jobs1109 Posted August 17, 2011 Share Posted August 17, 2011 Hi, I am trying to send mysql query to another page by link. I would like users to be able to click on a link from mysql query result and then be able to see more details on a second page for the row selected in the link. Here is an example of what I am trying to do. http://www.jobberbase.com/demo/. In my code when I go to second page it does not show the correct row of data when link is clicked. Here is my code . first page with mysql query results. <?php session_start(); ?> // Post variables <?php $_SESSION['Keyword_Search'] = $_POST['Keyword_Search']; $_SESSION['Jobcategory'] = $_POST['Jobcategory']; $Keyword_Search = $_SESSION['Keyword_Search']; $Jobcategory = $_SESSION['Jobcategory']; // echo $Keyword_Search; will be used in future // echo $Jobcategory; will be used in future include'../../DB-Connect101/DB-Connect101.php'; // getting data from table "Jobs-table" $query = "SELECT * FROM Jobs-table"; $result = mysql_query(" SELECT 'id', `JobTitle`, `Company`, `Salary`, `Description`, `CompanyURL`, `PhoneNumber`, `Requirements`, `JobCategory`, `JobType`, `Apply_To`, `Email`, `State`, `Address`, `Country`, `Zipcode` FROM `Jobs-table` ") or die(mysql_error());while ($row = mysql_fetch_array($result)) { echo 'Job Title: <a href="http://www.hiringinhilo.com/Hilo-Jobs/Job-Search-Form/Getting-Data-From-Database/Job-Details/Job-Details.php">'.$row['JobTitle'].'</a><br /><br />'; } ?> Here is the second page when results for selected row should be displayed. <?php session_start(); ?> <?php include'../../../DB-Connect101/DB-Connect101.php'; $Keyword_Search = $_SESSION['Keyword_Search']; $Jobcategory = $_SESSION['Jobcategory']; // echo $Keyword_Search; // echo $Jobcategory; ?> <br> <br> <br> <?php $id = $_SESSION['id']; $query = "SELECT * FROM Jobs-table where id=$id"; $result = mysql_query(" SELECT `id`, `JobTitle`, `Company`, `Salary`, `Description`, `CompanyURL`, `PhoneNumber`, `Requirements`, `JobCategory`, `JobType`, `Apply_To`, `Email`, `State`, `Address`, `Country`, `Zipcode` FROM `Jobs-table` ORDER BY id ASC") or die(mysql_error()); $row = mysql_fetch_array($result); echo "ID: ".$row['id']; echo "<br>Job Title: ".$row['JobTitle']; echo "<br> Company: ".$row['Company']; echo " <br>Salary: ".$row['Salary']; echo "<br> Description: ".$row['Description']; echo "<br> Company-URL: ".$row['CompanyURL']; echo "<br> Phone-Number: ".$row['Phone-Number']; echo "<br> Address: ".$row['Address']; echo "<br> Requirements: ".$row['Requirements']; echo "<br> Job-Type: ".$row['JobType']; echo "<br> JobCategory: ".$row['JobCategory']; echo "<br><br><br><br><br>"; ?> why is my code not working ? please help Quote Link to comment Share on other sites More sharing options...
dougjohnson Posted August 17, 2011 Share Posted August 17, 2011 I think your link needs to pass the appropriate "ID" to the second page? Where does $_SESSION['id'] get it's value? Quote Link to comment Share on other sites More sharing options...
xyph Posted August 17, 2011 Share Posted August 17, 2011 You want your url to look like this http://www.hiringinhilo.com/Hilo-Jobs/Job-Search-Form/Getting-Data-From-Database/Job-Details/Job-Details.php?id=$row['id'] Then on the display page, get that ID number using $_GET['id'] On another note, you should only select columns from your database that you plan on using. You can shorten your initial query to "SELECT `id`, `JobTitle` FROM `Jobs-table`" Quote Link to comment Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.