table view to detail view

Question

Motifaithed 3 Newbie Poster

12 Years Ago

Hi everyone,

I'm new to php and I'm working on a project, it's very typical that when you type in a search box and press the submit button, it will link you to a table with minimal content, but upon clicking a list on the table, it will link you again to the full detail information of your database,

here is my form code

]<html>
<form action="fetchDBAssets.php" method="POST">
search:<input type="text" name="assetSearch" />
<input type="submit" />
</form>
</html>

and here is my first form to show the brief table

<?php session_start();
error_reporting (E_ALL ^ E_NOTICE);



$con=mysql_connect("localhost","root","");
    if(!$con)
        {
            die('connection error' . mysql_error());
        }
    mysql_select_db("pcassets_db",$con);

$result=mysql_query("SELECT * FROM data WHERE (userFirstName='$_POST[assetSearch]' || pcName='$_POST[assetSearch]')");



    echo "<table border='1' name='table1'>
<tr>
<th>Firstname</th>
<th>Lastname</th>
</tr>";


    while ( $row = mysql_fetch_array($result) )
       {
        echo "<tr>";
        echo "<td>" . "<a href='fetchDBAssetsWhole.php'>" . $row['userFirstName'] . "</td>" . "<td>" . $row['userLastName'] . "</td>";
         echo "<br/>";
       }
echo "</table>";
mysql_close($con);

What I want is when i click to firstname, it will link again to the full detail of my database which is firstname, lastname, department, position and so on. Can somebody help me with this? Hope to read your reply soon.

Thank you

php

Edited 11 Years Ago by Nick Evan because: Fixed formatting

2 Contributors
11 Replies
334 Views
2 Days Discussion Span
Latest Post 12 Years Ago Latest Post by rpv_sen

rpv_sen 59 Junior Poster

12 Years Ago

Hi
if you are having id in your table

<a href='fetchDBAssetsWhole.php'>

Replace the above URL as below

<a href='fetchDBAssetsWhole.php?id=id='".$row['id']."''>

rpv_sen 59 Junior Poster

12 Years Ago

using GET

$id = $_GET;

select * from tablename where id='$id'

Like this

rpv_sen 59 Junior Poster

12 Years Ago

Hi,

here is what I did on fetchDBAssetsWhole.php

<?php 

//error_reporting (E_ALL ^ E_NOTICE);




$con=mysql_connect("localhost","root","");
	if(!$con)
		{
			die('connection error' . mysql_error());
		}
	mysql_select_db("pcassets_db",$con);
	
	$ID=$_GET['id'];

	$result=mysql_query("SELECT * FROM data WHERE (pcID='$ID')");

	echo $row['userFirstName'] . " " . $row['userLastName'] . " " . $row['department'] . " " . $row['position'] ;
	
	
	/*while ( $row = mysql_fetch_array($result) )
	   {
		
		echo $row['userFirstName'] . " " . $row['userLastName'] ;
		 echo "<br/>";
	   }*/

mysql_close($con);


?>

Here is the error

Notice: Undefined variable: row in C:\wamp\www\DBASSETS\fetchDBAssetsWhole.php on line 19

I'm doing it wrong, please help me more and thank you for your support

Hi Please Use the below code to test.

<?php session_start();
error_reporting (E_ALL ^ E_NOTICE);

$con=mysql_connect("localhost","root","");
if(!$con)
{
die('connection error' . mysql_error());
}
mysql_select_db("pcassets_db",$con);

$result=mysql_query("SELECT * FROM data WHERE (userFirstName='$_POST[assetSearch]' || pcName='$_POST[assetSearch]')");



echo "<table border='1' name='table1'>
<tr>
<th>Firstname</th>
<th>Lastname</th>
</tr>";


while ( $row = mysql_fetch_array($result) )
{
echo "<tr>";
echo "<td>" . "<a href='fetchDBAssetsWhole.php?id='".$row['pcID']."''>" . $row['userFirstName'] . "</td>" . "<td>" . $row['userLastName'] . "</td>";
echo "<br/>";
}
echo "</table>";
mysql_close($con);

fetchDBAssetsWhole.php

<?php 

//error_reporting (E_ALL ^ E_NOTICE);




$con=mysql_connect("localhost","root","");
	if(!$con)
		{
			die('connection error' . mysql_error());
		}
	mysql_select_db("pcassets_db",$con);
	
	$ID=$_GET['id'];

	$result=mysql_query("SELECT * FROM data WHERE pcID='$ID'");
	$row = mysql_fetch_array($result);

	echo $row['userFirstName'] . " " . $row['userLastName'] . " " . $row['department'] . " " . $row['position'] ;
	
	
	/*while ( $row = mysql_fetch_array($result) )
	   {
		
		echo $row['userFirstName'] . " " . $row['userLastName'] ;
		 echo "<br/>";
	   }*/

mysql_close($con);


?>

Reply to this topic

Be a part of the DaniWeb community

We're a friendly, industry-focused community of developers, IT pros, digital marketers, and technology enthusiasts meeting, networking, learning, and sharing knowledge.

