HTML link with a php action - php

Question

I am creating a website that stores information of homes like address, city etc. I got the upload to database part done and I got the search the database and display the information done.
But now my plan with that is you get the search results and from there your suppose to get the option to view the full profile of that home via link. Displaying the full profile with the information gathered from the database is already done through a separate php file. Now when I link the php file it will only display the last column in the database. How can I link it so when I click on "view full profile" it will connect to the correct data/profile that corresponds to the different address or price. Or if i cant use an html link what can I use? Is that impossible?
here is the code that displays the search results I have successfully been able to search the database and display it
<?php
if($sql->num_rows){
while ($row = $sql->fetch_array(MYSQLI_ASSOC)){
echo '<div id="listing">
<div id="propertyImage">
<img src="images/'.$row['imageName1'].'" width="200" height="150" alt=""/>
</div>
<div id="basicInfo">
<h2>$'.$row['Price'].'</h2>
<p style="font-size: 18px;"># '.$row['StreetAddress'].', '.$row['City'].', BC</p>
<p>'.$row['NumBed'].' Bedrooms | '.$row['NumBath'].' Bathrooms | '.$row['Property'].'</p>
<br>
<p>View Full Details | Get Directions
</div>
</div>';
height="150" alt=""/>';
}
}
else
{
echo '<h2>0 Search Results</h2>';
}
?>
Here is the php to display the information outputtest.php
<?php
mysql_connect("localhost","root","31588patrick");
#mysql_select_db("test") or die( "Unable to select database");
$query="SELECT * FROM propertyinfo";
$result=mysql_query($query);
$num=mysql_numrows($result);
mysql_close();
$i=0;
while ($i < $num) {
$varStreetAddress=mysql_result($result,$i,"StreetAddress");
$varCity=mysql_result($result,$i,"City");
$varProperty=mysql_result($result,$i,"Property");
$varNumBed=mysql_result($result,$i,"NumBed");
$varNumBath=mysql_result($result,$i,"NumBath");
$varPrice=mysql_result($result,$i,"Price");
$varEmail=mysql_result($result,$i,"Email");
$varPhone=mysql_result($result,$i,"Phone");
$varUtilities=mysql_result($result,$i,"utilities");
$varTermLease=mysql_result($result,$i,"TermLease");
$varNotes=mysql_result($result,$i,"Notes");
$image1=mysql_result($result,$i,"imageName1");
$image2=mysql_result($result,$i,"imageName2");
$image3=mysql_result($result,$i,"imageName3");
$image4=mysql_result($result,$i,"imageName4");
$i++;
}
?> html code will go after this
Thanks
edit its working
<?php
////////////using mysqli to connect with database
$mysqli = new mysqli("localhost","root","31588patrick", "test");
if ($mysqli->connect_errno) {
echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error;
}
///////////set variables
$record_id = $_GET['record_id'];
$sql = $mysqli->query("select * from propertyinfo where StreetAddress like '%$record_id%'");
if($sql === FALSE) {
die(mysql_error()); // TODO: better error handling
}
if($sql->num_rows){
while ($row = $sql->fetch_array(MYSQLI_ASSOC)){
$varStreetAddress=$row['StreetAddress'];
$varCity=$row['City'];
$varProperty=$row['Property'];
$varNumBed=$row['NumBed'];
$varNumBath=$row['NumBath'];
$varPrice=$row['Price'];
$varEmail=$row['Email'];
$varPhone=$row['Phone'];
$varUtilities=$row['utilities'];
$varTermLease=$row['TermLease'];
$varNotes=$row['Notes'];
$image1=$row['imageName1'];
$image2=$row['imageName2'];
$image3=$row['imageName3'];
$image4=$row['imageName4'];
}
}

Answer 1

If the link you are using is only fetching the last record in the database table, then the underlying query that is used must be fetching the last record instead of the intended record. Am I correct in understanding the problem you are having? That any link you use is just returning the last record in the database?
Posting a code snippet will help the community take a closer look.
in your php file, you run a query for all records, but ...
$record_id = mysql_real_escape_string($_GET["record_id"]);
$query="SELECT * FROM propertyinfo WHERE id=$record_id";
your link needs to attach a parameter that will allow you to query for a specific record.
a href="outputtest.php&record_id=8"
AND you would also want to check what is in the GET parameter before sticking it into the query! I know mysql_real_escape_string is outdated.