将来自MySQL的数据存储到数组中，以后再访问

Question

Eh.. I am sorry for asking such a basic question but I couldn't find a good answer neither on google or stackoverflow... here is what I want to do.

I save announcements in MySQL and then I show them to the users using

<?php while($announcerow = mysqli_fetch_assoc($announceResult)): ?>

then there is a button for admins to edit them, a modal opens and the admin is able to select which announcement he wants to edit, well, I try to show the announcements same way there too but if I do that the announcements wont show anymore, so I can use mysqli_fetch_assoc only once on a query as I see...

I thought to save them in an array and use them later, but then I do not know how to access the array later.

for example, I save them like this:

while($row = mysql_fetch_assoc($query)){

  // add each row returned into an array
  $array[] = $row;


}

but how do I show the announcements later for the users? also, my announcements have an ID, a title, an URL and a description, this is how I show them right now.

<?php while($announcerow = mysqli_fetch_assoc($announceResult)): ?>
                <a data-toggle="tooltip" title="<?php echo $announcerow['Description']; ?>" target="_blank" href="<?php echo $announcerow['URL']; ?>"><p class="text-light-blue"><span class="fa fa-external-link"></span> <?php echo $announcerow['Title']; ?><span class="text-muted"> - by <?php echo $announcerow['Author']; ?> (<?php echo $announcerow['Date']; ?>)</span></p></a>
            <?php endwhile; ?>

how will I show them from the array?

Sorry for this noobish question... couldn't find an answer.

I even tried mysql_data_seek(); so I can use fetch more times but I got this error:

Warning: mysql_data_seek(): supplied argument is not a valid MySQL result resource in 

I've tried this

while($row = mysql_fetch_assoc($announceResult)){
            $array[] = $row;
        }

and

<?php foreach($array as $datum): ?>
                 <a data-toggle="tooltip" title="<?php echo $datum['Description']; ?>" target="_blank" href="<?php echo $datum['URL']; ?>">
                      <p class="text-light-blue">
                           <span class="fa fa-external-link"></span>
                           <?php echo $datum['Title']; ?>
                           <span class="text-muted"> - by <?php echo $datum['Author']; ?> (<?php echo $datum['Date']; ?>)
                           </span>
                      </p>
                </a>
                <?php endforeach; ?>

but I get the following errors

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /home/a8438725/public_html/index.php on line 56

Notice: Undefined variable: array in /home/a8438725/public_html/index.php on line 777
Warning: Invalid argument supplied for foreach() in /home/a8438725/public_html/index.php on line 777

Here is the whole code of the sql and array

$announceSql = "SELECT `ID`, `Author`, `Title`, `Date`, `URL`, `Description` FROM `Announcements` ORDER BY ID DESC";
    $announceResult = mysqli_query($db, $announceSql);
    $announcerow_cnt = mysqli_num_rows($announceResult);

    while($row = mysql_fetch_assoc($announceResult)){
        $array[] = $row;
    }

I've also added this

if (!$announceResult) {
            printf("Error: %s\n", mysqli_error($db));
            exit();
        }

and there is no mysql error, or atleast no mysql error appeard

SOLVED! I was using mysql instead of mysqli_fetch_assoc

Answer 1

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /home/a8438725/public_html/index.php on line 56

That means the line you haven't shown us containing something like

$announceResult=mysqli_query($db_conn, $sql);

failed or $announceResult is in a different scope, or $announceResult has been changed since it was populated. The first of these would likely have triggered a warning too. So the most likely cause is that you've got a scope issue or bug somewhere in the 750+ lines of code you've not shown us here.

(hint: if you run into a problem like this and can't fix it in situ, it is advisable to try writing as small a program as necessary to reproduce the fault and use that to undertand the problem / illustrate your issue on Stack Overflow).

Notice: Undefined variable: array in /home/a8438725/public_html/index.php on line 777

Warning: Invalid argument supplied for foreach() in /home/a8438725/public_html/index.php on line 777

Since the loop populating $array() is driven by mysqli_fetch_assoc(), $array() was never initialized and populated.

Answer 2

You can use like this

<?php while($announcerow = mysqli_fetch_assoc($announceResult)){ ?>
           $array[] = $announcerow;
<?php }?>



    <?php foreach($array as $datum) {?>
         <a data-toggle="tooltip" title="<?php echo $datum['Description']; ?>" target="_blank" href="<?php echo $datum['URL']; ?>">
              <p class="text-light-blue">
                   <span class="fa fa-external-link"></span>
                   <?php echo $datum['Title']; ?>
                   <span class="text-muted"> - by <?php echo $datum['Author']; ?> (<?php echo $datum['Date']; ?>)
                   </span>
              </p>
        </a>
    <?php } ?>