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Aug 18th, 2009

PHP variable in link

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I think I have read the answer reading through several forums and many posts, but I am not getting it. Please help... I am a novice.

I have a page that displays some info for all the records in a database. It's for real estate listings, so the page displays the street address, subdivision and a description of the property. This page is created dynamically from a database. This page is called allListings.php.

I would like to add a link to the street address that would take you to a detail page -- called inventory.php -- where more information about the property can be viewed.

Help?

Here is what my last try looked like:
<a href="\inventory.php?id=" . $id . "\"><?php echo $row_rsAllListings['element_1_1']; ?></a>

While inventory.php displays the information I want, it is pulling the first record in the database regardless of the property I click.

I have attached my two pages.

Attached Files

	allListings.php (6.3 KB, 17 views)
	inventory.php (6.4 KB, 20 views)

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isak

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Aug 18th, 2009

Re: PHP variable in link

<a href="\inventory.php?id=" . $id . "\"><?php echo $row_rsAllListings['element_1_1']; ?></a>
Why do you echo $row_rsAllListings['element_1_1'] but you don't echo $id? Check the link href value after the page is loaded. I'm sure you'll spot the problem.

buddylee17

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Aug 18th, 2009

Re: PHP variable in link

php Syntax (Toggle Plain Text)
<div class="content">
<table width="600">
<?php $counter = 0; // initialize counter outside loop 
do { echo '<tr';
if ($counter++ % 2) {echo 'class="hilite"';} 
echo '><td width="162"><img src="'.$row_rsAllListings['element_16'].'"></td>';
echo '<td width="426"><h2><a href="\inventory.php?id="'.$id.'">'.$row_rsAllListings['element_1_1'];
echo '</a> |  '.$row_rsAllListings['element_2'].'<br />';
echo $row_rsAllListings['element_3'].'</h2><br />';
echo $row_rsAllListings['element_4'].'<br />';
echo $row_rsAllListings['element_5'].'Beds / '.$row_rsAllListings['element_6']; 
echo $row_rsAllListings['element_7'].'Baths / '.$row_rsAllListings['element_8'].'sf per '.$row_rsAllListings['element_9'].'<br /></td></tr>'; } while ($row_rsAllListings = mysql_fetch_assoc($rsAllListings)); ?>
</table>
</div>
<div class="content">
<table width="600">
<?php $counter = 0; // initialize counter outside loop 
do { echo '<tr';
if ($counter++ % 2) {echo 'class="hilite"';} 
echo '><td width="162"><img src="'.$row_rsAllListings['element_16'].'"></td>';
echo '<td width="426"><h2><a href="\inventory.php?id="'.$id.'">'.$row_rsAllListings['element_1_1'];
echo '</a> |  '.$row_rsAllListings['element_2'].'<br />';
echo $row_rsAllListings['element_3'].'</h2><br />';
echo $row_rsAllListings['element_4'].'<br />';
echo $row_rsAllListings['element_5'].'Beds / '.$row_rsAllListings['element_6']; 
echo $row_rsAllListings['element_7'].'Baths / '.$row_rsAllListings['element_8'].'sf per '.$row_rsAllListings['element_9'].'<br /></td></tr>'; } while ($row_rsAllListings = mysql_fetch_assoc($rsAllListings)); ?>
</table>
</div>

this code is from clicking 'suggestions' devPHP
original code

php Syntax (Toggle Plain Text)
<div class="content">
  <table width="600">
  <?php $counter = 0; // initialize counter outside loop ?>
<?php do { ?>
<tr <?php if ($counter++ % 2) {echo 'class="hilite"';} ?>>
      <td width="162"><img src="<?php echo $row_rsAllListings['element_16']; ?>></td>
      <td width="426">
<h2><a href="\inventory.php?id=" . $id . "\"><?php echo $row_rsAllListings['element_1_1']; ?></a> | <?php echo $row_rsAllListings['element_2']; ?><br />
        <?php echo $row_rsAllListings['element_3']; ?></h2>
        <br />
        <?php echo $row_rsAllListings['element_4']; ?><br />
        <?php echo $row_rsAllListings['element_5']; ?> Beds / <?php echo $row_rsAllListings['element_6']; ?>-<?php echo $row_rsAllListings['element_7']; ?> Baths / <?php echo $row_rsAllListings['element_8']; ?>sf per <?php echo $row_rsAllListings['element_9']; ?><br />
      </td>
    </tr>
    <?php } while ($row_rsAllListings = mysql_fetch_assoc($rsAllListings)); ?>
  </table>
</div>
 
<div class="content">
  <table width="600">
  <?php $counter = 0; // initialize counter outside loop ?>
<?php do { ?>
<tr <?php if ($counter++ % 2) {echo 'class="hilite"';} ?>>
      <td width="162"><img src="<?php echo $row_rsAllListings['element_16']; ?>></td>
      <td width="426">
<h2><a href="\inventory.php?id=" . $id . "\"><?php echo $row_rsAllListings['element_1_1']; ?></a> | <?php echo $row_rsAllListings['element_2']; ?><br />
        <?php echo $row_rsAllListings['element_3']; ?></h2>
        <br />
        <?php echo $row_rsAllListings['element_4']; ?><br />
        <?php echo $row_rsAllListings['element_5']; ?> Beds / <?php echo $row_rsAllListings['element_6']; ?>-<?php echo $row_rsAllListings['element_7']; ?> Baths / <?php echo $row_rsAllListings['element_8']; ?>sf per <?php echo $row_rsAllListings['element_9']; ?><br />
      </td>
    </tr>
    <?php } while ($row_rsAllListings = mysql_fetch_assoc($rsAllListings)); ?>
  </table>
</div>

Last edited by almostbob; Aug 18th, 2009 at 7:13 pm.

almostbob

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Aug 18th, 2009

Re: PHP variable in link

Okay, so now I have the correct id appearing at the top of the next page by editing the link to be:

PHP Syntax (Toggle Plain Text)
<a href="\inventory.php?id=<?php echo $row_rsAllListings['id']; ?>"\><?php echo $row_rsAllListings['element_1_1']; ?></a>
<a href="\inventory.php?id=<?php echo $row_rsAllListings['id']; ?>"\><?php echo $row_rsAllListings['element_1_1']; ?></a>

But the record being displayed in that next page is always the first record in the database.

