How to use PHP foreach with MySql [duplicate]

Question

I want to loop through each result from the mysql table and generate html div for each rows. The database table looks like following:

-------------------------------- id | color | width | img -------------------------------- 1 | red | 550 | url -------------------------------- 2 | black | 650 | url --------------------------------

I have an idea of how this might work but not sure about the syntax. Below is my attempt.

<?php // connect to the database $rows = // get rows from the db table foreach ($rows) { <div style="background-color:$color;width:$width;"> <img src="$img"> </div> }

So what's the difficulty there? (Don't say you want an example. There are billions of them all around. See the manual.) — mario
– mario, Commented Aug 27, 2015 at 23:17
Unless your sample was intentionally oversimplified, you need to echo(); the content in the for each loop to get it to output to the screen. — scunliffe
– scunliffe, Commented Aug 27, 2015 at 23:21
@scunliffe I would image that an sql command and a function call to issue that query to the database server and a function call to retrieve the result set would also come in quite handy as well. — RiggsFolly
– RiggsFolly, Commented Aug 27, 2015 at 23:28
Come back when you've fixed the syntax errors. In the meantime I'm voting to close this question. — Mike
– Mike, Commented Aug 27, 2015 at 23:52

Maverick · Accepted Answer · 2015-08-27 23:41:42Z

This is a simple example with a simple query.

<?php $mysqli = new mysqli("db_host", "db_user", "db_password", "db_name"); // check connection if ($mysqli->connect_errno) { die("Connect failed: ".$mysqli->connect_error); } $query = "SELECT * FROM imgs"; $result = $mysqli->query($query); while($row = $result->fetch_array()){ echo '<div style="background-color:'.$row[color].';width:'.$row[width].';"><img src="'.$row[img].'"></div>'; }

Subash · Accepted Answer · 2015-08-28 03:17:38Z

Assuming that the output of the database is same as $rows, your code should look like follows. Obviously you can improve on the concatenation of the echo statements,.

<?php $rows = [ 'row1' => [ 'color' => 'red', 'width' => '550px', 'img' => 'http://www.google.com.au/images/srpr/logo11w.png' ], 'row2' => [ 'color' => 'black', 'width' => '650px', 'img' => 'http://www.google.com.au/images/srpr/logo11w.png' ], ]; foreach ($rows as $key => $row) { $color = $row['color']; $image = $row['img']; $width = $row['width']; echo "<div style=\"background-color:$color;width:$width;\">"; echo "<img src=\"$image\">"; echo "</div>"; } ?>

Here is a working phpfiddle

imdba · Accepted Answer · 2015-08-27 23:33:29Z

foreach($result as $key=>$val){ var_dump($val) ; }

You can try it ; and you must be sure mysql $row is array !

WaduNoop · Accepted Answer · 2015-08-28 02:32:28Z

Here is a simple loop that will display images inside a div.

<?php for($i = 0; $i < 3; $i++){ echo "<div style='border: 1px solid black;'>"; echo "<img src='http://screenrant.com/wp-content/uploads/James-Earl-Jones-Darth-Vader-Star-Wars-Rebels.jpg'>"; echo "</div>"; }?>

Advise : Beware when to use double quotes and single quotes in a echo statement when you want to display a HTML script using echo or else it wouldn't display correctly. Also, use the correct syntax when handling result queries using foreach.

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