displaying search info on webpage

[toney]

I am trying to have the data from my database display on a webpage the problem I am having is two fold one the

1.  picture number is not displaying in order 

2. how do I get the birth date to display in D- M -  Y  on webpage        the output displays  = 2007-05-11   

 

<?php

$con = mysql_connect("localhost","","");

if (!$con)

  {

  die('Could not connect: ' . mysql_error());

  }

 

mysql_select_db("sacredfa_sacred", $con);

 

$query = "SELECT picture_number, first_name, middle_name, first_family_name, second_family_name, birthdate, gender

          FROM child_info";

 

$result = mysql_query($query);

 

 

if(!$result)

{

  echo "There was a problem getting the data";

}

else if(!$result)

{

echo "There were no results";

}

else

{

    echo "<b><center>Children to be sponsored</center></b><br><br>\n";

    while($row = mysql_fetch_assoc($result))

    {

 

 

echo "<table border='1'>

<tr>

<th>Picture Number</th>

<th>First Name</th>

<th>Middle Name</th>

<th>First Family Name</th>

<th>Second Family Name</th>

<th>Birthdate</th>

<th>Gender</th>

</tr>";

 

while($row = mysql_fetch_array($result))

  {

  echo "<tr>";

  echo "<td>" . $row['picture_number'] . "</td>";

  echo "<td>" . $row['first_name'] . "</td>";

  echo "<td>" . $row['middle_name'] . "</td>";

  echo "<td>" . $row['first_family_name'] . "</td>";

  echo "<td>" . $row['second_family_name'] . "</td>";

  echo "<td>" . $row['birthdate'] . "</td>";

  echo "<td>" . $row['gender'] . "</td>";

  echo "</tr>";

  }

echo "</table>";

}

}

mysql_close();

?>

[toney]

<?php
$con = mysql_connect("localhost","","");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("", $con);

$query = "SELECT picture_number, first_name, middle_name, first_family_name, second_family_name, birthdate, gender
          FROM child_info";
        
$result = mysql_query($query);


   if(!$result)
      {
             echo "There was a problem getting the data";
      }
   else if(!$result)
      {
         echo "There were no results";
      }
   else
      {
          echo "<b><center>Children to be sponsored</center></b><br><br>\n";
    while($row = mysql_fetch_assoc($result))
       {
      
         
echo "<table border='1'>
<tr>
<th>Picture Number</th>
<th>First Name</th>
<th>Middle Name</th>
<th>First Family Name</th>
<th>Second Family Name</th>
<th>Birthdate</th>
<th>Gender</th>
</tr>";

while($row = mysql_fetch_array($result))
  {
        echo "<tr>";
        echo "<td>" . $row['picture_number'] . "</td>";
        echo "<td>" . $row['first_name'] . "</td>";
        echo "<td>" . $row['middle_name'] . "</td>";
        echo "<td>" . $row['first_family_name'] . "</td>";
        echo "<td>" . $row['second_family_name'] . "</td>";
        echo "<td>" . $row['birthdate'] . "</td>";
        echo "<td>" . $row['gender'] . "</td>";
        echo "</tr>";
  }
echo "</table>";
      }
}
mysql_close();
?>

[ManiacDan]

Please post formatted code, and surround it with [ php ] tags.

 

...and change your database password.  Right now.  Stop reading and go change it.  You've posted it in public.

[Muddy_Funster]

make this change:

$query = "SELECT picture_number, first_name, middle_name, first_family_name, second_family_name, DATEFORMAT(birthdate, '%e-%c-%Y') as birthdate, gender
          FROM child_info ORDER BY picture_number ASC";

 

[toney]

I tried the suggestion and it did not work

[Pikachu2000]

"it did not work" isn't very descriptive.

[ManiacDan]

A number of things:

1)  You removed your db password from this page, but I bet you didn't change it in your actual database.  You have told (according to the counter here), 35 strangers your database password, and it's indexed by google.  You must change it, now.

 

2)  Adding the ORDER BY that Muddy_Funster suggested will fix your ordering problem unless your picture number isn't stored properly in your database, but you'll have to define "didn't work" first.

 

3)  Formatting your date output can be done with date() and strtotime(), like so: date('d-m-Y', strtotime($row['birthdate']));

[Muddy_Funster]

...

3)  Formatting your date output can be done with date() and strtotime(), like so: date('d-m-Y', strtotime($row['birthdate']));

but I formated the date for them too, i just forgot the _ in the middle of DATE_FORMAT() was all 

[ManiacDan]

...

3)  Formatting your date output can be done with date() and strtotime(), like so: date('d-m-Y', strtotime($row['birthdate']));

but I formated the date for them too, i just forgot the _ in the middle of DATE_FORMAT() was all 

I didn't even notice that.  Yes, everything muddy_funster said will fix all your problems if you put that underscore there.

[toney]

ok putting the  _ in  DATE_FORMAT worked  another problem is for some reason  the first record is not showing it starts from number 2

[Muddy_Funster]

yeah, your php is a bit of a mess: you seem to be doing everything twice: you have an elseif checking the same condition as the first if (not sure if you expect things to change from one line to the next, but they won't in this case) and then you nest your while loop for disply inside....well....a while loop for display, which is re-assigning different values to the same variables. I'm curious, cold you post a screen cap of what your actual output page looks like?

[toney]

http://www.sacredfamilysponsor.org/selectdata.php

[ManiacDan]

Once you change the password to your database, you must put the new password in your code.

[toney]

i forgot to upload it back to the server so i did that the link will work now

[ManiacDan]

    while($row = mysql_fetch_assoc($result))

 

That line appears twice in your code.

 

If you indent your code properly, you'll be better able to notice redundancies like this (and the elseif mentioned earlier).  Remove the outer    while($row = mysql_fetch_assoc($result))

and your code will stop skipping the first result.

[toney]

I removed the extra  while($row = mysql_fetch_assoc($result))  it now shows the first rcord.  now how do I but a link at the end of each record that will allow the searching person to view that particular record

[ManiacDan]

Write that page.  Put a link to it at the end of every row.

 

I mean, I know you're new to PHP, but "write the page" is about as close as we can get.

[toney]

I have 860 entries in the database I need a better way to display them

[ManiacDan]

Write a page to display a single entry based on ID.  Use $_GET['id']

 

Link the page you already have to that page.  Add "?id={$row['picture_number']}" to the URL.

 

Come back when you're stuck on either of those steps.

 

[toney]

please explain further

 

Write a page to display a single entry based on ID.  Use $_GET['id']

 

Link the page you already have to that page.  Add "?id={$row['picture_number']}" to the URL.

[ManiacDan]

please explain further

No.  if you need more help than this, you need to be taking a class or reading a book, not asking random people on the internet.  You've already made a page to display records, where did that come from?