Trouble displaying MySQL table in HTML/PHP

Question

I'm having a hard time figuring out how to get my table to display as a web page.

Everything else is working fine. I've been able to use a form to write records TO the table, but I'm simply unable to display the table.

Here's my code:

$host = "***";
$userName = "***";
$passWord = "***";
$db = "doctorWho";

mysql_connect($host,$userName,$passWord);
mysql_select_db($db) or die( "Unable to access database");
$query = "SELECT * FROM patients";

$result = mysql_query($query);

echo "<table border='1'>
    <tr>
    <th>Last Name</th>
    <th>First Name</th>
    <th>Address</th>
    <th>Age</th>
    <th>Sex</th>
    <th>Marital Status</th>
    <th>Medication</th>
    <th>Date Rx'd</th>
    <th>Quantity</th>
    </tr>";

while($row = mysqli_fetch_array($result))
  {
  echo "<tr>";
  echo "<td>" . $row['lastName'] . "</td>";
  echo "<td>" . $row['firstName'] . "</td>";
  echo "<td>" . $row['address'] . "</td>";
  echo "<td>" . $row['age'] . "</td>";
  echo "<td>" . $row['sex'] . "</td>";
  echo "<td>" . $row['maritalStatus'] . "</td>";
  echo "<td>" . $row['medication'] . "</td>";
  echo "<td>" . $row['medsWhen'] . "</td>";
  echo "<td>" . $row['medsQuant'] . "</td>";
  echo "</tr>";

  }
echo "</table>";

Do not use mysql_*? It is deprecated. mysqlpi or PDO is the replacement and the newest beast in town. — Ed Heal
– Ed Heal, Commented Apr 1, 2013 at 7:36
and if you are using mysql than please dont do it .. reason is exactly what Mr. @Ed Heal is telling form more information check stackoverflow.com/questions/12859942/… — Mayank K. Swami
– Mayank K. Swami, Commented Apr 1, 2013 at 7:38

Daedalus · Accepted Answer · 2013-04-01 07:35:12Z

4

You want mysql_fetch_array(). Currently, you are mixing two different extensions. Each takes something different, relevant to the extension.

That aside, you should consider either PDO or MySQLi, which is where the wrong function you attempted to use is from. This article should help you decide which you could possibly use.

answered Apr 1, 2013 at 7:35

Daedalus

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8 Comments

Pachi Over a year ago

@NullPonyPointer actually, I'm just a noob at this, slowly learning. Thanks for your help though!

Daedalus Over a year ago

@Pachi If this answer solved your problem, please consider accepting it by clicking the checkmark under the down arrow next to it, when it lets you.

Mayank K. Swami Over a year ago

@Pachi please noob doesnot mean we dont do debuting i am sure you are getting error and not trying(give enough attention ) to solve it also dont use @ operator to ignore error its always bad you must solve them

Pachi Over a year ago

@Daedalus I'm working through this. Thank you very much for the information you've given me!

Mayank K. Swami Over a year ago

ignoring error is very bad .. and caused UN-identified defects and can rune whole project so its good to start solving and handling them properly

|

Mayank K. Swami · Accepted Answer · 2013-04-01 07:40:20Z

0

From php ref.

mixed mysqli_fetch_array ( mysqli_result $result [, int $resulttype = MYSQLI_BOTH ] )

mysqli_fetch_array need to mysqli_result, but you are using mysql_query, you should try mysql_fetch_array

edited Apr 1, 2013 at 7:40

Mayank K. Swami

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answered Apr 1, 2013 at 7:38

dev.meghraj

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Comments

Jhonathan H. · Accepted Answer · 2013-04-01 07:48:53Z

0

// I think your connection should look like this
$conn = mysql_connect($host,$userName,$passWord);
mysql_select_db($conn,$db) or die( "Unable to access database");

// another is your using mysql as connection so you can use mysqli functions on that rofl.
//Instead use mysql functions
mysql_fetch_array();

note: Another one is instead of using @ sign to handle error. Its much more best practice to use.. ini_set() or configuring php.ini in handling displaying of errors.

edited Apr 1, 2013 at 7:48

answered Apr 1, 2013 at 7:37

Jhonathan H.

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1 Comment

Mayank K. Swami Over a year ago

please don't use @ its bad practice to use this

Vivek Sadh · Accepted Answer · 2013-04-01 08:48:23Z

0

Just change:-

while($row = mysqli_fetch_array($result))

to

while($row = mysql_fetch_array($result))

answered Apr 1, 2013 at 8:48

Vivek Sadh

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