PHP - Why Doesnt My 2nd Drop Down Have Distinct Values?

I am trying to make a Web Form for people to fill out that has two drop down menus where i want the first drop down menu's selection changes the values of the selections in the second drop down. ((IE. If the first drop down value is "product 01" then the second drop down shows values "Red, Green, Blue" while if the first drop down value is "product 02" then the second drop down shows values "Yellow, Green"))
Anyone have any ideas? Thanks in Advance.

Hi guys, I had a problem before which is similar to the problem im about to describe, however Pikachu helped me solve the last problem so a big thanks to him! However this problem is slightly different and im struggling to find a solution. I will first describe the problem I had which pikachu helped me solve as it would be easier for me to describe this past problem.

I have an update form, so a user can edit existing records which are pulled from a database. When the user changes the values in a text field then submits the form, the values in the text fields updates fine. However if they change the values in the text box and then tries to submit the form but there was an error, the values in the text fields go back to the values from the database and the user has to change the text fields again.

Well this problem was solved with the help of pikachu and the this is what I did...

<?php if(isset($_POST['title'])){echo htmlspecialchars($_POST['title']);} else if (isset($title)) {echo htmlspecialchars($title);}?>

Again, the above works fine however the problem I have now is the same problem as described above however it relates to drop down menus and not text fields. I can't seem to figure out how to retain the values in a drop down menu the same way I have for text fields. Below is the code that im using that I thought would work however it isnt working, the values of the drop down menu (if there was an error) are going back to the values from the database. Heres the code...

<option value="Psychopathic" <?php if ((isset($_POST['category']))&&($_POST['category'] == 'Psychopathic')) { echo ' selected=selected'; }  else if ((isset($category))&&($category == 'Psychopathic')) { echo ' selected=selected'; } ?>>Psychopathic</option>

Does anybody know what i am doing wrong and how i can fix this problem? I understand what i need to do but i cant seem to get the code to do what i want to do and thats to retain values of a drop down menu when the page is returned with an error.

Any help would be much appreciated.

AJay

I have several drop down boxes. When submit is clicked it execute query on another page. Problem is when submit is clicked selected values are return to defaults. Is it possible for the values to be remembered after submitting them? Part of the drop down code:
Code: [Select]
  <?php
            $result = mysql_query("SELECT *
 FROM product
GROUP BY Rim");
            ?>
           
  <select id="Rim" name="Rim"  selected="selected" style="width:158px;position:static;z-index:-1;">
  <option value="rim" selected="selected"><?php echo "Rim"; ?></option>

 <?php
 
 while ($row = mysql_fetch_array($result)) {
  echo "<option>" . $row['Rim']  . "</option>";
  echo "<br />";
  } ?>
 
       </select>  

Basically,

I have a database table called 'users' and I would like to populate a drop down box with these values of 'users'.

How?? - to call upon the values is this: 'upduser2'

Right now, all I am using is a text box, in where you have to type in the users name manually (this is so an admin can change variables and settings according to that current user). This is what I am using so far:


<h3>Update User Level</h3>
                                                    <? echo $form->error("upduser"); ?>
                                                    <table>
                                                      <form action="adminprocess.php" method="post">
                                                        <tr>
                                                          <td> Username:<br />
                    <input type="text" name="upduser" maxlength="30" value="<? echo $form->value("upduser"); ?>" /></td>
                                                          <td> Level:<br />
                                                          <select name="updlevel">
                                                              <option value="1">1 </option> //example settings
                                                              <option value="9">9 </option>
                                                            </select></td>
                                                          <td><br />
                                                            <input type="hidden" name="subupdlevel" value="1" />
                                                            <input type="submit" value="Update Level" /></td>
                                                        </tr>
                                                      </form>
                                                    </table></td>
                                                </tr>


Much help would be appreciated.

Hi guys,
Im currently doing a project for college and i have experienced some problems with saving my php drop down values to my database. im am getting no error but my values are not being stored to my db.If anyone could help me or point me in the correct direction then id be grateful. 
Ive attached my code.

