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June 3rd, 2005, 02:06 AM

Kiwi_Dunc

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Join Date: Jun 2005

Location: Sydney Australia

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Help with PHP script

Hi

I am new to PHP and what follows is one of my first scripts.

Everything here works except I can not post to the next php form the value selected in the list option. Any ideas explained in newbie language, would be greatly appreciated.

for obvious reasons I have removed my real db connections details.

Regards Duncan

PHP Code:


		
			
<?php
 ini_set ('display_errors', 1); // this forces any errors to be displayed
 // is a helpful prompt while I am learning
 
 $db="db_name";
 $link = mysql_connect("host","user","pwd");
 if (! $link)
 die("Couldn't connect to MySQL");
 mysql_select_db($db , $link)
 or die("Couldn't open $db: ".mysql_error());
 $result = mysql_query( "SELECT FISJobNumber, JobTitle, StaffName FROM tbljobs INNER JOIN tbllookupstaffname ON JobAssignedTo = record" )
 or die("SELECT Error: ".mysql_error());
 $num_rows = mysql_num_rows($result);
 
 print "<h1>There are $num_rows active Jobs.";
 
 ?>
 <FORM NAME="list" ACTION="ViewDetails.php" METHOD="post">
 <INPUT TYPE=SUBMIT NAME="list" Value="View Job Details" >
 <SELECT NAME="list">
 
   <?php    
 while ($row = mysql_fetch_array($result))
 { 
 
 echo "\n<OPTION VALUE='".$row['FISJobNumber']."' >{$row['FISJobNumber']}  {$row['JobTitle']}";
 }
 
 mysql_close($link);
 ?>

June 4th, 2005, 02:56 PM

edwinbrains

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Location: UK

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Well you don't have </select> or </form> tags, and there's no submit button. Do you have these in the real form or not?

June 5th, 2005, 07:07 PM

Kiwi_Dunc

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Location: Sydney Australia

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Tags

The form and select tags are in both this example and the code I am using.

You can see the start for both tags in my example in the line that follows

PHP Code:


		
			
print <h1>There are $num_rows active Jobs.";

I just did not copy enough code for you to see the end of the tags.

Here is that sections I left out the top to save repartition.

Regards Duncan

PHP Code:


		
			
 ?>
 <FORM NAME="list" ACTION="" METHOD="GET">
 <INPUT TYPE="button" NAME="button" Value="View Job" onClick="testSelect(this.form)">
 <SELECT NAME="list" id={$row['FISJobNumber']}>
 
   <?php    
 while ($row = mysql_fetch_array($result))
 { 
 
 echo "\n<OPTION value='".$row['FISJobNumber']."'>{$row['FISJobNumber']}  {$row['JobTitle']}";
 }
 
 mysql_close($link);
 ?>
       
 
 </SELECT>
 </FORM>

June 6th, 2005, 08:05 AM

MadCowDzz

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Is there a reason you changed that button from submit, to a regular button?

Maybe your Javascript is causing an error.

The ID in your <SELECT> is likely set to null...
If you check your rendered HTML you're likely to see:
<SELECT NAME="list" id=>

June 7th, 2005, 12:55 AM

Kiwi_Dunc

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Join Date: Jun 2005

Location: Sydney Australia

Posts: 3

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Sorry I posted an older version of the code by mistake that had a JavaScript thrown in, it was a previous attempt to solve my problem using Java.

I will repost the two pages completely

The first page called fhq14.php runs a predefined query and displays the results as a list.

Code:

<html>
 Display all Jobs<head>
 <link href="fhp1.css" rel="stylesheet" type="text/css">
 <link href="menustyles.css" rel="stylesheet" type="text/css">
 <SCRIPT >
 
 </SCRIPT>
 <style type="text/css">
 	<!--
 	.testcell	{padding:1px;padding-right:20px;font-family:Courier New;}
 	.testtable	{width:100%;}
 </style>
 <body>
 <div id="mainwrap">
 <div id="header">
 	<!--  Comment feild: This is where the Logo sits -->
 	<div id="logo">
 	<img src="logo[1].jpg" width="287" height="72">  
 	</div>
 </div>
 
