foreach mysql row from query echo out html with row values

Question

So basicly what I'm trying to accomplish is that foreach row in mysql query it prints out the html with the data from that row. Here's what I have, it keeps giving me an error on my foreach.

<?php

$shots = mysql_query("SELECT * FROM shots") or die(mysql_error());

while($row=mysql_fetch_array($shots))
$data[]=$row;


foreach($shots as $data)

  if (!empty($data)){

  $id = $data["id"];
  $shotby = $data["shot"];
  $passby = $data["pass"];
  $time = $data["time"];
?>


<div class="feedbody">

    <div class="title"><?php echo $shotby; ?></div>
    <div class="feed-data">: gets a pass from <span><?php echo $passby; ?</span> and he takes a shot!</div>
    <img class="dot" src="images/dot.png" />

</div>
<?php
}

}
?>

Or something like that. Can anybody help point me in the right direction. I've been trying to find the answer.

EDIT: adding the error as requested.

Warning: Invalid argument supplied for foreach() in /home/content/93/7527593/html/fusionboard/includes/feed.php on line 7

Also, I know it's a coding style thing, but please use curly braces, even on one-line bodies like your while loop and for loop. Makes everything much more readable and one day not using them will come back to bite you with a hard to track down bug. — Parrots
– Parrots, Commented Jan 16, 2012 at 18:55

Parrots · Accepted Answer · 2012-01-16 19:10:27Z

11

~~First, if you want to access the data by name (instead of index), you need to include MYSQL_ASSOC as a second parameter to mysql_fetch_array, or use mysql_fetch_assoc.~~

Not really sure why you were copying the MySQL results to a second array just to loop through that later - you can loop through the results directly:

<?php
$shots = mysql_query("SELECT * FROM shots") or die(mysql_error());
while($row = mysql_fetch_assoc($shots)) { ?>
<div class="feedbody">
    <div class="title"><?php echo $row["shot"]; ?></div>
    <div class="feed-data">: gets a pass from <span><?php echo $row["pass"]; ?></span> and he takes a shot!</div>
    <img class="dot" src="images/dot.png" />
</div>
<?php } ?>

Update after you posted the error message: the error from your original code was that you first went through and copied each result row into $data, but then in your foreach you tried to loop on $shots (again) and have it call each item $data.

What you probably wanted to do was have foreach ($data as $item) and then copy the properties from $item.

edited Jan 16, 2012 at 19:10

answered Jan 16, 2012 at 18:45

Parrots

27k14 gold badges61 silver badges78 bronze badges

Sign up to request clarification or add additional context in comments.

4 Comments

Biotox Over a year ago

You could just make it while($row = mysql_fetch_assoc($shots)) for a associative array. :) - upvoted though

Parrots Over a year ago

@biotox thanks, updated answer to reflect that (been a while since I've used native PHP mysql calls, ha).

Nick Fury Over a year ago

It's cause I'm not really sure what I'm doing half the time. Thanks a lot for the help.

user610217 Over a year ago

Agreed. mysql_fetch_assoc() doesn't require anything more.

user610217 · Accepted Answer · 2012-01-16 18:43:58Z

2

Something like this?

<?php
$shots = mysql_query("SELECT * FROM shots") or die(mysql_error());

while($row=mysql_fetch_assoc($shots))
{
  $id = $row["id"];
  $shotby = $row["shot"];
  $passby = $row["pass"];
  $time = $row["time"];
?>

<div class="feedbody">

    <div class="title"><?php echo $shotby; ?></div>
    <div class="feed-data">: gets a pass from <span><?php echo $passby; ?</span> and he takes a shot!</div>
    <img class="dot" src="images/dot.png" />

</div>
<?php
}
?>

answered Jan 16, 2012 at 18:43

user610217

Comments

PaulJ · Accepted Answer · 2012-01-16 18:43:09Z

1

Perhaps you need another closing brace? (Another "}" at the end, I mean).

answered Jan 16, 2012 at 18:43

PaulJ

1,7165 gold badges34 silver badges58 bronze badges

Comments

Daniel Castro · Accepted Answer · 2012-01-16 18:43:21Z

0

You saved your data in the $data variable, but your foreach uses the $shots variable.

Just change it to foreach($data as $something) and $something["id"] (for example) to retrieve a value

answered Jan 16, 2012 at 18:43

Daniel Castro

6336 silver badges16 bronze badges

Comments

Your Common Sense · Accepted Answer · 2012-01-16 18:45:32Z

-2

you are using one variable instead of another.
and many useless code.

while youneed only

<?php foreach($data as $row) { ?> 
<div class="feedbody"> 
    <div class="title"><?php echo $row['shotby'] ?></div> 
    <div class="feed-data">: 
        gets a pass from <span><?php echo $row['passby'] ?</span> and he takes a shot!
    </div> 
    <img class="dot" src="images/dot.png" /> 
</div> 
<?php } ?>

answered Jan 16, 2012 at 18:45

Your Common Sense

158k42 gold badges226 silver badges373 bronze badges

1 Comment

user610217 Over a year ago

you intend to loop through this for every column of every row?

Collectives™ on Stack Overflow

foreach mysql row from query echo out html with row values

5 Answers 5

4 Comments

Comments

Comments

Comments

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

5 Answers 5

4 Comments

Comments

Comments

Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related