I'm simply trying to set up a form where, if when a user clicks 'Submit', and then 'Back', the values from the form are preserved. My question is, how do I preserve the values of drop down menus. The following is a snippet of my code:

Code: [Select]
<select name="dropdown_dept" id="dept_list">
<option value=0><?php echo "Please select one..."?></option>
 <?php
    $dropdown_dept = "select dept_name from departments";
    $result_dept = $db_conn->query($dropdown_dept);
      
    if (!$result_dept)
    {
      echo '<p>Unable to get department data.</p>';
      return false;
    }
  
    for($i=0; $i<$result_dept->num_rows; $i++)
    { 
      $app_name_row = $result_dept -> fetch_array();
      ?>
        <option><?php echo($app_name_row[0]); ?></option>
      <? }

      ?>
      </select>


Above is where I have set up a drop down menu of departments. Given that code, how can I preserve the department name after a user clicks 'Submit'?

GREAT this forum - JUST GREAT !...

Issue: All data entered into my online form was lost (blanked out) and the form returned correctly with message "wrong verification code", when submitted with the wrong verification code.  However, going through this great forum I managed to get all - manually entered - data back ! I placed 
value="<?php echo $_GET['the_field_name'];?>"/  after each input field.

BUT... not so with input fields entered from drop-down menu !
How do I put a similar string for the field "Payment by" in this sample:

<tr>
<td class="table-inquire" width="47%">
<font face="Verdana" size="1" color="#000042">
Payment by:</font></td>
<td class="table-inquire" width="51%" colspan="2">
<font color="#400000" face="Verdana">
<select name="payment" size="1">
<option value="VISA">VISA</option>
<option value="MASTER">MASTER</option>
<option value="CASH">CASH</option>
<option value="T/T Banktransfer">T/T Banktransfer</option>
<option selected>Please select</option>
</select></font><font size="2" color="#400000" face="Verdana"></font></td>
</tr>

Any advise greatly appreciated. Thanks.

Hi i have this drop down list for date which contain 3 selects DAY MONTH YEAR hwo can i make so that when update form select keeps the same value has before help please

<?php
$months = array('','January','February','March','April','May','June','July','August','September','October','November','December');
echo '<select name="month_of_birth">';
for ($i=1;$i<13;++$i) {
   echo '<option value="' . sprintf("%02d",$i) . '">' . $months[$i] . '</option>';
}
echo '</select>';
echo '<select name="day_of_birth">';
for ($i=1;$i<32;++$i) {
   echo '<option value="' . sprintf("%02d",$i) . '">' . $i . '</option>';
}
echo '</select>';
echo '<select name="year_of_birth">';
$year = date("Y");
for ($i = $year;$i > $year-50;$i--) {
   $s = ($i == $year)?' selected':'';
   echo '<option value="' . $i . '" ' . $s . '>' . $i . '</option>';
}
echo '</select>';
?>

Hello all 

Ok here is the problem...
I want when a user inputs the requested data to the text fields , the script to insert those data in the prope table depending on the choise the user does by choosing one option from the drop down menu.
Below is the php code (apparently not working)


$editFormAction = $_SERVER['PHP_SELF'];
if (isset($_SERVER['QUERY_STRING'])) {
  $editFormAction .= "?" . htmlentities($_SERVER['QUERY_STRING']);
}
$site_type = $_REQUEST['category_selection'];
if ((isset($_POST["MM_insert"])) && ($_POST["MM_insert"] == "link_submission") && ($site_type = "Web_Sites")) {
  $insertSQL = sprintf("INSERT INTO partner_sites (url, url_title, anchor_text, `description`, webmaster_name, webmaster_email, category) VALUES (%s, %s, %s, %s, %s, %s, %s)",
                       GetSQLValueString($_POST['url_field'], "text"),
                       GetSQLValueString($_POST['title_field'], "text"),
                       GetSQLValueString($_POST['anchor_field'], "text"),
                       GetSQLValueString($_POST['description_field'], "text"),
                       GetSQLValueString($_POST['webmaster_nane_field'], "text"),
                       GetSQLValueString($_POST['webmaster_email_field'], "text"),
                       GetSQLValueString($_POST['category_selection'], "text"));

  mysql_select_db($database_content_conn, $content_conn);
  $Result1 = mysql_query($insertSQL, $content_conn) or die(mysql_error()); 

