displaying array values in a table

Question

i want to display values from an array in a table, i tried searching and none of it works, here is the snippet

$host="localhost";
$username="root";
$password="";
$db_name="sampledb";
$tbl_name="tblsched";
$cd1 = date("Y-m-d");
mysql_connect("$host", "$username", "$password")or die("cannot connect"); 
mysql_select_db("$db_name")or die("cannot select DB");

$sql = "SELECT * FROM tblteacher WHERE teacherName='$uname'";
$result=mysql_query($sql);
$row = mysql_fetch_assoc($result);

$teacherid  = $row['teacherID'];

function mymainfunc()
{
global $uname;
global $teacherid;
$bgcolor="#ADDFFF";
$tdali="center";
dateto();
datefrom();
$arrayto[] = array();
$arrayfrom[] = array();
$arrayto[] = dateto();
$arrayfrom[] = datefrom();
$x=0;
while($x<count($arrayto[]))
{
echo "<tr>";
echo "<td align=$tdali bgcolor=$bgcolor>$arrayfrom[$x] - $arrayto[$x]</td>";
echo "<td align=$tdali> </td>";
echo "<td align=$tdali> </td>";
echo "<td align=$tdali> </td>";
echo "<td align=$tdali> </td>";
echo "<td align=$tdali> </td>";
echo "<td align=$tdali> </td>";
echo "<td align=$tdali> </td>"; 
echo "</tr>";

$x++;
}
}
function dateto()
{
global $uname;
global $cd1;
global $teacherid;
$sql="SELECT SchedTimeTo FROM tblsched WHERE SchedTeacherID=$teacherid and SchedDateFrom<='$cd1' and SchedDateTo>='$cd1' ";
$result=mysql_query($sql);

if($result === FALSE) {
die(mysql_error());
}

$count=mysql_num_rows($result);
$row = mysql_fetch_array($result);

$STF = array();
while($row=mysql_fetch_array($result))
  {
  $STF[] = $row;
}

return $STF;
}

function datefrom()
{
global $uname;
global $cd1;
global $teacherid;
$sql="SELECT SchedTimeFrom FROM tblsched WHERE SchedTeacherID=$teacherid and SchedDateFrom<='$cd1' and SchedDateTo>='$cd1' ";
$result=mysql_query($sql);

if($result === FALSE) {
die(mysql_error());
}

$row = mysql_fetch_array($result);

$STF = array();
while($row=mysql_fetch_array($result))
  {
  $STF[] = $row;
  }

return $STF;
}

any suggestion on how can i display each array value in a different cell, every cell will contain different values from the array

Not part of the question but you should be using mysqli or PDO extensions to connect to your database and retrieve data instead of mysql. — pivot
– pivot, Commented Sep 20, 2012 at 14:06

Undrium · Accepted Answer · 2012-09-20 12:58:19Z

1

Remove the first:

$row = mysql_fetch_array($result);

Then add a:

var_dump($STF);

after each while loop in your functions. This will help you understand how it works better and hopefully the variable is populated with lots of goodies!

answered Sep 20, 2012 at 12:58

Undrium

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Comments

Yogesh Suthar · Accepted Answer · 2012-09-20 12:59:03Z

1

use like this

for($x=0;$x<count($arrayto);$x++)
{
echo "<tr>";
echo "<td align=$tdali bgcolor=$bgcolor>$arrayfrom[$x] - $arrayto[$x]</td>";
echo "<td align=$tdali> </td>";
echo "<td align=$tdali> </td>";
echo "<td align=$tdali> </td>";
echo "<td align=$tdali> </td>";
echo "<td align=$tdali> </td>";
echo "<td align=$tdali> </td>";
echo "<td align=$tdali> </td>"; 
echo "</tr>";

}

don't use $array[] for assigning, it take whatever value pass to it.

answered Sep 20, 2012 at 12:59

Yogesh Suthar

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Comments

Nirav Ranpara · Accepted Answer · 2012-09-20 13:09:34Z

1

    $sql=mysql_query("select  * from tblname");
    while($row=mysql_fetch_array($sql))
        {
            echo "<td>".$row["id"];
            echo "<td>".$row["name"];
        }

answered Sep 20, 2012 at 13:09

Nirav Ranpara

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Comments

SimonPeacock · Accepted Answer · 2012-09-20 13:01:31Z

0

loop through the $STF array with a for loop echoing out the result.

for ($STF as $r) 
{ 
  echo '<td>' .$r["whatever data you want to display here"] .'</td>';
}

answered Sep 20, 2012 at 13:01

SimonPeacock

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