PHP - Populate Table - Mysql

Hi everyone.

I have a combo box which lists usernames and onchange, the username value is passed to a textbox. however, I have 3 textboxes i need to populate based on the selection of the combobox: username. department and email. i have the username going into a textbox, but i'm not sure how to pass department and email into two other textboxes.

I'd appreciate any help you could provide. Thanks.

Code: [Select]
<script>
function CBtoTB()
{document.getElementById("username").value=document.getElementById("usernameselect").value}
 </script>

<?php  
$result=mysql_query("select Username, EMail, Department from users"); 
$options=""; 
while ($row=mysql_fetch_array($result)) { 
$username=$row["Username"]; 
$options.="<OPTION VALUE=\"$username\">".$username.'</option>'; 
} 
?>
<select name="usernameselect" id="usernameselect" onchange="CBtoTB()">
<option value="">&lt; select user &gt;<?php echo $options ?></option>
</select>
 <input name="username" type="text" id="username" value="<?php echo $username ?>" size="25" readonly="readonly" />

Hi Good day Can someone pls help me regarding with this matter.
I have a page for inserting records. And I have a combobox that is fetched through the specific field in database. I also have a table of records, for each row i have an href edit which throws the values to its respective texbox and combobox. 
THE PROBLEM is that when i click the href edit, the value from a selected row was not displayed from the combobox

Hello,

I am trying to populate the value="" of my form with the data from the mysql database so users can see what is already completed, trial and error have failed me. what am i doing wrong?

<?php
session_start();
include 'english.php';

if (!isset($_SESSION['user']))
   die("<br /><br />You need to login to view this page");
$user = $_SESSION['user'];

echo "<h3>Edit your Profile Search</h3>";

//opend database
$connect = mysql_connect("$dbhost","$dbuser","$dbpass");
mysql_select_db("$dbname"); //select database

// Get all the data from the "housing" table
$result = mysql_query("SELECT * FROM users WHERE user='$user'")
or die(mysql_error());

// keeps getting the next row until there are no more to get
while($row = mysql_fetch_array( $result ))
   // Print out the contents of each row into a table

?>
<form action="/includes/insert_profile.php" method="post">
<table border="0" cellspacing="0" cellpadding="3">

</td></tr>
<tr><td>First Name:</td><td><input type="text" name="first" value="$first">

<tr><td>Middle Name:</td><td><input type="text" name="middle" value="$middle">

<tr><td>Last Name:</td><td><input type="text" name="last" value="$last">

<tr><td>Email:</td><td><input type="text" name="email" value="$email">

</td></tr>
<tr><td>Date of Birth:</td><td>
<select id="day" name="day" class="short">
<option value="1" selected="selected">1</option>
<option value="2">2</option>
<option value="3">3</option>

</select>
<select id="month" name="month" class="medium">
<option value="January" selected="selected">January</option>
<option value="February">February</option>

</select>
<select id="year" name="year" class="medium">
<option value="2010" selected="selected">2010</option>
<option value="2009">2009</option>
<option value="2008">2008</option>

</td></tr>
<tr><td>
<input type="Submit" value="Save"/>
</td></tr>
</table>
</form>

create table mimi (mimiId int(11) not null, mimiBody varchar(255) );
<?php
//connecting to database
include_once ('conn.php');
$sql ="SELECT mimiId, mimiBody FROM mimi";

$result = mysqli_query($conn, $sql );

$mimi = mysqli_fetch_assoc($result);

$mimiId ='<span>No: '.$mimi['mimiId'].'</span>';

$mimiBody ='<p class="leading text-justify">'.$mimi['mimiBody'].'</p>';
?>
//what is next?

i want to download pdf or text document after clicking button or link how to do that

Hello,

I need some help. Say that I have a list in my MySQL database that contains elements "A", "S", "C", "D" etc... Now, I want to generate an html table where these elements should be distributed in a random and unique way while leaving some entries of the table empty, see the picture below. But, I have no clue how to do this... Any hints?