}

if ((isset($_POST["MM_insert"])) && ($_POST["MM_insert"] == "link_submission") && ($site_type = "Blogs")) {
  $insertSQL = sprintf("INSERT INTO partner_blogs (url, url_title, anchor_text, `description`, webmaster_name, webmaster_email, category) VALUES (%s, %s, %s, %s, %s, %s, %s)",
                       GetSQLValueString($_POST['url_field'], "text"),
                       GetSQLValueString($_POST['title_field'], "text"),
                       GetSQLValueString($_POST['anchor_field'], "text"),
                       GetSQLValueString($_POST['description_field'], "text"),
                       GetSQLValueString($_POST['webmaster_nane_field'], "text"),
                       GetSQLValueString($_POST['webmaster_email_field'], "text"),
                       GetSQLValueString($_POST['category_selection'], "text"));

  mysql_select_db($database_content_conn, $content_conn);
  $Result1 = mysql_query($insertSQL, $content_conn) or die(mysql_error());

}

if ((isset($_POST["MM_insert"])) && ($_POST["MM_insert"] == "link_submission") && ($site_type = "Directories")) {
  $insertSQL = sprintf("INSERT INTO partner_directories (url, url_title, anchor_text, `description`, webmaster_name, webmaster_email, category) VALUES (%s, %s, %s, %s, %s, %s, %s)",
                       GetSQLValueString($_POST['url_field'], "text"),
                       GetSQLValueString($_POST['title_field'], "text"),
                       GetSQLValueString($_POST['anchor_field'], "text"),
                       GetSQLValueString($_POST['description_field'], "text"),
                       GetSQLValueString($_POST['webmaster_nane_field'], "text"),
                       GetSQLValueString($_POST['webmaster_email_field'], "text"),
                       GetSQLValueString($_POST['category_selection'], "text"));

  mysql_select_db($database_content_conn, $content_conn);
  $Result1 = mysql_query($insertSQL, $content_conn) or die(mysql_error());
}

And the html form

<form action="<?php echo $editFormAction; ?>" method="POST" enctype="multipart/form-data" name="link_submission" id="link_submission">

<table width="630" border="0" align="center" cellpadding="5" cellspacing="5">
  <tr>
    <td width="76">URL:*</td>
    <td width="519"><label for="url_field"></label>
      <span id="sprytextfield1">
      <label for="url_field"></label>
      <input name="url_field" type="text" id="url_field" size="50" />
      <span class="textfieldRequiredMsg">A value is required.</span><span class="textfieldInvalidFormatMsg">Invalid format.</span></span></td>
    </tr>
  <tr>
    <td>Anchor Text:*</td>
    <td><label for="anchor_field"><span id="sprytextfield2">
      <input type="text" name="anchor_field" id="anchor_field" />
      <span class="textfieldRequiredMsg">A value is required.</span></span></label></td>
    </tr>
  <tr>
    <td>URL Title:*</td>
    <td><label for="title_field"><span id="sprytextfield3">
      <input type="text" name="title_field" id="title_field" />
      <span class="textfieldRequiredMsg">A value is required.</span></span></label></td>
    </tr>
  <tr>
    <td>Description:*</td>
    <td><span id="sprytextarea1">
    <label for="description_field"></label>
    <textarea name="description_field" id="description_field" cols="45" rows="3"></textarea>
    <span id="countsprytextarea1">&nbsp;</span><span class="textareaRequiredMsg">A value is required.</span><span class="textareaMaxCharsMsg">Exceeded maximum number of characters.</span></span></td>
    </tr>
  <tr>
    <td>Webmaster Name:*</td>
    <td><label for="textfield2"><span id="sprytextfield4">
      <input type="text" name="webmaster_nane_field" id="webmaster_nane_field" />
      <span class="textfieldRequiredMsg">A value is required.</span></span></label></td>
    </tr>
  <tr>
    <td>Webmaster E-mail:*</td>
    <td><label for="textfield3"><span id="sprytextfield5">
    <input name="webmaster_email_field" type="text" id="webmaster_email_field" size="40" />
    <span class="textfieldRequiredMsg">A value is required.</span><span class="textfieldInvalidFormatMsg">Invalid format.</span></span></label></td>
    </tr>
  <tr>
    <td>Category:*</td>
    <td><span id="spryselect1">
      <label for="category_selection"></label>
      <select name="category_selection" id="category_selection">
        <option>Select An Option</option>
        <option value="Web_Sites">Web Sites</option>
        <option value="Blogs">Blogs</option>
        <option value="Directories">Directories</option>
      </select>
<span class="selectRequiredMsg">Please select an item.</span></span></td>
  </tr>
  <tr>
    <td>&nbsp;</td>
    <td><label for="select"></label>
      <input type="submit" name="button" id="button" value="Url Submission" /></td>
    </tr>
</table>
<input type="hidden" name="MM_insert" value="link_submission" />
</form>

