PHP mysql display query in 4 columns - php

Question

How to display this query result in 4 columns
<?php
$direction = $_POST['direction'];
$sumword = $_POST['sumword'];
$length = $_POST['length'];
$con = mysql_connect("localhost","elsha","12q(5PSZ.");
$db = mysql_select_db("elsha",$con);
$query = "SELECT answer FROM words WHERE direction = '$direction' AND sumword = '$sumword' AND sumletter = '$length'";
$result = mysql_query($query);
if(mysql_error()) {
//check that no error has occurred first; take this out in production or make more graceful handling
die(mysql_error());
}
if(mysql_num_rows($result) == 0) {
echo "No Results";
} else {
while($row = mysql_fetch_array($result)) {
echo '<img src="/images/'. $row['0'].'.jpg"><br> '. $row['0'].'<br>';
}
}
?>
like this :
result1 result2 result3 result4
result5 result6 result7 result8

Answer 1

Try this
$i = 0;
while($row = mysql_fetch_array($result)) {
echo '<img src="/images/'. $row['0'].'.jpg"> ';
$i++;
if($i % 4 == 1 && $i!=1){
echo '<br>';
}
}

Answer 2

Use flag like
$flag=0;
while($row = mysql_fetch_array($result)) {
if(($flag%4)==0) {
echo '<tr>';
}
echo '<td><img src="/images/'. $row['0'].'.jpg"><br> '. $row['0'].'</td>';
if(($flag%4)==3) {
echo '</tr>';
$flag=-1;
}
$flag++;
}

Answer 3

replace your while with the following code. try this
$a=1;
while($row = mysql_fetch_array($result)) {
if($a%4==0) {
echo '<img src="/images/'. $row['0'].'.jpg">'. $row['0'].'<br>';
} //4th line with break
else {
echo '<img src="/images/'. $row['0'].'.jpg"> '. $row['0'].' '; // prints first 3 lines
}
$a=$a+1;
}

Answer 4

I think you are trying to display the image name too below the image. Try this:
$i = 1;
while($row = mysql_fetch_array($result)) {
echo '<div class="imgs">';
echo '<img src="/images/'. $row['0'].'.jpg"><br> '. $row['0'];
echo '</div>';
if ($i === 4) {
echo '<div class="clear"></div>';
$i = 1;
} else {
$i++;
}
}
and then style div's
.imgs {
float: left;
/* other styles */
}
.clear {
clear: both;
}