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1 Year Ago
0

PHP won't execute in html form

I am running a form that pulls data from multiple mysql tables. However whenever the php/mysql code is in the html form tags it will not read the php/mysql nor will it read any php for exampl echo "here";. I have done this several times before so I imagine it is just an issue with a config file. Any thoughts are much appreciated! Code below.

<td>Education</td>
<td><select name="education">
		<option value="<?echo $education;?>"><?echo $edu_desc;?></option>
		<? echo "here";
		$edusql = ("SELECT unique_id, desc FROM education order by unique_id asc");
		   echo $edusql;
		   $edurslt = mysql_query($edusql) or die("Error: ".mysql_errno().":- ".mysql_error());
while($erow = mysql_fetch_array($edurslt))	
		{
		$unique_id = $erow['0'];
		$desc = $erow['1'];
		?>
		<option value="<?echo $unique_id;?>"><?echo $desc;?></option>
	  <?}
		?>
	</select>
camerondane
Newbie Poster
6 posts since Oct 2010
Reputation Points: 10
Solved Threads: 0
 
1 Year Ago
0

Hi there,

Change your php start tags from <? to <?php as code below.

That should solve the problem.

Please remember to mark the thread as solved if it worked.

Regards,

Max

<td>Education</td><td>

<select name="education">		
<option value="<?php echo $education;?>"><?php echo $edu_desc;?></option>
<?php echo "here";		$edusql = ("SELECT unique_id, desc FROM education order by unique_id asc");
		  echo $edusql;
		  $edurslt = mysql_query($edusql) or die("Error: ".mysql_errno().":- ".mysql_error());
while($erow = mysql_fetch_array($edurslt))			
                           {
			$unique_id = $erow['0'];
			$desc = $erow['1'];
?>		
<option value="<?php echo $unique_id;?>"><?php echo $desc;?></option>	  
<?php 
}		
?>	
</select>
trilithon
Light Poster
38 posts since Jun 2010
Reputation Points: 10
Solved Threads: 4
 
1 Year Ago
0

Thanks Max! Much appreciated, unfortunately that did not solve it. I should mention that it just stops the page entirely, so the rest of the html form does not output as well. Any other thoughts? Thanks!

camerondane
Newbie Poster
6 posts since Oct 2010
Reputation Points: 10
Solved Threads: 0
 
1 Year Ago
0

One other detail it seems only to not execute under a , I am creating a dynamic drop down where the options pull from mysql table. Thanks

camerondane
Newbie Poster
6 posts since Oct 2010
Reputation Points: 10
Solved Threads: 0
 
1 Year Ago
0

I tested it with the below code and it displays the following:

Education "and the dropdown box next to it"

Regards,

Max

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>

<body>

<table>
<tr>
<td>Education</td><td>

<select name="education">		
<option value="<?php echo $education;?>"><?php echo $edu_desc;?></option>
<?php echo "here";		$edusql = ("SELECT unique_id, desc FROM education order by unique_id asc");
		  echo $edusql;
		  $edurslt = mysql_query($edusql) or die("Error: ".mysql_errno().":- ".mysql_error());
			while($erow = mysql_fetch_array($edurslt))			{
			$unique_id = $erow['0'];
			$desc = $erow['1'];
											?>		
<option value="<?php echo $unique_id;?>"><?php echo $desc;?></option>	  
<?php 
}		
?>	
</select>

</td>
</tr>
</table>

</body>
</html>

If it still does not work, please post more code, I think the problem lies in the HTML code surrounding the PHP

trilithon
Light Poster
38 posts since Jun 2010
Reputation Points: 10
Solved Threads: 4
 
1 Year Ago
0

Here is the entire page below. And the link to the beta page: http://www.opcionesinternacionales.com/fasty75/Userport/admin/ap_app_detail.php
Much appreciated!