Im begging for your help.....

Hello,

Hoping someone can help...

I am pulling records from a db of cities and state abbrevs.  My db has many duplicate city names but i want to echo out only the distinct ones.  My query is based on a radius around a city and i think this might be what is tripping up the DISTINCT mysql query.  Here is what i have:
Code: [Select]
$teachradius = 50;
$add_under = array(" " => "_");
$query = sprintf("SELECT DISTINCT city, state_abbrev, ( 3959 * acos( cos( radians('%s') ) * cos( radians( lat ) ) * cos( radians( lng ) - radians('%s') ) + sin( radians('%s') ) * sin( radians( lat ) ) ) ) AS distance FROM cities HAVING distance < '%s' ORDER BY distance LIMIT 0 , 20",
  mysql_real_escape_string($lat_i),
  mysql_real_escape_string($lng_i),
  mysql_real_escape_string($lat_i),
  mysql_real_escape_string($teachradius));
$result = mysql_query($query);
while ($row = @mysql_fetch_assoc($result)){
echo '<li><a href="/' . strtolower($row['state_abbrev']) . '/' . strtolower(strtr($row['city'],$add_under)) . '.html">' . ucwords($row['city']) . ', ' . $row['state_abbrev'] . ' Dogs</a></li>',"\n";
}

Can you see what I might be doing incorrectly in my query so that i can echo out distinct city names / state abbrevs only?

Thanks in advance...

Hi am having a few problems solving this code with select distinct clause. None of what i tryed works.
Can anyone help please thanks.

this is just some of the query's i tryed
$query7 = yasDB_select("SELECT DISTINCT * FROM useronline WHERE id;");
$query7 = yasDB_select("SELECT DISTINCT ip FROM useronline WHERE id;");
$query7 = yasDB_select("SELECT DISTINCT ip FROM useronline WHERE ip;");
$query7 = yasDB_select("SELECT DISTINCT ip,timestamp FROM useronline WHERE id;");
$query7 = yasDB_select("SELECT DISTINCT id,ip,timestamp FROM useronline WHERE id;");
$query7 = yasDB_select("SELECT DISTINCT ip FROM useronline;");

and again but without ";"

$query7 = yasDB_select("SELECT DISTINCT * FROM useronline WHERE id");
$query7 = yasDB_select("SELECT DISTINCT ip FROM useronline WHERE id");
$query7 = yasDB_select("SELECT DISTINCT ip FROM useronline WHERE ip");
$query7 = yasDB_select("SELECT DISTINCT ip,timestamp FROM useronline WHERE id");
$query7 = yasDB_select("SELECT DISTINCT id,ip,timestamp FROM useronline WHERE id");
$query7 = yasDB_select("SELECT DISTINCT ip FROM useronline");


this is the code am working on.
Code: [Select]
$query7 = yasDB_select("SELECT DISTINCT * FROM useronline WHERE id;");
$visitors_online = $query7->fetch_array(MYSQLI_ASSOC);
$visitors_online = $query7->num_rows;
$query7->close();

visitors online : <?php echo $visitors_online;?><br/>

How can I stop duplication in the below code? Where do I implement the DISTINCT function?