Progress for sure, but not there yet.

isak

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Aug 18th, 2009

Re: PHP variable in link

Click to Expand / Collapse Quote originally posted by almostbob ...

try
php Syntax (Toggle Plain Text)
<tr <?php if ($counter++ % 2) { echo 'class="hilite"'; }
echo '><td width="162"><img src='.$row_rsAllListings['element_16'].'></td><td width="426"><h2><a href=/"/inventory.php?id='.$row_rsAllListings['element_1_1'].'</a> | '.$row_rsAllListings['element_2'].'<br />'; ?>
<tr <?php if ($counter++ % 2) { echo 'class="hilite"'; } echo '><td width="162"><img src='.$row_rsAllListings['element_16'].'></td><td width="426"><h2><a href=/"/inventory.php?id='.$row_rsAllListings['element_1_1'].'</a> | '.$row_rsAllListings['element_2'].'<br />'; ?>
in the original code sample the php variable $id is not within a <?php ?> block, =>cant be accessed
not sure what editor you are using,
but
when I pasted your code into notepad++ $id in this link was hilighted red immediately, v.easy to find

I think you are missing a closing tag for the table row?

I have added the $id back into the code, but getting the same record on the next page regardless of the link I click.

Last edited by isak; Aug 18th, 2009 at 7:05 pm.

isak

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Aug 18th, 2009

Re: PHP variable in link

inventory.php
there is no 'select * from database where id='.$id
there is no reference to '$id' in the file
the href calling the inventory script
inventory.php?id=$id
if you don't use it, you lose it, and id is getting lost

almostbob

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Aug 18th, 2009

Re: PHP variable in link

almostbob,

This is where I mention "novice" again. Can you explain more simply what and where this works? I would LOVE to use it.

isak

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Aug 18th, 2009

Re: PHP variable in link

your link to inventory.php has a parameter ?id=$id
assuming $id is the id link of the record in the database
that id is available in inventory.php as $id
somewhere in the inventory.php file there should be (could be, usually is) an sql query that selects the row defined by that id
something like

sql Syntax (Toggle Plain Text)

"select * from database where id=$id"

I could not find such query in the file
I could find many queries, but none that selected on the basis of $id
it may be as simple as finding the variable used in inventory.php to select and
change that named variable to $id /* or */
change the parameter name passed by the link to that used in inventory.php /* or */
add a sql select with the right format
the sql in the files is convoluted, almost as if written to hide its intent

Last edited by almostbob; Aug 18th, 2009 at 9:27 pm.

almostbob

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Aug 18th, 2009

Re: PHP variable in link

I edited the inventory.php file to include the $id. I just put it at the beginning of the street address however I really would rather it not appear in the inventory page when viewed. I have attached the new inventory.php file.

So while it *works* it is still only pulling the first record in the database.

The sql written in the file is written by Dreamweaver.

BTW... nice clean code that you wrote. Thanks.

Attached Files

inventory.php (6.5 KB, 19 views)

isak

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Aug 18th, 2009

Re: PHP variable in link

without debugging the code, don't have the db structure, don't know what table refers to what, you have to end up with something like

php Syntax (Toggle Plain Text)
mysql_select_db($database_rsGunn, $rsGunn);
$query_rsInventory1 = "SELECT ap_form_1.element_1_1, ap_form_1.element_1_2, ap_form_1.element_1_3, ap_form_1.element_1_4, ap_form_1.element_1_5, ap_form_1.element_1_6, ap_form_1.element_2, ap_form_1.element_3, ap_form_1.element_16, ap_form_1.element_4, ap_form_1.element_5, ap_form_1.element_57, ap_form_1.element_6, ap_form_1.element_7, ap_form_1.element_8, ap_form_1.element_9, ap_form_1.id FROM ap_form_1 where id=".$id;
$rsInventory1 = mysql_query($query_rsInventory1, $rsGunn) or die(mysql_error());
mysql_select_db($database_rsGunn, $rsGunn);
$query_rsInventory1 = "SELECT ap_form_1.element_1_1, ap_form_1.element_1_2, ap_form_1.element_1_3, ap_form_1.element_1_4, ap_form_1.element_1_5, ap_form_1.element_1_6, ap_form_1.element_2, ap_form_1.element_3, ap_form_1.element_16, ap_form_1.element_4, ap_form_1.element_5, ap_form_1.element_57, ap_form_1.element_6, ap_form_1.element_7, ap_form_1.element_8, ap_form_1.element_9, ap_form_1.id FROM ap_form_1 where id=".$id;
$rsInventory1 = mysql_query($query_rsInventory1, $rsGunn) or die(mysql_error());

horrible data structure ap_form.element_57, not at all intuitive

Last edited by almostbob; Aug 18th, 2009 at 11:40 pm.

almostbob

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