[SOLVED] Creating an Array

[Wildhalf]

It's been a while since i looked at php and SQL nad my head is hoping at the minute...

 

I know this isnt to hard but how do i turn returned results into an array if my query is as follows

 

$getData = mysql_query('SELECT * FROM name');

$result = mysql_query($getData);

 

Thanks for you help

[discomatt]

mysql_fetch_assoc() or mysql_fetch_row()

[papaface]

$getData = mysql_query('SELECT * FROM name');
while ($result = mysql_fetch_assoc($getData))
{
echo $result['fieldname'] . "<br />";
}

[Wildhalf]

$getData = mysql_query('SELECT * FROM name');
while ($result = mysql_fetch_assoc($getData))
{
echo $result['fieldname'] . "<br />";
}

 

I get the following errors

 

mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource

 

I am using XAMPP on my local machine to test

[papaface]

Have you actually opened a connection to mysql?

Show us all your code.

[marcus]

$getData = mysql_query("SELECT * FROM `name`") or die(mysql_error());

while($result = mysql_fetch_assoc($getData)){
     echo $result['fieldname'] . "<br>\n";
}

 

error checking.

[Wildhalf]

Have you actually opened a connection to mysql?

Show us all your code.

 

 

Heres my code

 

 <?php
$con = mysql_connect("localhost","pagination","abc123");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }


$rowData = array();
$getData = mysql_query('SELECT * FROM test');
while ($rowData = mysql_fetch_assoc($getData))
{
echo $rowData['name'] . "<br />";
}                                                

    // Include the pagination class
    include 'pagination.class.php';
    // Create the pagination object
    $pagination = new pagination;
    // parse the data
    $dataPages = $pagination->generate($rowData, 20);

    foreach ($dataPages as $dataKey => $dataArray) {
    // Show the information about the item
    echo '<p><b>'.$dataArray['fieldname'].'</b> '.$dataArray['fieldname2'].'</p>';
    } 

      ?>

[wildteen88]

You have an erro in your sql query if your are getting that error, what does the following return:

$query = 'SELECT * FROM name';
$getData = mysql_query($query) or die('SQL Query Error: ' . mysql_error() . '<br />Query: <pre>' . $query . '</pre>');
while ($result = mysql_fetch_assoc($getData))
{
    echo $result['fieldname'] . "<br />";
}

[Wildhalf]

You have an erro in your sql query if your are getting that error, what does the following return:

$query = 'SELECT * FROM name';
$getData = mysql_query($query) or die('SQL Query Error: ' . mysql_error() . '<br />Query: <pre>' . $query . '</pre>');
while ($result = mysql_fetch_assoc($getData))
{
    echo $result['fieldname'] . "<br />";
}

 

Came back with

 

SQL Query Error: No database selected

 

 

[Wildhalf]

got database connection sorted just going to check original code now...

[Wildhalf]

After getting connection problems solved i figured out what was wrong with the rest of the code...