Thanks in advance,
Vero

Hello everyone,

Sorry if this has been answered but if it has I can't find it anywhere. So, from the begining then.  Lets say I had a member table and in it I wanted to store what their top 3 interests are.  Their$ row has all the usual things to identify them userID and password etc.. and I had a further 3 columns which were labled top3_1 top3_2 & top3_3 to put each of their interests in from a post form.

If instead I wanted to store this data as a PHP Array instead (using 1 column instead of 3) is there a way to store it as readable data when you open the PHPmyadmin?

At the moment all it says is array and when I call it back to the browser (say on a page where they could review and update their interests) it displays 'a' as top3_01 'r' as top3_02 and 'r' as top3_03 (in each putting what would be 'array' as it appears in the table if there were 5 results. Does anyone know what I mean?

For example -

If we had a form which collected the top 3 interests to put in a table called users,

Code: [Select]
<form action="back_to_same_page_for_processing.php" method="post" enctype="multipart/form-data">
   
    <input name="top3_01" type="text" value="enter interest number 1 here" />
    <input name="top3_02" type="text" value="enter interest number 2 here" />
    <input name="top3_03" type="text" value="enter interest number 3 here" />
    <input type="submit" name="update_button" value=" Save and Update! " />

   
    </form> // If my quick code example for this form is not correct dont worry its not the point im getting at :)
And they put 'bowling' in top3_01, 'running' in top3_02 and 'diving' in top3_03 and we catch that on the same page with some PHP at the top -->

Code: [Select]
        if (isset($_POST)['update_button']) {
   
    $top3_01 = $_POST['top3_01']; // i.e, 'bowling' changing POST vars to local vars
    $top3_02 = $_POST['top3_02']; // i.e, 'running'
    $top3_03 = $_POST['top3_03']; // i.e, 'diving'

With me so far? If I had a table which had 3 columns (1 for each interest) I could put something like -

  Code: [Select]
   include('connect_msql.php');
   
    mysql_query("Select * FROM users WHERE id='$id' AND blah blah blah");
   
    mysql_query("UPDATE users SET top3_01='$top3_01', top3_02='$top3_02', top3_03='$top3_03' WHERE id='$id'");


And hopefully if ive got it right, it will put them each in their own little column. Easy enough huh? But heres the thing, I want to put all these into an array to be stored in the 1 column (say called 'top3') and whats more have them clearly readable in PHPmyadmin and editable from there yet still be able to be called back an rendered on page when requested.

Continuing the example then, assuming ive changed the table for the 'top3' column instead of individual colums,  I could put something like this -


Code: [Select]
            if (isset($_POST)['update_button']) {
       
        $top3_01 = $_POST['top3_01']; // i.e, 'bowling' changing POST vars to local vars
        $top3_02 = $_POST['top3_02']; // i.e, 'running'
        $top3_03 = $_POST['top3_03']; // i.e, 'diving'
   
    $top3_array = array($top3_01,$top3_02,$top3_03);
   
    include('connect_msql.php');
   
    mysql_query("UPDATE members SET top3='$top3_array' WHERE id='$id' AND blah blah blah");
But it will appear in the column as 'Array' and when its called for using a query it will render the literal string. a r r in each field instead.  Now I know you can use the 'serialize()' & 'unserialize()' funtcions but it makes the entry in the database practically unreadable. Is there a way to make it readable and editable without having to create a content management system?

If so please let me know and I'll be your friend forever, lol, ok maybe not but I'd really appreciate the help anyways.  The other thing is, If you can do this or something like it, how am I to add entries to that array to go back into the data base?

I hope ive explained myself enough here, but if not say so and I'll have another go.  Thanks very much people,  L-PLate (P.s if I sort this out on my own ill post it all here)

I know I'm doing it something right, but can someone tell me why only one table is showing up? Can you help me fix the issue?

