MySQL (phpMyAdmin) & PHP

iWant2Know · 08-22-2010, 01:15 PM

I have written a program in PHP that will execute SQL queries and display the results in my browser. The program is not performing the functions when information is input and query button is submitted. The page returns with no information in the text box. I am usig a remote server, MySQL for my server is managed by phpMyAdmin but this is not specified in the program. Could this be part of the issue?

I am a newbie and need some clarity on how to use php programming to request and display results from MySQL which is managed by phpMyAdmin from a remote server.

Thanks.

paulsm4 · 08-22-2010, 01:59 PM

Hi -

1. Yes, you should be able to do exactly what you're describing.

2. No, phpAdmin has nothing to do with it.

SUGGESTION:
Please post your PHP code, or a short, equivalent test case.
It should be fairly short (under 50 lines).
Please use code tags.

Once we see the code, we should be able to figure out what's going wrong.

Thank you in advance .. PSM

PS:
Typically, your web server and your MySQL server are both co-located on the same host.
Is this your scenario, or are HTTP and MySQL on two different servers?

jlinkels · 08-22-2010, 01:59 PM

You should approach this problem logically:

Have you tried to connect using the mysql client?
Code:
```
mysql -h your.mysql.host -u user -ppass
```
. If that fails, connecting using PHP is not much use. You should check why mysql is not allowing remote connections. You might want to login on the mysql box and try the same, omitting the -h your.mysql.host
Have you tried this example? http://www.php.net/manual/en/functio...etch-array.php

jlinkels

paulsm4 · 08-22-2010, 10:06 PM

Hi -

Here's another good "hello world" type example:
http://www.daniweb.com/forums/thread87999.html

Code:

<html>
  <form method="post" action="test.php">
    User Name : <input name="UserName" type="text" id="UserName" 
      value="User Name" size="20" maxlength="30" />
    <input type="submit" name="submit" value="Submit">
  </form>
</html><?php
  if (isset($_POST['submit'])) {
    mysql_connect("Localhost","username","password")
      or die("Failure to communicate");
    mysql_select_db("mydb")
      or die("Could not connect to Database");
    if ($_REQUEST['UserName'] == NULL) {
      echo "Please Enter a User Name";
    }
    else {
      $p="INSERT INTO mytable(UserName)VALUES('$UserName')";
      $q=mysql_query($p);
    }
  }
  else {
    echo "Enter a User Name";
  }
 ?>

iWant2Know · 08-23-2010, 12:41 AM

<?php
/*Program: mysql_send.php
*Desc: PHP program that sends an SQL query to the
* MySQL server and displays the results.
*/

echo "<html>
<head><title>SQL Query Sender</title></head>
<body>";
if(ini_get("magic_quotes_gpc") == “1”)
{
$_POST[‘query’] = stripslashes($_POST[‘query’]);
}
$host=" ";
$user=" ";
$password=" ";

/* Section that executes query and displays the results */
if(!empty($_POST[‘form’]))
{
$cxn = mysqli_connect($host,$user,$password,
$_POST['database']);
$result = mysqli_query($cxn,$_POST[‘query’]);
echo "Database Selected: <b>{$_POST[‘database’]}</b><br>
Query: <b>{$_POST[‘query’]}</b>
<h3>Results</h3><hr>";
if($result == false)
{
echo "<h4>Error: ".mysqli_error($cxn)."</h4>";
}
elseif(@mysqli_num_rows($result) == 0)
{
echo "<h4>Query completed.
No results returned.</h4>";
}
else
{
/* Display results */
echo "<table border=’1’><thead><tr>";
$finfo = mysqli_fetch_fields($result);
foreach($finfo as $field)
{
echo "<th>".$field->name."</th>";
}
echo "</tr></thead>
<tbody>";
for ($i=0;$i < mysqli_num_rows($result);$i++)
{
echo "<tr>";
$row = mysqli_fetch_row($result);
foreach($row as $value)
{
echo "<td>".$value."</td>";
}
echo "</tr>";
}
echo"</tbody></table>";
}
/* Display form with only buttons after results */
$query = str_replace("‘","%&%",$_POST[‘query’]);
echo "<hr><br>
<form action=’{$_SERVER[‘PHP_SELF’]}’ method=’POST’>
<input type=’hidden’ name=’query’ value=’$query’>
<input type=’hidden’ name=’database’
value={$_POST[‘database’]}>
<input type=’submit’ name=’queryButton’
value=’New Query’>
<input type=’submit’ name=’queryButton’
value=’Edit Query’>
</form>";
exit();
}
/* Displays form for query input */
if (@$_POST[‘queryButton’] != "Edit Query")
{
$query = " ";
}
else
{
$query = str_replace("%&%","’",$_POST[‘query’]);
}
?>
<form action="<?php echo $_SERVER[‘PHP_SELF’] ?>"
method="POST">
<table>
<tr><td style=’text-align: right; font-weight: bold’>
Type in database name</td>
<td><input type="text" name="database"
value=<?php echo @$_POST[‘database’] ?> ></td>
</tr>
<tr><td style=’text-align: right; font-weight: bold’
valign="top">Type in SQL query</td>
<td><textarea name="query" cols="60"
rows="10"><?php echo $query ?></textarea></td>
</tr>
<tr><td colspan="2" style=’text-align: center’>
<input type="submit" value="Submit Query"></td>
</tr>
</table>
<input type="hidden" name="form" value="yes">
</form>
</body></html>

Xyro · 08-24-2010, 03:44 PM

As paulsm4 mentioned, phpMyAdmin would not be the cause of your problems. It is simply a web-based front-end to manage your DB.

Try this code and report results:

Code:

<?php
  $user = '<INSERT MYSQL USERNAME HERE>';
  $pass = '<INSERT MYSQL PASSWORD HERE>';
  $host = '<INSERT MYSQL HOSTNAME HERE>';
  $dbnm = '<INSERT MYSQL DATABASE NAME>';

  $link = mysql_connect( $host, $user, $pass ) or die( mysql_error() );
  mysql_select_db( $dbnm ) or die( mysql_error() );
  mysql_close( $link );
  echo "All seems OK\n";
?>

Also, where did you get the 'mysqli_*' commands? Did you mean 'mysql_*' (no 'i')?

PS: These are very good tutorials.