$sql="SELECT * FROM ((resource l inner join resource_skill ln on l.Resource_ID = ln.Resource_ID) inner join skill n on ln.Skill_ID = n.Skill_ID) WHERE First_Name LIKE '%" . $name .  "%' OR Last_Name LIKE '%" . $name ."%' OR Skill_Name LIKE '%" . $name ."%'";
   
//-run  the query against the mysql query function
     $result=mysql_query($sql);
   
//-create  while loop and loop through result set
     while($row=mysql_fetch_array($result)){
             $First_Name  =$row['First_Name'];
             $Last_Name=$row['Last_Name'];
             $Resource_ID=$row['Resource_ID'];
   
//-display the result of the array
     echo "<ul>\n";
     echo "<li>" . "<a  href=\"a.php?id=$Resource_ID\">"   .$First_Name . " " . $Last_Name .  "</a></li>\n";
     echo "</ul>";
     }
     }

This topic has been moved to MySQL Help.

http://www.phpfreaks.com/forums/index.php?topic=343149.0

Hello.

I am trying to display only one instance of records that have the same memberid in my db. I am using the following statement but it continues to show all of the records that have the same memberid. Any ideas what I may be doing wrong?

Code: [Select]
$sql = "select DISTINCT memberid, event, category, date, enddate, locality, location, address, city, state, zip, contact, phone, notes, doc1, doc2, doc3, doc4, doc5 from event where date >= '$datenow' ORDER by date ASC";
Thanks for any help!

I am getting a little frustrated, and I need some help. I've working on this all day (I am new... otherwise I'd fly through it). This is a simple problem I just can't seem to put a ' in the right spot or something...

Here's the gist. In file myphotos.php is the following:
<?PHP
include('functions.php');
$link
$title = 'Light Graspers Certification';
$linktent = '<div class="titler">Photo Review</div><div class="contentm">'.$data[index].'</div>';
$excess = '<div class="titlerex">Testimonials</div><div class="contentmex">'.$displayoptions.'</div>';
echo getphoto($title, $linktent, $excess);
?>

Now my goal is under $displayoptions the following will happen.

SELECT DISTINCT name FROM homework WHERE uid='$loggedin' ORDER BY id";
With proper code I should see a distinct list of all the "name" rows. The user clicks on it to be directed to just those rows... etc...

Here is was some old code I tried manipulating to get the same result... but I couldn't seem to get it to work. Of course I only need one column in my new code.
Any help would be a appreciated.

Adam
$columns = 5;
//change the query to get another field from the database
$query = "SELECT DISTINCT city,state FROM church WHERE state='$state' ORDER BY city";
$result = mysql_query($query);
$num_rows = mysql_num_rows($result);
$rows = ceil($num_rows / $columns);
while($row = mysql_fetch_array($result)) {
$data[] = $row['city'];
//store the other field into an array

}
for($i = 0; $i < $rows; $i++) {
{
echo "<tr valign=bottom>";
echo "<td bgcolor=#2172A1 colspan=10><img src=img/blank.gif width=1 height=1></td>";
echo "</tr>";
}
echo "<TR valign=center>";
for($j = 0; $j < $columns; $j++) {
if(isset($data[$i + ($j * $rows)])) {
echo "<td class=tabval><b><a href='findnew.php?state=".$state."&city=" . $data[$i + ($j * $rows)] . "'>" . $data[$i + ($j * $rows)] . "</a></b></td>";
}
}
echo "</TR>";
}
echo "<tr valign=bottom>";
    echo "<td bgcolor=#2172A1 colspan=10><img src=img/blank.gif width=1 height=8></td>";
    echo "</tr>";

echo "$state represents $num_rows3 \n of our $num_rows2 churches\n that we serve. ";

}