PHP - Foreach With Mysql Field Names

Hi all.  I've got a little function to generate a simple form depending on the input.  What I'd like to do is alias the field names that get output to make them a little more human readable.  However, I'm not entirely sure how to do that with the output that I get.

The function....
function build_form($table_name){
$result=mysql_query("SELECT * FROM $table_name");
$num=mysql_num_rows($result);
$i=1;
echo "<table>";
while ($i < mysql_num_fields($result)) {
$fields=mysql_fetch_field($result,$i);
echo "<tr><td>" . $fields->name . "</td><td><input type=\"text\" size=\"30\" name=\"" . $fields->name . "\" /></td></tr>";
$i++;
};
echo "</table>";
};

Q:  Where would I have to modify the output from mysql_fetch_field?

My modified function...
function build_form($table_name){
$result=mysql_query("SELECT * FROM $table_name");
$num=mysql_num_rows($result);
$i=1;
echo "<table>";
while ($i < mysql_num_fields($result)) {
$fields=mysql_fetch_field($result,$i);
                $field_name=str_replace("_"," ",$fields->name);
echo "<tr><td>" . $field_name . "</td><td><input type=\"text\" size=\"30\" name=\"" . $field_name . "\" /></td></tr>";
$i++;
};
echo "</table>";
};

Not sure if that will work, but am I heading in the right direction?  Any help is much appreciated!

I would like to echo field names in my table on webpage one by one. I know mySQL has "describe" function which will list the complete table, I am looking for a way to display each field name one by one with other stuff in between them like input field or description.

I have the following mysql query:
Code: [Select]
        $lead_query=$this->db->query("
   SELECT `first_name`, `last_name` , `state` FROM leads WHERE `lead_id`='$lead_id'
   ");
   $this->view->lead_query=$lead_query->fetchALL();
Now when I retrieve the above details  I need to return the first_name last_name together(separated as space) and the name of the key as client_name in the array.

I need it that way because when i return a json_encode($lead_query), I want to return the first_name.last_name as client_name.

Hi all.

The title pretty much says it all.  I really have no idea if this is even possible or where to start.  A recent dilemma caused me to have to change a field name with some spaces in it.  I have a database with ~65 tables, a majority of which were converted from Access.  So I'm curious as to how many others have bad field names.  Is there a way to use PHP to go through all the field names in a database and remove spaces?

Okay, so I'm trying to run a query to pull the information from a specific item that they click on in order to edit it.

When I run an echo $query, though, it shows the field names rather than the information from the table.

How can I make it so that it pulls the information rather than the field names?

Here's what I have...


<?php
include("lib.php");
define("PAGENAME", "Edit Equipment");

if ($player->access < 100)

$msg1 = "<font color=\"red\">"; //name error?
$error = 0;

$query = $db->execute("SELECT * FROM items WHERE id=?", array($_POST['id']));

echo "$id";

$result = mysql_query($query);

echo "$query";

$data = mysql_fetch_assoc($result); 

$msg1 .= "</font>"; //name error?


?>


Thanks!!

This topic has been moved to MySQL Help.

http://www.phpfreaks.com/forums/index.php?topic=318312.0

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>

Hello,

I'm having some issues with PHP thinking that the variables that I send it are the actual columns in my database.  First, I pull off Quadratic_Functions and introductory_problem from http://localhost:8888/algebra_book/Chapters/Quadratic_Functions/introductory_problem.php using the code below:

Code: [Select]
$chapter_page =  $_SERVER['PHP_SELF'];
$chapter_page_array = explode('/',$chapter_page);
$size =count($chapter_page_array);
$chapter = $chapter_page_array[$size-2];
$page = $chapter_page_array[$size-1];
$page_array = explode('.', $page);
$page = $page_array[0];
Based on my printing of the variables $chapter and $page I think that it's doing what I want it to do.  I then use the following function:

Code: [Select]
$supplemental_id = getSupplementalId($dbRead,$chapter,$page);
to check out if the there's a supplemental_id for the Quadratic_Function chapter and introductory_problem page name via:

Code: [Select]
function getSupplementalId($read,$user_id,$chapter,$page)
  {
  $sql = "SELECT supplemental_id
  FROM instructors_supplemental
WHERE page_name = $page
AND chapter_name='$chapter";

return $read->fetchRow($sql);
  }
If I stick in actual values, as seen below, the thing runs fine.