Motifaithed 3 Newbie Poster · Answer 1 · 2011-11-08T13:52:54+00:00

Hi,

thanks for the reply, and how is the code will be on the fetchDBAssetsWhole.php then?

Motifaithed 3 Newbie Poster · Answer 2 · 2011-11-09T08:48:31+00:00

Hi,

here is what I did on fetchDBAssetsWhole.php

<?php 

//error_reporting (E_ALL ^ E_NOTICE);




$con=mysql_connect("localhost","root","");
	if(!$con)
		{
			die('connection error' . mysql_error());
		}
	mysql_select_db("pcassets_db",$con);
	
	$ID=$_GET['id'];

	$result=mysql_query("SELECT * FROM data WHERE (pcID='$ID')");

	echo $row['userFirstName'] . " " . $row['userLastName'] . " " . $row['department'] . " " . $row['position'] ;
	
	
	/*while ( $row = mysql_fetch_array($result) )
	   {
		
		echo $row['userFirstName'] . " " . $row['userLastName'] ;
		 echo "<br/>";
	   }*/

mysql_close($con);


?>

Here is the error

Notice: Undefined variable: row in C:\wamp\www\DBASSETS\fetchDBAssetsWhole.php on line 19

I'm doing it wrong, please help me more and thank you for your support

Motifaithed 3 Newbie Poster · Answer 3 · 2011-11-09T08:59:00+00:00

Motifaithed 3 Newbie Poster

12 Years Ago

Hi again,

pcID is the ID on my table

Motifaithed 3 Newbie Poster · Answer 4 · 2011-11-09T11:09:37+00:00

Hi,

I tried the code, and here is the error now

Notice: Undefined variable: _result in C:\wamp\www\DBASSETS\fetchDBAssetsWhole.php on line 18

Warning: mysql_fetch_array() expects parameter 1 to be resource, null given in C:\wamp\www\DBASSETS\fetchDBAssetsWhole.php on line 18

rpv_sen 59 Junior Poster · Answer 5 · 2011-11-09T15:16:39+00:00

Hi sorry for the script.

I need some information, you are getting the first name by POST method and check the field(pcName) name in table

rpv_sen 59 Junior Poster · Answer 6 · 2011-11-09T15:31:56+00:00

po1.php

<html>
<form action="oo.php" method="POST">
search:<input type="text" name="assetSearch" />
<input type="submit" />
</form>
</html>

oo.php

<?php session_start();
error_reporting (E_ALL ^ E_NOTICE);

$con=mysql_connect("localhost","root","123");
if(!$con)
{
die('connection error' . mysql_error());
}
mysql_select_db("test",$con);
?>
<table border='1' name='table1'>
<tr>
<th>Firstname</th>
<th>Lastname</th>
</tr>
<?php
$result="SELECT * FROM data WHERE (userFirstName='$_POST[assetSearch]' || pcName='$_POST[assetSearch]')";
$q = mysql_query($result,$con);
while( $row = mysql_fetch_array($q))
{
?>
<tr>
<td><a href="fetchDBAssetsWhole.php?id=<?php echo $row['pcID'];?>"><?php echo $row['userFirstName']; ?></a></td>
<td><?php echo $row['userLastName']; ?></td>
</tr>
<?php } mysql_close($con); ?>
</table>

fetchDBAssetsWhole

<?php 

//error_reporting (E_ALL ^ E_NOTICE);




$con=mysql_connect("localhost","root","123");
	if(!$con)
		{
			die('connection error' . mysql_error());
		}
	mysql_select_db("test",$con);
	
	$ID=$_GET['id'];

	$result=mysql_query("SELECT * FROM data WHERE pcID='$ID'");
	$row = mysql_fetch_array($result);

	echo $row['userFirstName'] . " " . $row['userLastName'] . " " . $row['dob'] . " " . $row['birthplace'] ;
	
	
	/*while ( $row = mysql_fetch_array($result) )
	   {
		
		echo $row['userFirstName'] . " " . $row['userLastName'] ;
		 echo "<br/>";
	   }*/

mysql_close($con);


?>

Note: check pcName field is the table

Motifaithed 3 Newbie Poster · Answer 7 · 2011-11-10T12:40:32+00:00

Thanks Man, all is working now, you're a genius, many thanks!!!

rpv_sen 59 Junior Poster · Answer 8 · 2011-11-10T12:55:28+00:00

Thanks Man, all is working now, you're a genius, many thanks!!!

Hi Don't use you're a genius for me Man