如何从MYSQL数据库获取1条记录的ID，然后将其传递到详细信息页面以使用动态php和mysqli查看

Question

I have a html page listing all the jobs in my database, I managed to get everything displaying on www.mysite/alljobs.php (or the list all records page), great, but I cannot get hold of the different id's from this page to display on a dynamic link to showing the job details for each job ie www.mysite/thejob.php?job_id=8 (the specific record details page).

I have recently had to upgrade my PHP version to PHP7 and need to use MYSQLI with PHP. Can any help with the code I need for both the list page and the details page so that dynamic pages are created when I click on the alljobs.php links.

Answer 1

如果对每条记录使用<a> ，则可以将id放在href中，如下所示：

<a href="www.mysite/thejob.php?job_id=<?php echo $id;?>"> Link </a>

Answer 2

您可以在PHP报价中使用echo传递您的id变量

<a href="www.mysite/thejob.php?job_id=<?php echo $id;?>">View</a>

Answer 3

You can loop over your query results and print out the ID for each one in a link, like this

<?php
  foreach ($myQueryResults as $item){
?>

<a href="www.mysite/thejob.php?job_id=<?php echo $item->id;?>"> Link </a>

<?php
  }
?>

Answer 4

you can send the id on link url and the get with $jobID = <?php $_GET['id']?> on that page. Don't forget to sanitize.

<a href="www.mysite/thejob.php?id=<?php echo $jobId;?>">Details</a>