 <div id="nav_bar">
 	
 	<!--  here is the set of drop down menus -->
 	<!-- 
 
 <div id="menus">
 			<ul>
 				<li>Close Window
 				</ul>
 			</div>
 </div>
 <!-- 
 
 
 <!--  Comment feild: This is Main Page sits -->
 <div id="mainpage">
 	
 		<div id="maincontentblock">
 			
 			<?php
 ini_set ('display_errors', 1); // this forces any errors to be displayed
 // is a helpful prompt while I am learning
 
 $db="db_name";
 $link = mysql_connect("host","user","pwd");
 if (! $link)
 die("Couldn't connect to MySQL");
 mysql_select_db($db , $link)
 or die("Couldn't open $db: ".mysql_error());
 $result = mysql_query( "SELECT FISJobNumber, JobTitle, StaffName FROM tbljobs INNER JOIN tbllookupstaffname ON JobAssignedTo = record" )
 or die("SELECT Error: ".mysql_error());
 $num_rows = mysql_num_rows($result);
 
 print "<h1>There are $num_rows active Jobs.";
 
 ?>
 <FORM NAME="list" ACTION="ViewDetails.php" METHOD="post">
 <INPUT TYPE=SUBMIT NAME="list" Value="View Job Details" >
 <SELECT NAME="list">
 
   <?php	
 while ($row = mysql_fetch_array($result))
 { 
 
 echo "\n<OPTION VALUE='".$row['FISJobNumber']."' >{$row['FISJobNumber']}  {$row['JobTitle']}";
 }
 
 mysql_close($link);
 ?>
    
 
 </SELECT>
 </FORM>
 
   </div> 
 
 
 	</div>
 	
 </div>
 
 
 </div>
 
 </body>
 </html>
 </body>
 
 </html>

The page it calls is "ViewDetails.php and its code is here.

Code:

<html>
 View Job Details<head>
 <link href="fhp1.css" rel="stylesheet" type="text/css">
 <link href="menustyles.css" rel="stylesheet" type="text/css">
 <style type="text/css">
 </style>
 <body>
 <div id="mainwrap">
 <div id="header">
 	<!--  Comment feild: This is where the Logo sits -->
 	<div id="logo">
 	<img src="logo[1].jpg" width="287" height="72">  
 	</div>
 </div>
 
 <div id="nav_bar">
 	
 	<!--  here is the set of drop down menus -->
 	<!-- 
 
 <div id="menus">
 			<ul>
 				<li>Close Window
 				</ul>
 			</div>
 </div>
 <!-- 
 
 
 <!--  Comment feild: This is Main Page sits -->
 <div id="mainpage">
 	
 		<div id="maincontentblock">
     	<!--  Comment feild: This is where the content for the Main Info window is located -->
 	  
 <?php
 
 
 ini_set ('display_errors', 1); // this forces any errors to be displayed
 // is a helpful prompt while I am learning
 
 $db="db_name";
 $link = mysql_connect("host","user","pwd");
 if (! $link)
 die("Couldn't connect to MySQL");
 mysql_select_db($db , $link)
 or die("Couldn't open $db: ".mysql_error());
 !isset ($_POST['list'])) {
 echo $id;
 $result = mysql_query( "SELECT FISJobNumber, JobTitle, DateCommenced, Name, Notes FROM tbljobs INNER JOIN tblcompnaydetails ON companyid = CompanyID WHERE FISJobNumber = $list";
 or die("SELECT Error: ".mysql_error());
 while ($row = mysql_fetch_array($result))
 { 
 print "<h1>FIS Number -- {$row['FISJobNumber']}-- Job Title --{$row['JobTitle']}.";
 print "<p>Date Job commenced --{$row['DateCommenced']}
 
  
 ;
 print "<p>The Client is  --{$row['Name']}
 
  
 ;
 print "<p>Scope of the job
 
  
 ;
 print "<p>{$row['Notes']}
 
  
 ;
 print <p>Edit Details
 
 \n";
 }
 }
 mysql_close($link);
 ?>
 	</div>
 	
 </div>
 
 
 </div>
 </div>
 
 </body>
 </html>
 </body>
 
 </html>

Sorry about the confusion I hope this is not too much code to post.

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