PHP - Select Then Insert Into Another Table

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.

could anyone please help me with the code which is i have already displayed data as a multi select list but now i need to select one or more from them and insert into another database table. would be appreciate your help. thanx

I am simply trying to insert a value generated from an array in a while loop, but it seems the value is not global and I can't pass it as I like. I did some research on this, but could not find an answer that solved my issue...

Here is my select box code which is working perfect:

<select name="city">
<?php
$sql = "SELECT id, city_name FROM cities ".
"ORDER BY city_name";
$results_set = (mysqli_query($cxn, $sql)) or die("Was not able to produce the result set!");
while($row = mysqli_fetch_array($results_set)) {
echo "<option value=$row[id]>$row[city_name]</option>";
}
?>
</select>

Are any variables defined in a while loop global to the while loop only?

Here is my SQL which you can see my $row[id] being passed thru field city_id... The value is being generated as supposed to based on value like: value="1", value="2" etc.. for the select options for each city name. So the values are there... But I CANNOT get that numerical id to pass to the database when submitting my form.

Any ideas for a workaround to get this value passing as normal?

if (isset($_POST['addPosting'])) {
      $query = "INSERT INTO Postings (id, city_id, title, description) VALUES
('','$row[id]','$_POST[title]','$_POST[description]')";

I have 2 queries that I want to join together to make one row  

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";

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

Hi there,
I have a form which I want to submit data into my tables.
There are going to be 4 tables involved with this form, and these 4 tables should relate to one another in some sort of way.

My problem is either PHP or MySQL, but I keep getting a warning which I can't figure out.
I remember this warning appearing even if the code before it is wrong, therefore I am not relying on it.

This is what the error says:
Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in G:\xampp\htdocs\xampp\dsa\wp3.php on line 40

Here's my code:
Code: [Select]
<html>
<head>
<title>WP3</title>
</head>
<body>
<form id="search" name="search" id="search" method="get" action="searchresults.php" />
<input type="text" name="terms" value="Search..." />
<input class="button" type="submit" name="Search" value="Search" />
</form>

<?php

include_once('connectvars.php');

$dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);


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

$firstname = mysqli_real_escape_string($dbc, trim($_POST['firstName']));
$middlename = mysqli_real_escape_string($dbc, trim($_POST['middleName']));
$lastname = mysqli_real_escape_string($dbc, trim($_POST['lastName']));
$image = mysqli_real_escape_string($dbc, trim($_POST['image']));
$phone = mysqli_real_escape_string($dbc, trim($_POST['phone']));
$organisation = mysqli_real_escape_string($dbc, trim($_POST['organisation']));
$street = mysqli_real_escape_string($dbc, trim($_POST['street']));
$town = mysqli_real_escape_string($dbc, trim($_POST['town']));
$city = mysqli_real_escape_string($dbc, trim($_POST['city']));


if (!empty($firstname) && !empty($middlename) && !empty($lastname) && !empty($image) && !empty($phone) && !empty($organisation) && !empty($city)) {

$query = "INSERT INTO report (organisation, phoneNo) VALUES ('$organisation', '$phone');
  INSERT INTO person (firstName, middleName, lastName) VALUES ('$firstname', '$middlename', '$lastname');
  INSERT INTO identification (image) VALUES ('$image');
  INSERT INTO location (street, town, city) VALUES ('$street', '$town', '$city')";

$data = mysqli_query($dbc, $query);

if (mysqli_num_rows($data) == 0) {

mysqli_query($dbc, $query);
echo "Thank you, your report has been received.";

}

else {
        // An account already exists for this username, so display an error message
        echo '<p>This report already exists.</p>';
        $username = "";
}
}
else echo "Please enter all of the fields";
}

?>

<form id="report_sighting" action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
<h2>Report a sighting</h2>
<table>
<tr>
<td>
<label>First name:</label>
</td>
<td>
<input type="text" id="firstname" name="firstName" value="<?php if (!empty($firstname)) echo $firstname; ?>" />
</td>
</tr>
<tr>
<td>
<label>Middle name:</label>
</td>
<td>
<input type="text" id="middlename" name="middleName" value="<?php if (!empty($middlename)) echo $middlename; ?>" />
</td>
</tr>
<tr>
<td>
<label>Last name:</label>
</td>
<td>
<input type="text" id="lastname" name="lastName" value="<?php if (!empty($lastname)) echo $lastname; ?>" />
</td>
<tr>
<td>
<label>Upload Identification:</label>
</td>
<td>
<input type="file" id="image" name="image" />
</td>
<tr>
<tr>
<td>
<label>Contact phone number: </label>
</td>
<td>
<input type="text" id="phone" name="phone" />
</td>
<tr>
<tr>
<td>
<label>Organisation: </label>
</td>
<td>
<input type="text" id="organisation" name="organisation" />
</td>
</tr>
<tr>
<td>
<label>Street seen: </label>
</td>
<td>
<input type="text" id="street" name="street" />
</td>
</tr>
<tr>
<td>
<label>Town seen: </label>
</td>
<td>
<input type="text" id="town" name="town" />
</td>
</tr>
<tr>
<td>
<label>City seen: </label>
</td>
<td>
<input type="text" id="city" name="city" />
</td>
</tr>
<tr>
<td>
&nbsp;
</td>
<td>
<input type="submit" value="Report" name="report" />
</td>
</tr>
</table>
</form>

</body>
</html>

I've checked out the SQL statement and it's alright, so that leaves me with the PHP.

I would very much appreciate if anyone could help me out here,
thanks.

Hey guys,
it seems that so far in my little script everything is working except for the MySQL insert
Nothing seems to appear in the table it's supposed to insert too but i dont see any errors anywhere.

