PHP - Selecting An Individual Field

This topic has been moved to MySQL Help.

http://www.phpfreaks.com/forums/index.php?topic=313049.0

I have been banging my head against a wall for a few days over this and think that I have read so many alternative ways, I have got myself confused!
 
What I am trying to do is post some information into the database and on success, provide certain variables back to my script (for example last_insert_id) as I would like to show a success message an add an option to a select on that page.
 
I am trying to get this part working without the mysql first just so I understand it. Currently I can post the data, have that received successfully in the processing page (addDetail.php) and send back output from that page........
 
insert.php 

$("#sub").click(function() {
  var name = $("#name").val();
  var town = $("#town").val();
        
  jQuery.ajax({
    type: "POST",
    url: "postScripts/addDetail.php",
   
    data: 'name='+name+'&town='+town,
    
    success: function(html) {

      $(".modal-body").prepend(
          "Returned: " +html
        );
    }
  });
  return false
}) 

addDetail.php

$name = $_POST['name'];
$town = $_POST['town'];
 
//Check $_POST data
echo "<pre>"; print_r($_POST) ;  echo "</pre>";
 
$return["name"] = $name;
$return["town"] = $town;

echo json_encode($return); 

 
 Result in insert.php

Returned:
Array
(
    [name] => The Name
    [town] => The Town
)
{"name":"The Name","town":"The Town"}

This works fine to bring back everything on the processing page in one go but I want to bring back separated variables so I can use them individually. I thought that changing 

 "Returned: " +html

to

 "Returned: " +html.name

would allow me to bring back just that variable but it comes back undefined. This is the closest I have got as other methods I tried either brought nothing back or just [objectObject].    How would I be able to bring back both 'name' and 'town' as separated values so I can use them individually in my script?   Thanks in advance   Steve
Edited by MargateSteve, 29 September 2014 - 08:41 AM.

Hey everyone
 I'm pulling data from my db and then trying to store each value to its own var. the problem is the amount of values is always changing.

here is what I have that isnt working

$result = mysql_query("SELECT thismonth FROM list");

while($row = mysql_fetch_array($result))
  {
echo $row[0] . "<br>";  // this prints out all the values perfectly
     
  }
  


no I just dont know how to store them so i can use them later.

I've got my code working OK in that the correct results are drawn from the database. What I have here in the code is an array consisting of an image, accompanied by its title and thirdly a link to activate a quiz associated with the image. Everything works fine, except because the page is dynamic, I always have an unknown number of sets of data (image, title and link) which I want to display so that each set of data takes up a new row.
The code I have here, while displaying the correct data, places all of the results into the same row.
Any suggestions are most welcome.
Code: [Select]
<?php
// Query the database
$query1 = mysql_query("SELECT title FROM topics WHERE managerId='".$managerId."' AND egroup1='"."1"."' ORDER BY title ASC"); 
$query2 = mysql_query("SELECT url_small FROM topics WHERE managerId='".$managerId."' AND egroup1='"."1"."' ORDER BY title ASC"); 
$query3 = mysql_query("SELECT title FROM topics WHERE managerId='".$managerId."' AND egroup1='"."1"."' ORDER BY title ASC");
while($row1 = mysql_fetch_array($query3))
{
$linkname .= $row1['title']."<br />\n";
}
?>
                  <table>
                    <tr>
                      <td>
                      <?php
                      while ($row2 = mysql_fetch_array($query2)) 
                      { 
                      $thumbnail.= $row2['url_small'];
                      echo "<img src='wood_tool_images/{$row2['url_small']}' alt='' /><br />\n";
                      }
                      ?>
                      </td>
                      <td height="200">
                      <?php
                      echo $linkname 
                      ?>
                      </td>
                      <td>
                      <?php
                      while ($row1 = mysql_fetch_array($query1)) 
                      { 
                      $quizname.= $row1['title'];
                      echo "<a href='../{$row1['title']} Safety Quiz/{$row1['title']} Safety Quiz.php'>Take This Quiz</a><br />\n";
                      }
                      ?>
                      </td>
                     </tr>
                  </table>

Hello!   I have five websites on a single server, and each website uses the same geographic data, for instance US zip codes and city/state latitudes and longitudes. Each website uses it's own copy of these "large" tables.   I would like to avoid having to maintain the same data in 5 databases, but I am more concerned with database performance, e.g., the speed at which the data is retrieved for each website.   Question: Should I set up a new separate database that contains the shared tables for all websites and allow each website to connect to this new database for shared data? Or is it better for each website have it's own copy of the duplicate tables?   Thanks for any info!   Cheers

Dear all,

I would like to find the total number of individual record in MySQL with php.

For example:
I have different department name stored in a column in MySQL.
I would like to find out the total number of each department.

But I do not know the names of the department

How can I do it?? Counting the total of each department without knowing their names

Thanks!

I have a database query set up that returns an array. I then cycle through the rows in a foreach statement that wraps each value in a <td> tag to output it to a table.

