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The following code contain some errors which im unable to locate. The problem is.....Im doing search and update. Im able to search the database and update Animal_Type and Animal_Breed BUT cannot update Premium_Habitat and Age......


Can some one locate that fatal error please

Thanks

<?php
	include("svrconfig.php");	
	$svr_connect = mysql_connect("localhost","root","magicpass");
	
	if(!$svr_connect)
	{
		die("Could not connect to server: ". mysql_error());
	}
	else
	{
		echo "Connected to server...<br/>";
		$db_connect = mysql_select_db("test_update", $svr_connect);
		
		if(!$db_connect)
		{
			die("Could not locate table: ". mysql_error());
		}
		else
		{
			echo "Connected to table: \"$table\"<br/><br/>";
		}
	}
	
	if(!isset($pet_code))
	{
		echo "Opps! No match found!<br/>";
	}
	
	$pet_code = $_POST['Pet_Code'];
	
	if(isset($pet_code) && $pet_code != '')
	{
		echo "You are UPDATING $pet_code.<br/>";
	}
	else
	{
		echo"No PET CODE entered...<br/>";
		die("<a href = \"pet_search.php\"> <strong>GO BACK</strong>");
	}
	
	//write search query
		
	$search = "SELECT * FROM $table WHERE Pet_Code = '$pet_code'";
	$result = mysql_query($search);
	$num = mysql_num_rows($result);
	
	mysql_close($svr_connect);
//--------------------------------------------------------------------------------------------------------------------------------------------//
	$i = 0;
	
	while($i < $num)
	{
		$pet_code = mysql_result($result,$i,"Pet_Code");
		$a_type = mysql_result($result,$i,"Animal_Type");
		$a_breed = mysql_result($result,$i,"Animal_Breed");
		$p_habitat = mysql_result($result,$i,"Premium_Habitat");
		$a_age = mysql_result($result,$i,"Age");
		
		echo "<form action = 'update.php' method = 'post'>
		<h3> $pet_code </h3>
		<input type='hidden' name='Pet_Code' value='$pet_code'>
		Animal Type: <br/> <input type = 'text' maxlength = '20' name = 'ud_a_type' value = '$a_type'><br/><br/>
		Animal Breed: <br/><input type = 'text' maxlength = '30' name = 'ud_a_breed' value = '$a_breed'><br/><br/>
		Premium Habitat: <br/><input type = 'text' maxlength = '30' name = 'ud_p_habitat' value = '$p_habitat'><br/><br/>
		Animal Age <input type = 'text' name = 'ud_age' value = '$a_age' >
		<input type = 'submit' value = 'UPDATE PETS'></forms>";
		$i++;
	}
	
	
	
	@$ud_a_type = $_POST['ud_a_type'];	
	@$ud_a_breed = $_POST['ud_a_breed'];
	@$ud_p_habitat = $_POST['ud_b_habitat'];
	@$ud_age = $_POST['ud_age'];
	
	include("svrconfig.php");	
	$svr_connect = mysql_connect("localhost","root","magicpass");// I assigned my password
	$db_connect = mysql_select_db("test_update", $svr_connect);
	$update_query = "UPDATE pets SET Animal_Type = '$ud_a_type', Animal_Breed = '$ud_a_breed', Premium_Habitat = '$ud_p_habitat', Age = 'ud_age' WHERE Pet_Code = '$pet_code'";
	mysql_query($update_query);
	mysql_close($svr_connect);
?>

Edited by peter_budo: Do not use CODE SNIPPED to post your question! Use DISCUSSION THREAD

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Contributors
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5 Years
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Last Post by bossie09
0

Line 80:

Age = 'ud_age' WHERE Pet_Code =

^Forgot the $ before ud_age


Line 74:

@$ud_p_habitat = $_POST['ud_b_habitat'];

You misspelled the key. It's 'ud_p_habitat'

Also, why have a @ when those variables need to be set correctly?

Edited by makman99: n/a

0

Im getting errors, so i have to suppress them.... Moreover its doing its job

Thanks

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