CODEWITHSUNDEEP
phphtmlmysql
Eli Skywalker
Eli Skywalker

Reputation: 43

How to create a table in HTML using PHP

I have 5 pictures stored in a folder and their links stored on the database. I want to put them in a table of three columns on each row.

<body>
<center>
<table border='1'>
<?php
$host="";
$username="";
$password="";
$db_name="fruits_db";
$tbl_name="fruits_tbl";
$connection=mysqli_connect("$host","$username","$password","$db_name");
if (mysqli_connect_errno())
{
    echo "The application has failed to connect to the mysql database server: " .mysqli_connect_error();
}
$result = mysqli_query($connection, "SELECT * FROM fruits_tbl")or die("Error: " . mysqli_error($connection));
$num_rows=mysqli_num_rows($result);
$rows =  $num_rows/3;

for($i=1; $i<=$rows ; $i++)
{
    echo "<tr>";
    for($j=1; $j<=3; $j++)
    {
        while($row = mysqli_fetch_array($result))
        {
            echo
            ("<td width='180px' height='200px'>"
             ."<div class = 'fruit_image'>"
             ."<img src='"
             .$row['fruit_image']
             ."'/>"
             ."</div>"
             ."<div class = 'fruit_title'>"
             .$row['fruit_name']
             ."</div>"
             ."</td>"
            );
        }
    }
    echo "</tr>";
}

mysqli_close($connection);
?>
</table>
</center>
</body>
</html>

The above code I created, contains two FOR loops. The script should count the number of rows in the table, and then divide by 3(the number of columns on each row in the HTML table). I wonder where I'm going wrong wit this code.

Upvotes: 1

Views: 72773

Answers (5)

Mishari
Mishari

Reputation: 1

<?php
function studentTable($name,$grade){
echo "<tr> <td>$name</td><td>$grade</td></tr>";
}
?>
    <table style="width:100%" border="solid">
<tr>
<th>Name</th>
<th>Grade</th> 
 </tr>
 <?php studentTable("Sarah", 90) ?>

Upvotes: 0

rajratna maitry
rajratna maitry

Reputation: 508

    print "<center><table border=1>
        <tr>
        <td>id</td>
        <td>name</td>
        <td>company</td>
        <td>branch</td>
        <td>job title</td>
        <td>contact</td>
        <td>email</td>
        <td>mgs</td>
        </tr>";

    while($row=mysql_fetch_array($query))
    {
        print "<tr>";
       for ($i=0;$i<=(count($row)/2);$i++)
              {
               print "<td>$row[$i]</td>";
               } print"</tr>";
   }

    }
    else{echo " <p>No Records Found</p>";}

Upvotes: 0

Revent
Revent

Reputation: 2109

If you just want to format the display with a new row after every 3 records, you could use the modulus operator:

$cntr = 0;
echo '<tr>';
while($row = mysqli_fetch_array($result)) {
   $cntr++;
   echo '
      <td width="180px" height="200px">
         <div class="fruit_image">
            <img src="'.$row['fruit_image'].'" />
         </div>
         <div class="fruit_title">'.$row['fruit_name'].'</div>
      </td>';
   if ($cntr % 3 == 0 && $cntr != $num_rows)
      echo '</tr><tr>';
}
echo '</tr>';

Keep in mind however that all the solutions presented so far may leave you with a last row with one or two td elements. You can fill this if you desire with empty <td>&nbsp;</td> columns.

Upvotes: 0

Sean
Sean

Reputation: 12433

With your while($row = mysqli_fetch_array($result)){} inside your 1st for loop it will run through all your rows, before the outside loop runs 2nd/3rd time.

Here is another way to do it -

$counter = 1;
// start 1st row
echo "<tr>";
while($row = mysqli_fetch_array($result)){
// if the 4th cell, end last row, and start new row
    if ($counter%3==1){
        echo "</tr><tr>";
    }
    echo
        "<td width='180px' height='200px'>"
        ."<div class = 'fruit_image'>"
        ."<img src='"
        .$row['fruit_image']
        ."'/>"
        ."</div>"
        ."<div class = 'fruit_title'>"
        .$row['fruit_name']
        ."</div>"
        ."</td>";
    // increase the counter
    $counter++;
}
// close the last row
echo "</tr>";

Upvotes: 1

jcsanyi
jcsanyi

Reputation: 8174

You're looping through all the results in the first table cell. Try something like this instead:

for($i=1; $i<=$rows ; $i++) {
    echo "<tr>";
    for($j=1; $j<=3; $j++) {
        $row = mysqli_fetch_array($result);
        if ($row) {
            echo(
                "<td width='180px' height='200px'>"
                ."<div class = 'fruit_image'>"
                ."<img src='"
                .$row['fruit_image']
                ."'/>"
                ."</div>"
                ."<div class = 'fruit_title'>"
                .$row['fruit_name']
                ."</div>"
                ."</td>"
            );
        }
    }
    echo "</tr>";
}

Upvotes: 0

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