PHP- Fetch from database and store in drop down menu html

Question

I can't seem to get the following code to make a dropdown menu that contains data from a mysql database. The "include('connect.php');" connects to the mysql database and I know it works on separate pages. Any suggestions?

Below is the entire code. listCustomer

 <BODY>
 <H1>Find Customer's Albums Page</H1>
 From a dropdown list of customers, a user should be able to pick a customer and see a list of     albums (all fields in the CD table) purchased by that customer.
 <HR>
 <FORM ACTION="listCustomer.php" METHOD="POST"/>
 Customer:
 <select name="mydropdownCust">
 <option value="101">101</option>
 <option value="102">102</option>
 <option value="103">103</option>
 <option value="104">104</option>
 <option value="105">105</option>
 <option value="106">106</option>
 <option value="107">107</option>
 <option value="108">108</option>
 <option value="109">109</option>
 <option value="110">110</option>
 </select>
 <BR>

 <?php
 include('connect.php');

 $query = "SELECT Cnum, CName FROM Customer";
 $result = mysql_query ($query);
 echo "<select name=dropdown value=''>Dropdown</option>";
 while($r = mysql_fetch_array($result))
 {
 echo "<option value=$r["Cnum"]>$r["CName"]</option>"; 
 }
 echo "</select>";
 ?>

 <BR>
 <INPUT TYPE="SUBMIT" Value="Submit"/>
 </FORM>

 <FORM ACTION="listMenu.html" METHOD="POST"/>
 <INPUT TYPE="SUBMIT" Value="Main Menu"/>
 </FORM>
 </BODY>
 </HTML>

could you share what are you getting as a result? Furthermore, I think this Dropdown after the end of the — ilbesculpi, May 22 '13 at 21:17
It displays this "Dropdown"; while($r = mysql_fetch_array($result)){ echo ""; } echo ""; ?>" on the website. — Sean Cleveland, May 22 '13 at 21:23
I don't think the variable syntax is your issue (just tested the echo locally with the way you've constructed your echo, try a var_dump on $r inside your loop — Scuzzy, May 22 '13 at 21:28
`echo "";` prematurely terminates the echoed string. Is there an issue with `echo " — stslavik, May 22 '13 at 21:34
This is what displays: http://imageshack.us/photo/my-images/163/24256538.jpg/ — Sean Cleveland, May 22 '13 at 21:35
1. Use a good editor like Netbeans/Eclipse. 2. A best practice of writing string with variable is using `sprintf` — Thanh Trung, May 22 '13 at 22:36
When in doubt, start commenting out portions and try to run it. I commented out all but the `while` loop, found it worked, and then ran the `while` making changes as things broke. Usually line-by-line on small things is quick and easy. — stslavik, May 22 '13 at 23:11

score 5 · Answer 1 · edited Jun 20 '20 at 09:12

<?php
include('connect.php');

$query = "SELECT Cnum, CName FROM Customer";
$result = mysql_query ($query);
echo "<select name='dropdown' value=''><option>Dropdown</option>";
while($r = mysql_fetch_array($result)) {
  echo "<option value=".$r['Cnum'].">".$r['CName']."</option>"; 
}
echo "</select>";
?>

From the looks of things, you're missing an opening option tag, so it's just outputting "Dropdown" as a line of text.

Edit

Just to be completely transparent, because I did not have connect.php, I had to add my own DB connections. My whole page looked thusly:

<?
//Adding to display errors.
error_reporting(E_ALL);
ini_set('display_errors', '1');
?>
<HTML>
<HEAD>
</HEAD>
<BODY>
 <H1>Find Customer's Albums Page</H1>
 From a dropdown list of customers, a user should be able to pick a customer and see a list of albums (all fields in the CD table) purchased by that customer.
 <HR>
 <FORM ACTION="listCustomer.php" METHOD="POST"/>
 Customer:
 <select name="mydropdownCust">
 <option value="101">101</option>
 <option value="102">102</option>
 <option value="103">103</option>
 <option value="104">104</option>
 <option value="105">105</option>
 <option value="106">106</option>
 <option value="107">107</option>
 <option value="108">108</option>
 <option value="109">109</option>
 <option value="110">110</option>
 </select>
 <BR />
 <?php
  // BEGIN ADDED CONNECTION HACKY GARBAGE
  $con=mysql_connect("localhost","root","root");
  // Check connection
  if (mysqli_connect_errno($con)) {
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }
  $selected = mysql_select_db("sample",$con) 
    or die("Could not select examples");
  // END ADDED CONNECTION HACKY GARBAGE

