php query database print to html table with headers

Question

I have never programmed anything in php before and haven't touched html in 10 years. I could use some help. I am querying a postgresql database using php. I am trying to display my query results in a table format with headers like this:

first_name   last_name   employee_id
tom          jones       111
bob          barker      112
bill         davis       113

Sample code I am trying to get to work correctly:

echo("<table border=2");
while ($line = pg_fetch_array($result, null, PGSQL_ASSOC)) {
    foreach ($line as $col_value => $row_value) {
        echo("<tr><td>$col_value</td><td>$row_value</td></tr>\n");
    }
}
echo("</table>");

My formatting is being displayed like this:

first_name tom
last_name jones
employee_id 111
first_name bob
last_name barker
employee_id 112
first_name bill
last_name davis
employee_id 113

As you can see I am storing my query in an associative array.

Thanks for any help.

You might want to look into Symfony, if you've a programming background but are unfamiliar with PHP. Symfony2 combined with Doctrine2 is an excellent PHP framework. — Denis de Bernardy
– Denis de Bernardy, Commented Jul 1, 2011 at 7:45
Looking at it again, my answer fixes an html formatting issue that was there but the data would still be displayed like your example. I saw the missing > and thought that was it...my mistake. Dirk's answer looks like it will get you what except for the column headers which you would have to deal with manually. — Jason
– Jason, Commented Jul 1, 2011 at 10:07

Jason · Accepted Answer · 2011-07-01 00:15:22Z

3

Looks like you might be missing a bracket on the opening table tag:

Try changing this:

echo("<table border=2");

to this:

echo('<table border="2">');

and see if that helps.

answered Jul 1, 2011 at 0:15

Jason

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1 Comment

spaghettiwestern Over a year ago

I wish it were that easy ; ).

Dirk · Accepted Answer · 2011-07-01 07:11:54Z

1

echo("<table border=2><tr><td>first_name</td><td>last_name</td><td>employee_id</td></tr>");
while ($line = pg_fetch_array($result, null, PGSQL_ASSOC)) {
    echo("<tr>");
    foreach ($line as $col_value => $row_value) {
        echo("<td>$row_value</td>");
    }
    echo("</tr>\n");
}
echo("</table>");

Or:

echo("<table border=2><tr><td>first_name</td><td>last_name</td><td>employee_id</td></tr>");
while ($line = pg_fetch_array($result, null, PGSQL_ASSOC)) {
    echo("<tr><td>".$line[0]."</td><td>".$line[1]."</td><td>".$line[2]."</td></tr>\n");
}
echo("</table>");

answered Jul 1, 2011 at 7:11

Dirk

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2 Comments

spaghettiwestern Over a year ago

Thanks a bunch that is very close to what I am looking for. Do you have any ideas on how I could dynamically create the table column headers rather than hard coding them in?

Dirk Over a year ago

SELECT COLUMN_NAME FROM information_schema.`COLUMNS` C WHERE table_name = '<your table name>' ?

mori · Accepted Answer · 2022-04-25 06:11:33Z

0

echo "<table>\n";
echo("<table border=2");
while ($line = pg_fetch_array($result, null, PGSQL_ASSOC)) {
    echo "\t<tr>\n";
    foreach ($line as $col_value => $row_value) {
        echo "\t\t<td>$col_value</td>\n";
    }
    echo "\t</tr>\n";
    echo "\t<tr>\n";
    foreach ($line as $col_value) {
        echo "\t\t<td>$col_value</td>\n";
    }
    echo "\t</tr>\n";
    break;
}
while ($line = pg_fetch_array($result, null, PGSQL_ASSOC)) {
    echo "\t<tr>\n";
    foreach ($line as $col_value) {
        echo "\t\t<td>$col_value</td>\n";
    }
    echo "\t</tr>\n";
}
echo "</table>\n";

answered Apr 25, 2022 at 6:11

mori

1

Comments

jeni · Accepted Answer · 2011-07-01 07:12:47Z

-1

<?php
echo "<table width=100% border="1">";
while ($row = mysql_fetch_array($result))
 {  
  $id= $row["id"]; 
  $f_name= $row["f_name"]; 
  echo "<tr><td>";
  echo $id;
  echo"</td>";
  echo"<td>";
  echo $f_name;
  echo"</tr>";
 } 
echo"</table>";
?>

try this may help

edited Jul 1, 2011 at 7:12

answered Jul 1, 2011 at 7:05

jeni

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1 Comment

LuxZg Over a year ago

I know this is 13 year old answer but question explicitly said "querying a postgresql database" and you answer with mysql_fetch..

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