Any help would be appreciated.

I know it's very messy but it's my first actual 'project'
  Code: [Select]
<?php
$apiKey="<omitted>";
$playeritemsURL="http://api.steampowered.com/IEconItems_440/GetPlayerItems/v0001/";
$fullURL=$playeritemsURL."?key=".$apiKey."&SteamID=".$_GET["SteamID64"]."&format=xml";
$playerLocalName=$_GET["SteamID64"].".xml";
$ch=curl_init($fullURL);
$fh=fopen($playerLocalName, 'w');
?>

You entered ID: <?php echo $_GET["SteamID64"];?></br>

<?php
echo $fullURL;
curl_setopt($ch,CURLOPT_FILE,$fh);
curl_exec($ch);
curl_close($ch);
?>

</br>

<?php
$con = mysql_connect("localhost","root","");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("csv_db", $con);
$readXML=file_get_contents($playerLocalName);
$xml=simplexml_load_file($playerLocalName);
$output=$_GET["SteamID64"].".defindex.txt";
$fh=fopen($output, 'w') or die("can't open file");
//echo $xml->items->item->defindex[0];
foreach ($xml->xpath('//defindex') as $defindex) 
{
fwrite($fh, $defindex."\n");
echo $defindex."</br>";
mysql_query("INSERT INTO ".$_GET["SteamID64"]." (defindex) VALUES (".$defindex.")");
} 
?>

Code: [Select]
   $myusername = $_POST['myusername'];
  mysql_query("INSERT INTO logedin ( username ) VALUES ('$myusername')") OR die("Could not send the message: <br>".mysql_error());


wont insert anything no error's given even
just trying to insert username into the the loged in table yes i know i spelt loged in wrong but i did that on the db lol

Hi,

I'm quite new to php and not the greatest of coders, so urgently require your help.

I want to be able to take data from an un-normalised table and load into 2 normalised tables as below:

1) Table 1 - All Music Tracks (Un-normalised table)

Album Name
Artist Name
Track Name
Track Length
etc.

2) Table 2 - Albums Table (Normalised table)

ID
Artist Name
Album Name
etc.

3) Table 3 - Tracks Table (Normalised table)

ID
Track Name
Album ID (foreign key to Albums table)
etc.

Can someone please provide some sample code as a starting point? The All Music Tracks table includes both the album name and track name in each row, but I only want the album appearing once in the Albums tables, would I have to do an Distinct select first to get the album names into an array? I'm not sure how I would then have  nested loop to do the insert into the Tracks table.

Much Appreciated,

PD

I am trying to insert records from an foreach array, it only inserts the first record and I cannot work out how to get it to loop through all of the records.

Thank you


foreach($_SESSION['post_data_array'] AS $seat) {
     $rowId = substr($seat, 0, 1);
     $columnId = substr($seat, 1);
      echo $rowId . $columnId . ", "; 
}

$sql125 = "INSERT INTO booked_seats(booking_id, row_id, column_id) values ('$book_id', '$rowId', '$columnId')";

$result125 = mysql_query($sql125);

if ($result125)

{

echo "worked";

}

else

{

echo "didnt work";

}

Im trying to insert some values automatically into a table once the form loads, but Im getting an error. Here is the code

Code: [Select]
<?php
$aid = $_GET['aid'];
$sd = $_GET['sd'];


?>


<style>
#message {margin:20px; padding:20px; display:block; background:#cccccc; color:#cc0000;}
</style>

<div id="message">Your notification has been submitted.</div>

<div style="text-align:center ">
<?php
$connection = mysql_connect("localhost",
        "username",
        "password");

    mysql_select_db("articles", $connection); 

$query="INSERT INTO broken_links (articleid, article) VALUES ('$aid', '$sp')";
if (!mysql_query($sql,$con))
  {
  die('Error: ' . mysql_error());
  }
echo "Submitted";
mysql_close($con)
?>

<table border="0" cellpadding="3" cellspacing="3" style="margin:0 auto;" >
 
<input type="submit" id="Login" value="&nbsp;&nbsp;Thank you. Please press to close&nbsp;&nbsp;" onclick="tb_remove()"></td>
</tr>
</table>
</div>

Any help will be appreciated

Hello,

I'm working on a PHP register form, all I want to do, is to be able to insert an extra variable into the database:

Code: [Select]
// now we insert it into the database
$insert = "   INSERT INTO users (username, password) VALUES ('".$_POST['username']."', '".$_POST['pass']."')   ";

$add_member = mysql_query($insert);
right, so I have the username and password variable being passed. But I also want to pass a variable into a column called 'weaponAttachments', all I want for this variable to be is 5000.

So, in other words, something like this: 5000['weaponAttachments']

Thanks

Hi,   The following code was written by someone else. It allows me to upload images to a directory while saving image name in the mysql table.   I also want the code to allow me save other data (surname, first name) along with the image name into the table, but my try is not working, only the images get uploaded.   What am I missing here?

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

$path=$path.$_FILES['file_upload']['name'];

if(move_uploaded_file($_FILES['file_upload']['tmp_name'],$path))
{
echo " ".basename($_FILES['file_upload']['name'])." has been uploaded<br/>";
echo '<img src="gallery/'.$_FILES['file_upload']['name'].'" width="48" height="48"/>';
$img=$_FILES['file_upload']['name'];
$query="insert into imgtables (fname,imgurl,date) values('$fname',STR_TO_DATE('$dateofbirth','%d-%m-%y'),'$img',now())";
if($sp->query($query)){
echo "<br/>Inserted to DB also"; 
}else{
echo "Error <br/>".$sp->error; 
}
}
else
{
echo "There is an error,please retry or check path";
}
}

?>

joseph