Heres my code:


function showcoords()
{
echo"J3st3r's CoordVision";
$result=dbquery("SELECT alliance, region, coordx, coordy FROM ".DB_COORDFUSION."");
dbarray($result);
$fields_num = mysql_num_fields($result);

echo "<table border='1'>";
// printing table headers


    echo "<td>Alliance</td>";
    echo "<td>Region</td>";
    echo "<td>Coord</td>";


// printing table rows
while($row = mysql_fetch_array($result))
{
    

    // $row is array... foreach( .. ) puts every element
    // of $row to $cell variable
    foreach($row AS $Cell)
     echo "<tr>";     
     echo "<td>".$row['alliance']."</td>\n";
     echo "<td>".$row['region']."</td>\n";
     echo "<td>".$row['coordx'].",".$row['coordy']."</td>\n";
     echo "</tr>\n";
    
}
echo "</table>";
mysql_free_result($result);

}



I have 2 rows inserted into my coords table. Just frustrated and ignorant to php.

I have a table like this:
http://empirebuildingsestate.com/company.html

How can I separate the table values into variables so I can insert them into my mySQL database?
I have been trying DOMDocument but i need some help.

Hello,

So I need a bit of help with tables. I currently have my website set up with the following tables:

products (productId, productName, link, productPrice, productDesc)

productscents (scentId, scentName, scentDesc)

salves (productId, productName, productDesc)


So now most products in the products table can be made in any of the scents in the productScents table (except for the salves and reed diffuser jar).  I have a separate table of the salve types that can be ordered. How do I join each product with each productScent.

For example:

If I have the following products with their Id number:
1 Body Mist
2 Massage Oil
3 Reed Diffuser Jar


and the following scents with their id (id has leading zeros):
0001 Mango
0002 Passionfruit
0003 Grapefruit

How would I create a master products table that will display the following:
3        Reed Diffuser Jar
10001 Mango Body Mist
10002 Passionfruit Body Mist
10003 Grapefruit Body Mist
20001 Mango Massage Oil
20002 Passionfruit Massage Oil
20003 Grapefruit Massage Oil


I hope this makes sense. lol.

Any help would be greatly appreciated.

Hi

$eid2 = mysql_query("SELECT id FROM engines WHERE keyword = '%$tricker_engine%' LIMIT 1") or die(mysql_error());
$row = mysql_fetch_assoc($eid2);
$eid = $row['id'];

I want $eid to be the ID of the row where keyword = '%$tricker_engine%'.

What is wrong with my code above?

Greetings,   My current code logs into a database, opens a table named randomproverb, randomly selects 1 proverb phrase, and then SHOULD display the proverb in the footer of my web page. As of right now, the best I can do is get it to display "Array", but not the text proverb... this code below actually causes my whole footer to not even show up. Please help!

<?php

include("inc_connect.php"); //Connects to the database, does work properly, already tested

$Proverb = "randomproverb";
$SQLproverb = "SELECT * FROM $Proverb ORDER BY RAND() LIMIT 1";
$QueryResult = @mysql_query($SQLproverb, $DBConnect);
	while (($Row = mysql_fetch_assoc($QueryResult)) !== FALSE) {
	echo "<p style = 'text-align:center'>" . {$Row[proverb]} . "</p>\n";
	}
	$SQLString = "UPDATE randomproverb SET display_count = display_count + 1 WHERE proverb = $QueryResult[]";
	$QueryResult = @mysql_query($SQLstring, $DBConnect);
...
?>

Right, it's hard to explain, but I'll try my best.
Litterally, I have a 'tab' box on my homepage.


And a table called 'news'. Now, the table has the rows "shorttitle" "shortstory" and "image" in it.
I want the title the three latest titles to be on the three tabs, and the story and image in the respective content areas. (The table auto increments the ID, and so order by ID descending limit of 3 yes?)