<?php
require ("db_con.php");

$select = ("SELECT ap_info.ap_id, ap_info.f_name, ap_info.l_name, ap_info.dob, ap_info.email, ap_info.skype, ap_info.home_phone, ap_info.cell_phone, ap_info.home_add, ap_info.education, education.desc, ap_info.marital_status, marital_status.desc, nationality, start_date, end_date, driving, cycling FROM ap_info join education on education.unique_id = ap_info.education join marital_status on marital_status.marital_id = ap_info.marital_status where ap_info.ap_id = '2'");
echo $select;

$result = mysql_query($select) or die("Error: ".mysql_errno().":- ".mysql_error());
while($lrow = mysql_fetch_array($result))
		
		{
		$ap_id = $lrow['0'];
		$f_name = $lrow['1'];
		$l_name = $lrow['2'];
		$dob = $lrow['3'];
		$email = $lrow['4'];
		$skype = $lrow['5'];
		$home_phone = $lrow['6'];
		$cell_phone = $lrow['7'];
		$home_add = $lrow['8'];
		$education = $lrow['9'];
		$edu_desc = $lrow['10'];
		$marital_status = $lrow['11'];
		$marital_desc = $lrow['12'];
		$nationality = $lrow['13'];
		$start_date = $lrow['13'];
		$end_date = $lrow['14'];
		$driving = $lrow['15'];
		$cycling = $lrow['16'];
		}
		
?>
<form action="detail_submit.php" method="post"
<table cellpadding="2" cellspacing="2" border="1">
<tr>
<td>First Name</td>
<td><input type="text" name="f_name" value="<?php echo $f_name;?>"></td>
</tr>

<tr>
<td>Last Name</td>
<td><input type="text" name="l_name" value="<?php echo $l_name;?>"></td>
</tr>

<tr>
<td>DOB</td>
<td><input type="text" name="dob" value="<?php echo $dob;?>"></td>
</tr>

<tr>
<td>Email</td>
<td><input type="text" name="email" value="<?php echo $email;?>"></td>
</tr>

<tr>
<td>Skype</td>
<td><input type="text" name="skype" value="<?php echo $skype;?>"></td>
</tr>

<tr>
<td>Home Phone</td>
<td><input type="text" name="home_phone" value="<?php echo $home_phone;?>"></td>
</tr>

<tr>
<td>Cell Phone</td>
<td><input type="text" name="cell_phone" value="<?php echo $cell_phone;?>"></td>
</tr>

<tr>
<td>Address</td>
<td><textarea name="home_add" rows="3" cols="18"><?php echo $home_add;?></textarea></td>
</tr>

<tr>
<td>Education</td>
<td><select name="education">
		<option value="<?php echo $education;?>"><?php echo $edu_desc;?></option>
		<?php echo "here";
		$edusql = ("SELECT unique_id, desc FROM education order by unique_id asc");
		   echo $edusql;
		   $edurslt = mysql_query($edusql) or die("Error: ".mysql_errno().":- ".mysql_error());
while($erow = mysql_fetch_array($edurslt))	
		{
		$unique_id = $erow['0'];
		$desc = $erow['1'];
		?>
		<option value="<?php echo $unique_id;?>"><?php echo $desc;?></option>
	  <?php 
	  }
		?>
	</select>	
</td>
</tr>

<tr>
<td>Marital</td>
<td><select name="marital">
		<option value="<?php echo $marital;?>"><?php echo $marital_desc?></option></td>
		<?php $marsql = ("SELECT marital_id, desc FROM marital_status order by marital_id asc");
		   $marrslt = mysql_query($marsql) or die("Error: ".mysql_errno().":- ".mysql_error());
while($mrow = mysql_fetch_array($marrslt))	
		{
		$marital_id = $mrow['0'];
		$mar_desc = $mrow['1'];
		?>
		<option value="<?php echo $marital_id;?>"><?php echo $mar_desc;?></option>
	  <?php 
	  }
		?>
	</select>	
</tr>

<tr>
<td>Nationality</td>
<td><input type="text" name="nationality" value="<?php echo $nationality;?>"></td>
</tr>

<tr>
<td>Start Date</td>
<td><input type="text" name="start_date" value="<?php echo $start_date;?>"></td>
</tr>

<tr>
<td>End Date</td>
<td><input type="text" name="end_date" value="<?php echo $end_date;?>"></td>
</tr>

<tr>
<td>Driving</td>
<td><input type="text" name="driving" value="<?php echo $driving;?>"></td>
</tr>


<tr>
<td>Cycling</td>
<td><input type="text" name="cycling" value="<?php echo $cycling;?>"></td>
</tr>

<tr>
<td>Passport #</td>
<td><input type="text" name="pass_num" value="<?php echo $pass_num;?>"></td>
</tr>

