layer></ilayer>
<script>
var countDownTime=countDownInterval+1;
function countDown(){
countDownTime--;
if (countDownTime <=0){
countDownTime=countDownInterval;
clearTimeout(counter)
window.location.href="military3.php" //Redirection URL
return
}
var mins = Math.floor(countDownTime/60)
var secs = countDownTime-(mins*60)
if (document.all) //if IE 4+
document.all.countDownText.innerText = mins+" minutes "+secs+ " ";
else if (document.getElementById) //else if NS6+
document.getElementById("countDownText").innerHTML=mins+" minutes "+secs+ " "
else if (document.layers){
document.c_reload.document.c_reload2.document.write('Soldiers will be ready in... <span id="countDownText">'+countDownTime+' </span> seconds')
document.c_reload.document.c_reload2.document.close()
}
counter=setTimeout("countDown()", 1000);
}

function startit(){
if (document.all||document.getElementById)
document.write('Soldiers will be ready in <span id="countDownText">'+countDownTime+' </span> seconds')
countDown()
}

if (document.all||document.getElementById)
startit()
else
window.onload=startit
</script>
<? } ?>

I tried replacing the javascript vars with PHP echoes for unique variables, but then no timer shows up, even static.
So could anyone advice me on how I could use this code to apply for all MySQL records?

Thanks in advance,
Thauwa

P.S. If I am unclear with my quandary, do let me know. Thank you.

Hello guys, this is my first post so sorry if I made any mistake   I need select record from one table and move to another table   But I get this message saying "Warning: mysqli_query() expects at least 2 parameters, 1 given in" I had that on line 158, but now i get on line 156   I start to do PHP and mysql few weeks ago, only respond i get from teacher is search and search.

<?php

				if (isset($_POST['username'])) 
				{
					$searchq = $_POST['username'];
					
					mysqli_query("SELECT * FROM login WHERE username='$searchq'")or die ("could not search");
					
			while($row = mysqli_fetch_array($con, $query)) 
			{
				$username = $row['username'];
				$password = $row['password'];
				$age = $row['age'];
				$phonenumber = $row['phonenumber'];
				$nationality = $row['nationality'];
				
					mysqli_query("INSERT INTO admin SET username ='$username', password='$password', age='$age', phonenumber='$phonenumber', nationality='$nationality'" ) ;
					echo"Data successfully inserted";
			}

				}
	?> 

When i search i see this type of code "$data = mysqli_query" add variable before mysqli   What should I do, to make it work. And send record from one table to another. Thank you

hello, anybody able to see what im doing wrong?

Code: [Select]
$id2=mysql_insert_id();
$year=$_POST['y'];
$yr=substr($year,-2);
$mth=$_POST['m'];
$init=$_POST['initial'];
$jobnumber=$yr.$mth.'-'.$id2.$init;

$query = "INSERT INTO jobs VALUES (
'',
'$id2',
'$_POST[initial]',
'$_POST[y]-$_POST[m]-$_POST[d]',date
'$_POST[contact]',
'$_POST[contactphone]',
'$_POST[customer]',
'$_POST[address]',
'$_POST[city]',
'$_POST[postal]',
'$_POST[province]',
'$_POST[description]'
)";
mysql_query($query) or die('Error, adding new job failed.  Check you fields and try again.');

echo "You have successfully entered a new job.  The job number is $jobnumber";