Now, the code for the tabs is:
Code: [Select]
<?php
$query = "SELECT * FROM news where published = '1' ORDER by id DESC LIMIT 3";
$row_news = mysql_fetch_array($query);
?>
<div id="myTabs">
         <ul>
            <li><a href="#firsttab"></a><? echo $row_news['shorttitle']; ?></li>
            <li><a href="#secondtab">newsID2</a></li>           
            <li id="last"><a href="#thirdtab">newsID3 </a></li>           
        </ul>
<div id="firsttab" class="tab_content">
 <div class="rafat">
 <img src="<? echo $row_news['image']; ?>" />
 <p> <? echo $row_news['shortstory']; ?> </p>
 </div>
</div>
<div id="secondtab" class="tab_content">
 <div class="rafat">
 <img src="newsIMAGE2" />
 <p> newsSTORY2 </p>
 </div>
</div>
<div id="thirdtab" class="tab_content">
 <div class="rafat">
 <img src="newsIMAGE3" />
 <p> newsSTORY3 </p>
 </div>
</div>
</div>
Now, i've filled in the first tab with the MYSQL queries, but i'm unsure how to fill in the other two, because of the way it's coded.. I can't do a 'repeating' code, because of the IDs of the divs, and the three tabs are bunched together..

Anyone any ideas?
Much apreciated.

Thanks
Luke

I have a script that is supposed to allow a user to add their details to a members table in the database.

It acts as if the user has been added to the database by stating the thank you message, but there is no entry into the database.

I can connect to the database and have another script running OK that allows me to view all of the information in the members table with a loop (this data was entered using php my admin's GUI).

Here is the script that is supposed to add the member.

Any help would be greatly appreciated.

<?php

error_reporting(-1);

$user_name = "root";
$password = "";
$database = "test";
$server = "localhost";

$db_handle = mysql_connect($server, $user_name, $password);

$db_found = mysql_select_db($database, $db_handle);

if ($db_found)
{
$SQL = "INSERT INTO members (fname, sname, email, password, gender) VALUES ('$_POST[fname]','$_POST[sname]','$_POST[email]','$_POST[password]','$_POST[gender]')";
$result = mysql_query($SQL);

mysql_close($db_handle);

print "Thanks for joining us ".$_POST['fname'].".";
}
else
{
print "Database NOT Found";
mysql_close($db_handle);
}

?>

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

I am trying to place a unique number into a mysql table.

Currently, my code generates a random number, then is supposed to scan through the table for that number.
If the code finds that number already in the table, it generates a new random number and repeats the process.
I have commented my code for the purpose of this help forum:

Code: [Select]
        $result = mysqli_query($link,"SELECT * FROM testTable");

do
{
$end = true; //prepares end of loop
$idNum = rand(1,10); //rand(1,999999);  <-- for testing purposes I have reduced the number generated
$idNumTx = (string)$idNum;

                //loop through the rows
while ($row = mysqli_fetch_assoc($result))
{
if ($row['idNum'] = $idNum) //check if the random number equal to this row
{
$end = false; //prep end of loop repeat
echo $idNumTx; //display rand number that failed for testing purposes
echo " NO! "; //display error for testing purposes
}
}
} while(!$end);

I know I must be doing something wrong, as when I run this, it runs the if statement within the while loop always executes, and I get an output like:
Code: [Select]
1 NO! 1 NO! 1 NO! 1 NO! 1 NO! 1 NO! 5 Win

Win is when it places the value before it, in this case 5, into the table. However, the value of 5 might already be in the table and it doesn't seem to matter.
I execute the code multiple times, and it seems to increase the number of "# NO!" almost (but not every) time. However, each time ALL of the "# NO!" are the same #, and the "# Win" just seems to be random (as it should be, but not unique).

Checking the table after shows me random numbers between 1 and 10 (as it should) in the correct field, but the are not unique. (Ex/ Both row 1 and 5 could have the same value, say 6)

I'm hopefully doing something simple wrong, so someone please point it out to me