Code: [Select]
$sql = "SELECT supplemental_id
  FROM instructors_supplemental
WHERE page_name = 'introductory_problem'
AND chapter_name='Quadratic_Functions'";

But if I run it in the abstract version, with variables for page and chapter name (the first version), I get Fatal error: Uncaught exception 'PDOException' with message 'SQLSTATE[42S22]: Column not found: 1054 Unknown column 'introductory_problem' in 'where clause'' in...

It's almost as if it thinks that my variables are the names of the columns. 

Any thoughts would be appreciated....

I can get files to  upload to folder and add a new row for each in to a mysql db using the code below, how do I get all the filles in one row with multiple uploads

Code: [Select]
//Check if the form was submitted.
if(isset($_POST['uploadButton'])) {
  //Specify folder path where the file is going.
  $path = 'uploads/';

  //Upload file one by one.
  foreach($_FILES['file']['name'] as $key => $val) {
    if($val != '') {
      $file_name = $_FILES['file']['name'][$key]; //Get file name.
      $file_tmp  = $_FILES['file']['tmp_name'][$key]; //Get temporary file name.
      $file = $path . $file_name;
      mysql_query("INSERT INTO images (img1) VALUES('$file');") or trigger_error('Unable to INsert: ' . mysql_error());
   




 
      //Move uploaded file
      if(move_uploaded_file($file_tmp, $file)) {
        echo 'File was succesfully uploaded to: ' . $file . '<br />';
      } else {
        //Display error message if there was a problem uploading file.
        echo 'Error uploading file "' . $key . '."<br />';
      }
echo $target_path;


        }
     }
  }

?>

This topic has been moved to MySQL Help.

http://www.phpfreaks.com/forums/index.php?topic=344739.0

In querying my database, I'm Selecting nameFirst and nameLast, and it produces a name like Joe Smith.  I'm trying to match a photo with the name.  Right now I'm uploading photos into a folder naming the file (e.g. SmithJoe.jpg). 

For reasons that involve writers being able to upload and access photos, I'm trying to use an image plugin.  When uploading photos, it strips capital letters, so SmithJoe,jpg goes in as smithjoe.jpg, and it's not matching my database query. 

Here is the code I'm working with that works quite well with this one exception:


$query = 'SELECT * FROM wp_playerRank';
$results = mysql_query($query);
while($line = mysql_fetch_assoc($results)) {

if ($line['wpID'] == $wp_tagID)  {

echo '<div class="player">';


// Here is the code that produces the image.  I need to get rid of capital letters for ease of use

echo '<div><img src="/wp-content/uploads/' . $line['nameLast'] . $line['nameFirst'] . '.jpg"></div>';



echo '<div class="playerData">' . $line['height'] . ' ';
if ($line['position'] == 'PG')  {echo 'Point Guard';}
elseif ($line['position'] == 'SG')  {echo 'Shooting Guard';}
elseif ($line['position'] == 'SF')  {echo 'Small Forward';}
elseif ($line['position'] == 'PF')  {echo 'Power Forward';}
elseif ($line['position'] == 'C')  {echo 'Center';}
echo '</div>';
echo '<div class="playerData">' . $line['hschool'] . '</div>';
echo '<div class="playerData">' . $line['summer'] . '</div>';
echo '<div class="playerData">Class of ' .$line['year'] . '</div>';

if ($line['committed'] == 'y') {
echo '<div>
<br>Committed to <strong>'. $line['college'] . '</strong></div>
';}

}
}

Hi, I am new to php, mysql and have made some progress, but recently ran into a wall. I hope someone can help me. I am building a simple links script that has categories and a list of links with website name, url and a brief description.  I have two tables-categories and links.

I have gotten the back end finished and can add, edit and delete links and categories. The problem is in the front end.  I have the script set up so that when I click on a category name a page is dynamically created using ?id=idnumber.  and the category name is echoed on the page. But I can't seem to find the solution I need to get the links that have a matching category name to display. I have tried using JOIN, LEFT JOIN, INNER JOIN, and RIGHT JOIN and get the same result, ALL the links in the links table are displayed.

