如何从数据库中显示表中的数据？

Question

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This code to calculate the age of the date of birth of the database, but the problem is when shown in the table, the location of that age are outside the table, can someone help me, this code

<?php
    include 'koneksi.php';
    $sql=mysql_query("SELECT * FROM pasien");
    function umur($tgl_lahir){
        $parts = split('-', $tgl_lahir);
        $thn_lahir='Year: ' + $parts[0];
        $bln_lahir='Month: ' + $parts[1];
        $tgl_lahir='Day: ' + $parts[2];
        $tanggal_today = date('d');
        $bulan_today=date('m');
        $tahun_today = date('Y');
        $harilahir=gregoriantojd($bln_lahir,$tgl_lahir,$thn_lahir);
        $hariini=gregoriantojd($bulan_today,$tanggal_today,$tahun_today);
        $umur=$hariini-$harilahir;
        $tahun=$umur/365; 
        $sisa=$umur%365; 
        $bulan=$sisa/30;
        $hari=$sisa%30;
        echo floor($tahun).",".floor($bulan);
    }
    while ($row=mysql_fetch_array($sql)){ 

        echo"<table border='1px'>";
        echo "<tr>";
        echo "<td width='50px'>".umur($row['tgl_lahir'])."</td>";
        echo "<td>".$row['nama']."</td>";
        echo "</tr>";
        echo"</table>";
    }
?>

Answer 1

You're already echo the result of function umur . But in the function you are doing it one more time.

Change your umur function last line

echo floor($tahun).",".floor($bulan);

to

return floor($tahun).",".floor($bulan);

And move your <table> and </table> outside of while . Otherwise you would have a new table for every row.

Answer 2

Just move echo"<table border='1px'>"; before while loop and move echo"</table>"; after while loop.

Update your code to

<?php
include 'koneksi.php';
$sql=mysql_query("SELECT * FROM pasien");
function umur($tgl_lahir)
{
    $parts = split('-', $tgl_lahir);
    $thn_lahir='Year: ' + $parts[0];
    $bln_lahir='Month: ' + $parts[1];
    $tgl_lahir='Day: ' + $parts[2];
    $tanggal_today = date('d');
    $bulan_today=date('m');
    $tahun_today = date('Y');
    $harilahir=gregoriantojd($bln_lahir,$tgl_lahir,$thn_lahir);
    $hariini=gregoriantojd($bulan_today,$tanggal_today,$tahun_today);
    $umur=$hariini-$harilahir;
    $tahun=$umur/365; 
    $sisa=$umur%365; 
    $bulan=$sisa/30;
    $hari=$sisa%30;
    return floor($tahun).",".floor($bulan);
}


echo"<table border='1px'>";
while ($row=mysql_fetch_array($sql))
{ 

    echo "<tr>";
    echo "<td width='50px'>".umur($row['tgl_lahir'])."</td>";
    echo "<td>".$row['nama']."</td>";
    echo "</tr>";
}
echo"</table>";

It is recommended to use return instead of echo in function to return the value. You can use echo umur($row['tgl_lahir']); to print the returned value of function.

Answer 3

try this: You need to return the value from the function rather than echo it

<?php
    include 'koneksi.php';

    $sql=mysql_query("SELECT * FROM pasien");
    function umur($tgl_lahir){
        $parts = split('-', $tgl_lahir);
        $thn_lahir='Year: ' + $parts[0];
        $bln_lahir='Month: ' + $parts[1];
        $tgl_lahir='Day: ' + $parts[2];
        $tanggal_today = date('d');
        $bulan_today=date('m');
        $tahun_today = date('Y');
        $harilahir=gregoriantojd($bln_lahir,$tgl_lahir,$thn_lahir);
        $hariini=gregoriantojd($bulan_today,$tanggal_today,$tahun_today);
        $umur=$hariini-$harilahir;
        $tahun=$umur/365; 
        $sisa=$umur%365; 
        $bulan=$sisa/30;
        $hari=$sisa%30;
        return floor($tahun).",".floor($bulan);
    }
    echo"<table border='1px'>";
    while ($row=mysql_fetch_array($sql)){ 
        echo "<tr>";
        echo "<td width='50px'>".umur($row['tgl_lahir'])."</td>";
        echo "<td>".$row['nama']."</td>";
        echo "</tr>";
    }
    echo"</table>";
 ?>