It works really great, but now I need to access two of the values and add some info to them. The first field returned is an image filename and I want to wrap it in an image tag that also concatenates the image path location to the value. All the images are stored in the same location, so basically I want to take the first value form the array and add the path to where the images are and wrap the whole thing in an <img> tag.

The other field I need to modify is the last value in the array which is actually an email address. For that field I want to make them an active email link so I need to wrap an <a>href=mailto: </a> tag around the value.

How could I modify this code so I could add the necessary tags to these elements of the array?

Here is the code I am running:
Code: [Select]
$query = "Select
                 member_image as 'Image',
                 member_name as 'Name',
                 city as 'City',
                 phone as 'Phone',
                 email as 'Email'
                 FROM directory";

//connect to database
$conn=db_connect();

//call function do_query to run the query and output the table
do_query($conn, $querey);

The functions called are as follows:
Code: [Select]
function db_connect()
{

    $conn = new mysqli("localhost", "username", "password", "databasename");

}

function do_query($conn, $query);
{

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

   WHILE ($row = mysqli_fetch_assoc($result))
   {
     If (!$heading) //only do if header row hasn't been output yet
     {  $heading = TRUE; //so we only do it once
         echo "<table>\n<tr<>\n";
         foreach ($row as $key => $val)
           { echo "<th>$key</th>";
            }
          echo "</tr>\n";
      }
      echo "<tr>";
      foreach ($row as $val)
     {
          echo "<td> $val </td>";
      }
     echo "</tr>\n";
} //close the while
echo "</table>\n";

}

Hi
I am trying to build a site which has an image gallery with links on it. I want to list the the images in different ways and I have successfully made it work with a "shuffle"-script like this:

Code: [Select]
<?php 
$links[] = Array("../r/ahlens.html", "k/k.jpg");
$links[] = Array("../r/brothers.html", "k/k.jpg");
$links[] = Array("../r/hm.html", "k/k.jpg");
$links[] = Array("../r/kappahl.html", "k/k.jpg");
$links[] = Array("../r/mq.html", "k/k.jpg");
$links[] = Array("sida1.php", "k/l.jpg");
$links[] = Array("../r/kappahl.html", "k/k.jpg");
$links[] = Array("../r/mq.html", "k/k.jpg");
$links[] = Array("../r/nk.html", "k/k.jpg");

shuffle($links);

foreach($links as $link){
    echo "<a href=\"${link[0]}\" ><img src=\"${link[1]}\" class=\"randomImage\" /></a> ";  
}  
?>
It works fine and opens a new page when clicked.

Now I would like to make it listed by file name instead of shuffling. How do I do that without loosing the links? And can it be done with a Directory because there will be a lot more photos in the future?
(I would also like the link to pages to be opened in a nice popup. Can I implement that?)
Thanks in advance!

This is my first real jump into PHP, I created a small script a few years ago but have not touched it since (or any other programming for that matter), so I'm not sure how to start this.

I need a script that I can run once a day through cron and take the date from one table/filed and insert it into a different table/field, converting the human readable date to a Unix date.

Table Name: Ads
Field: endtime_value (human readable date)

to

Table Name: Node
Field: auto_expire (Converted to Unix time)

Both use a field named "nid" as the key field, so the fields should match each nid field from one table to the next.  Following a tutorial I have been able to insert into a field certain data, but I don't know how to do it so the nid's match and how to convert the human readable date to Unix time.  Thanks in advance!.

Hi,

I had a form which contains, 8 select boxes. In those 8 select boxes, 3 of the boxes are interlinked, like, I am able to generate data in other 2 select boxes, if one box is been selected. i.e. based upon selection of one select box, and i am able to generate data in other select box. So, 3 select boxes are interlinked. And remaining 5 selects are optional, if the user selects, then we need to generate data based upon selection of the user.

Here the main issue is how to grab data in the post back form, like, how to capture the number of selects boxes selected and what are those selected, and based upon that we need to generate data.

Here one more important thing is that, in the report generation form(postback form), I need to show the data in a table. So , in table I had 9 columns and these would be common to whatever the selection made, but, only the requeired data would be changed based upon the selection of the remaining 5 optional boxes.

Hope you got me!!!!

Hi:
I'm going crazy trying to do the following:
I'm making a job registration process where the user registers on one php page to the website, must acknowlege and email receipt using an activate php page, then is directed to upload their C.V. (resume) based on the email address they enter in the active page output. I then run an upload page to store the resume in teh MySQL db based on the users email address in the same record.

If I isolate the process of the user registering to the db, it works perfectly.
If I isolate the file upload process into the db, it works perfect.
I simply cannot upload teh file to the existing record based on teh email form field matching the user_email field in the db.
With the processes together, teh user is activated, but teh file is not uploaded.

Maybe I've simply been at this too long today, but am compeled to get through it by end day.
If anyone can help sugest a better way or help me fix this, I will soo greatly appreciate it.

My code is as follows for the 2 pages.