I think I need to create a new array and use the foreach loop to accomplish this, but I am not sure. Nor am I sure how to use the foreach even though I have read several tutorials on it. 

Here is the code I have so far. Any help would be appreciated. Thanks in advance.

Code: [Select]
<? include "db.php" ?>
<?php
$id = intval($_GET['id']); // force user input to an int
if(!empty($id)) // if the user input is not empty or 0
{
$query = "SELECT * FROM categories WHERE id =  $id  LIMIT 1"; // attempt to select the row for the given id
$result = mysql_query($query) or die(mysql_error());

if(mysql_num_rows($result) > 0) // if the categorie was found
{
$row = mysql_fetch_assoc($result); // fetch the resultset to an array called $row

echo $row['categories']; // echo the categories field from the row
// etc.
}
else
{
die('Error: Bad ID'); // the categories was not found
}
}
else
{
die('Error: Bad ID'); // the id given was not valid
}
mysql_close();
?>
<? include "db.php" ?>
<?php
$sql="SELECT * FROM links,categories ORDER BY links.categories LIMIT 10";
$result=mysql_query($sql);

?>
<table width="80%" align="center" border="0" cellspacing="1" cellpadding="0">
<tr>
<td>
<table width="100%" border="1" cellspacing="0" cellpadding="3">
<tr>
<td colspan="3"><strong>List data from mysql </strong> </td>
</tr>

<tr>
<td align="center"><strong>Name</strong></td>
<td align="center"><strong>URL</strong></td>
<td align="center"><strong>Content</strong></td>
</tr>
<?php
while($rows=mysql_fetch_array($result)){
?>
<tr>
<td><? echo $rows['name']; ?></td>
<td><a href="<? echo $rows['url']; ?>"><? echo $rows['url']; ?></a></td>
<td><? echo $rows['content']; ?></td>
</tr>
<?php
}
?>
</table>
</td>
</tr>
</table>
<?php
mysql_close();
?>

 

I'm trying to get a foreach loop to one variable that I can insert into MySQL - I've tried a number of things but can't get it working.  How do I create one single variable that is created from the foreach?

Thanks!

Jason

ok this may not be able to be done .... but here goes nothing ...

i have a roster that displays member names as a link to thier profile using a form and onclick submit() now what i didnt realize is that the form name will have to increment for each record returned from the query.  i am by no means a guru with php so i need some help adding the foreach statement and setting the form name to increment each time


what i need:
I need this to automatically go +1 for each record returned
Code: [Select]
<form name="view3" action="./viewinfo.php" method="post">

<a href="#" onclick="document['view3'].submit()">

and i need the correct way to use the foreach statement ... i have read the manual and it just confuzzeled me .


here is what i have so far.



<?php

include './clan_new/config.php';

include './clan_new/access.php';

$db = mysql_connect ($hostname, $username, $password) or die ('Failed to connect to database: ' . mysql_error());

mysql_select_db($database);

$query = "SELECT * FROM $member_table WHERE $member_table.rank = 6 ORDER BY $member_table.name ASC";

$result = mysql_query($query) or die ('Failed to query ' . mysql_error());

while ($row = mysql_fetch_assoc($result)) {

$steamid = $row['authid'];

$name = $row['name'];

$email = $row['email'];

$fid = $row['fid'];

$avatar = $row['avatar'];

echo "<tr>";

if ($avatar==""){

echo "<td>&nbsp;</td>";

} else{

echo "<td><img height='35' width='35' src='./clan_new/avatars/$avatar'></td>";

}

?>
        <form name="view3" action="./viewinfo.php" method="post">
        <input type="hidden" id="authid" name="authid" value="<?php  echo "$steamid" ; ?>" />
        </form>

<td align="center"><a href="#" onclick="document['view3'].submit()"> <?php  echo "$name" ; ?></a></td>
        <?php 

echo "<td align=\"center\"><a class=\"style2\" href=\"mailto:$email\" class=\"style2\">$email</a></td>";

if ($fid==""){

echo "";

} else{

echo "<td align=\"center\">+<a class=\"style2\" href=\"steam://friends/add/$fid\">Friend</a></td>";

}

echo "</tr>";

}

mysql_free_result($result);

mysql_close($db);

?>