  $query = "SELECT Cnum, CName FROM Customer";
  $result = mysql_query ($query);
  echo "<select name='dropdown' value=''><option>Dropdown</option>";
  while($r = mysql_fetch_array($result)) {
    echo "<option value=".$r['Cnum'].">".$r['CName']."</option>"; 
  }
  echo "</select>";
 ?>

 <BR />
 <INPUT TYPE="SUBMIT" Value="Submit"/>
 </FORM>

<FORM ACTION="listMenu.html" METHOD="POST"/>
<INPUT TYPE="SUBMIT" Value="Main Menu"/>
</FORM>
</BODY>
</HTML>

Also missing quotes around dropbox in the name attribute. For consistency at least. — Frederik Wordenskjold, May 22 '13 at 21:45
I can't make it work. Here is what it looks like. http://imageshack.us/photo/my-images/441/10517841.jpg/ — Sean Cleveland, May 22 '13 at 21:47
... Really? I created a 2 row table, with 2 columns Cnum (an int) and CName (a varchar 20 chars long), and the above (with a couple changes since I don't have your connect.php) pulled the info into a drop down. You uploaded this as printed? And you cleared your cache? — stslavik, May 22 '13 at 22:50
could it be that I copied a lot of the code instead of typed it out? — Sean Cleveland, May 22 '13 at 22:54

score 4 · Answer 2 · edited May 23 '17 at 11:50

First off, you are missing an option opening tag, as correctly mentioned by stslavik. But this is not causing the issue here as it seems (it's auto-corrected by the browser - in my tests atleast).

Secondly, this wont work (problem causer):

echo "<option value=$r["Cnum"]>$r["CName"]</option>";

You should use

echo "<option value=".$r["Cnum"].">".$r["CName"]."</option>";

or, as I always prefer single quotes to enclose echo or print output strings:

echo '<option value='.$r['Cnum'].'>'.$r['CName'].'</option>';

Third alternative (complex syntax: What does ${ } mean in PHP syntax?)

echo "<option value={$r["Cnum"]}>{$r["CName"]}</option>";

Leandro Villagran · Answer 3 · 2013-05-22T21:45:13.447

1

assuming you get data from the database try this

echo "<option value={$r['Cnum']}>{$r['CName']}</option>";

edited May 22 '13 at 21:45

answered May 22 '13 at 21:36

Leandro Villagran

300
1
8

What are you trying to do here? First line is identical with the same line in the question - it ends the string prematurely. – Frederik Wordenskjold May 22 '13 at 21:40
sorry the code got messed up what is what i was trying to say is that you need to use the {} for complex expressions – Leandro Villagran May 22 '13 at 21:47
Why would you like those parenthesis here!? That will just wrap the variable in {} strings. – Frederik Wordenskjold May 22 '13 at 21:51
@FrederikWordenskjold do you know php don't you see he is using double quotes so instead of using single quotes `echo " – Leandro Villagran May 22 '13 at 21:59
what do you mean it doesnt work what part doesnt work the code i gave you is perfectly ok you didn't even ask the question you ass @FrederikWordenskjold – Leandro Villagran May 22 '13 at 22:20
let us [continue this discussion in chat](http://chat.stackoverflow.com/rooms/30442/discussion-between-frederik-wordenskjold-and-leandro-villagran) – Frederik Wordenskjold May 22 '13 at 22:21

RonEskinder · Answer 4 · 2013-05-22T21:40:43.210

-1

try,

echo "<option value=' . $r['Cnum'] . '>' . $r['CName'] . '</option>";

instead of

echo "<option value=$r[Cnum]>$r[CName]</option>";

edited May 22 '13 at 21:40

answered May 22 '13 at 21:19

RonEskinder

527
8
24

Parse error: syntax error, unexpected '"', expecting T_STRING or T_VARIABLE or T_NUM_STRING in... – Scuzzy May 22 '13 at 21:30

PHP- Fetch from database and store in drop down menu html

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