---------activate.php-------
<?php
session_start();
include ('reg_dbc.php');
if (!isset($_GET['usr']) && !isset($_GET['code']) )
{
$msg = "ERROR: The code does not match..";
exit();
}
$rsCode = mysql_query("SELECT activation_code from subscribers where user_email='$_GET[usr]'") or die(mysql_error());
list($acode) = mysql_fetch_array($rsCode);
if ($_GET['code'] == $acode)
{
mysql_query("update subscribers set user_activated=1 where user_email='$_GET[usr]'") or die(mysql_error());
echo "<h3><center>Thank You! This is step 2 of 3. </h3>Your email is confirmed. Please upload your C.V. now to complete step 3.</center>";
} else
{ echo "ERROR: Incorrect activation code... not valid"; }

?>
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">

<head>
<meta content="text/html; charset=utf-8" http-equiv="Content-Type" />
<title>Job application activation</title>
</head>
<body>
<center>
<br/><br/><br/>
<p align="center">
        <form name="form1" method="post" action="upload.php" style="padding:5px;">
        <p>Re-enter you Email : <input name="email" type="text" id="email"/></p></form>
          <form enctype="multipart/form-data" action="upload.php" method="POST">
          <input type="hidden" name="MAX_FILE_SIZE" value="4000000">
          Upload your C.V.: <input name="userfile" type="file" id="userfile">
          <input name="upload" type="submit" id="upload" value="Upload your C.V."/></form>
</p>
</center>
</body>

</html>
--------upload.php----------
<?php
session_start();
if (!isset($_GET['usr']) && !isset($_GET['code']) )
{
$msg = "ERROR: The code does not match..";
exit();
}
if(isset($_POST['upload']) && $_FILES['userfile']['size'] > 0)
{
$fileName = $_FILES['userfile']['name'];
$tmpName  = $_FILES['userfile']['tmp_name'];
$fileSize = $_FILES['userfile']['size'];
$fileType = $_FILES['userfile']['type'];
$email = $_POST['email']['user_email'];

$fp      = fopen($tmpName, 'r');
$content = fread($fp, filesize($tmpName));
$content = addslashes($content);
fclose($fp);

if(!get_magic_quotes_gpc())
{
    $fileName = addslashes($fileName);
}

include 'reg_dbc.php';

$query = "UPDATE subscribers WHERE $email = user_email (name, size, type, content ) ". "VALUES ('$fileName', '$fileSize', '$fileType', '$content')";

mysql_query($query) or die('Error, query failed');
mysql_close($dbname);
}
?>
<center>
<br/>
<br/>
<br/>
<br/>
Thank you for uploading your <?php echo "$fileName"; ?> file, completing your registration, and providing us your C.V. for this position.
<br/>
<br/>
<br/>
We will contact you if your canditature qualifies.
</center>

Code: [Select]
$new_array2=array_diff($my_array,$itemIds);

$updatedb = mysql_query("UPDATE content_type_ads SET field_expire_value = '666' WHERE $new_array2 = field_item_id_value");
"$new_array2" has only 12 digit numbers for each item in the array, and I want to match them to the "field_item_id_value" field in the table, updating the "field_expire_value" field for the record where they match.  I know I am close because the UPDATE code works before I add the WHERE statement (it of course adds '666' to EVERY field, but for a noobie it's a start!).

(Another quite newbie...)
I have already an online booking form. The Form works fine now and I would like to add on one more issue, but the code ignores what I want to check. I have 4 fields: arrival, departure, no. of persons and comments to check.

Scenario 1:

All field mentioned above are emtpty: Workes fine and message appears: "You have not provided any booking details".

Scenario 2:

If arrival (date_start) and departure (date_end) is entered, there should be at least an entry either in the field "comment", or in the field "pax". If not, there should be a message: "You have not provided sufficient booking details".

INSTEAD: The booking request is sent, which should not be the case.

The code is currently:

      # all fields empty : arrival, departure, pax and comments - error

if(empty($data_start) && empty($data_end) && empty($pax)&& empty($comment)){
    exit("You have not specified any booking details to submit to us.
    <br>Please use your browser to go back to the form.
    <br>If you experience problems with this Form, please e-mail us");
    exit;
}
      #If arrival  and departure date is entered, there should be at least an entry

either in the field "comment", or in the field "pax".
      
if(!isset($data_start) && !isset($data_end) && empty($pax) && empty($comment)){
    exit("You have not provided sufficient booking details.
    <br>Please use your browser to go back to the form.
    <br>If you experience problems with this Form, please e-mail us");
    exit;
}

The form can be tested at http://www.phuket-beachvillas.com/contact-own/contact-it.php

Can someone please check and tell me what's wrong with the code ? Thanks to anyone for any hint.

Hi,
I'm developing an app using flash AS2 as the front end via PHP and a mySQL database at the backend on my server.

what i'm looking to do is update/insert into a table called 'cards' and at the same time update/insert into another table called 'jingle'. There is a field called 'cardID' in jingle that should be the same as the ID number created in 'cards' thus creating a link between entries in the different tables that can be called up as i choose.

hope i've been clear i just wouldn't know where to start any help would be appreciated.

MySQL client version: 5.0.91
PHPmyAdmin Version information: 3.2.4


thanks in advance