<tr>
<td>Passport Country</td>
<td><input type="text" name="pass_country" value="<?php echo $pass_country;?>"></td>
</tr>

<tr>
<td>Languages</td>
<td>
<?php
$sellang = ("SELECT languages_ap.lang_id, languages.language, languages_ap.speaks FROM languages_ap join languages on languages.lang_id = languages_ap.lang_id WHERE languages_ap.ap_id = '$ap_id' order by languages.lang_id ");
//echo $sellang;
$langrslt = mysql_query($sellang) or die("Error: ".mysql_errno().":- ".mysql_error());
while($lrow = mysql_fetch_array($langrslt))	
		{
		$lang_id = $lrow['0'];
		$language = $lrow['1'];
		$speaks = $lrow['2'];?>
		<input type="checkbox" name="<?php echo $language;?>" value="<?php echo $lang_id;?>" <?php if ($speaks == 'y'){echo "checked";}?>>&nbsp;<?php echo $language;?></br>
		
		<?php 
		}
?>
</td>
</tr>

<tr>
<td>First Aid</td>
<td><input type="text" name="first_aid" value="<?php echo $first_aid;?>"></td>
</tr>
<tr><td><input type="hidden" name="ap_id" value="<?php echo $ap_id;?>"></td>
<td><input type="Submit" value="Submit"></td>
</tr>
</table>
</form>
camerondane
Newbie Poster
6 posts since Oct 2010
Reputation Points: 10
Solved Threads: 0
 
1 Year Ago
0
("SELECT ap_info.ap_id, ap_info.f_name, ap_info.l_name, ap_info.dob, ap_info.email, ap_info.skype, ap_info.home_phone, ap_info.cell_phone, ap_info.home_add, ap_info.education, education.desc, ap_info.marital_status, marital_status.desc, nationality, start_date, end_date, driving, cycling FROM ap_info join education on education.unique_id = ap_info.education join marital_status on marital_status.marital_id = ap_info.marital_status where ap_info.ap_id = '2'")
("SELECT languages_ap.lang_id, languages.language, languages_ap.speaks FROM languages_ap join languages on languages.lang_id = languages_ap.lang_id WHERE languages_ap.ap_id = '$ap_id' order by languages.lang_id ");

Remove the curly brackets '()'. This may cause the SQL injection when the query run.

mysql_query($sellang)

This one will look like:

mysql_query(("SELECT languages_ap.lang_id, languages.language, languages_ap.speaks FROM languages_ap join languages on languages.lang_id = languages_ap.lang_id WHERE languages_ap.ap_id = '$ap_id' order by languages.lang_id "))

Try this one:

$select = "SELECT ap_info.ap_id, ap_info.f_name, ap_info.l_name, ap_info.dob, ap_info.email, ap_info.skype, ap_info.home_phone, ap_info.cell_phone, ap_info.home_add, ap_info.education, education.desc, ap_info.marital_status, marital_status.desc, nationality, start_date, end_date, driving, cycling FROM ap_info join education on education.unique_id = ap_info.education join marital_status on marital_status.marital_id = ap_info.marital_status where ap_info.ap_id = '2'";

Instead of

$select = ("SELECT ap_info.ap_id, ap_info.f_name, ap_info.l_name, ap_info.dob, ap_info.email, ap_info.skype, ap_info.home_phone, ap_info.cell_phone, ap_info.home_add, ap_info.education, education.desc, ap_info.marital_status, marital_status.desc, nationality, start_date, end_date, driving, cycling FROM ap_info join education on education.unique_id = ap_info.education join marital_status on marital_status.marital_id = ap_info.marital_status where ap_info.ap_id = '2'");

Hope this help ..!

Zero13
Practically a Master Poster
624 posts since Jan 2009
Reputation Points: 120
Solved Threads: 139
 
1 Year Ago
0

Z it doesn't alter the output in any way, dang it. I appreciate your help. Does anyone think it could be an Apache config file issue somewhere on the server? It is through godaddy and they do not offer technical support!

camerondane
Newbie Poster
6 posts since Oct 2010
Reputation Points: 10
Solved Threads: 0
 
1 Year Ago
0

It won't execute under a but outside of a the php will execute? It is very strange. Instead of using drop downs I will just use check boxes. Thanks for all your help!

camerondane
Newbie Poster
6 posts since Oct 2010
Reputation Points: 10
Solved Threads: 0
 

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